Differential Calculus - PDF

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```					Department of Mathematical Sciences

MA1002 Calculus Diﬀerential Calculus
Dr John Pulham

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September 13, 1999, Version 1.3 Copyright © 1999 by Ian Craw and the University of Aberdeen All rights reserved. Additional copies may be obtained from: Department of Mathematical Sciences University of Aberdeen Aberdeen AB9 2TY DSN: mth199-101462-2

Contents
1 The 1.1 1.2 1.3 1.4 Diﬀerential Calculus Introduction . . . . . . . . . . . . . . . . . . . . . . . . . The slope of a graph and the tangent . . . . . . . . . . Increasing and Decreasing . . . . . . . . . . . . . . . . . Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Rate of Change . . . . . . . . . . . . . . . . . . . 1.5 Some Basic Derivatives . . . . . . . . . . . . . . . . . . . 1.5.1 f (x) = xn where n is a positive integer. . . . . 1 . . . . . . . . . . . . . . . . . . . . . 1.5.2 f (x) = x 1.5.3 The Absolute Value f (x) = |x| . . . . . . . . . . 1.5.4 Derivatives of sin x and cos x . . . . . . . . . . 1.6 Rules of Diﬀerentiation . . . . . . . . . . . . . . . . . . . 1.6.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Examples of the Use of the Rules . . . . . . . . . . . . . 1.7.1 Polynomials . . . . . . . . . . . . . . . . . . . . . n = 1, 2, 3, 4 . . . 1.7.2 The function f (x) = x,n 1.7.3 Long Products . . . . . . . . . . . . . . . . . . . 1.7.4 Trigonometric Functions . . . . . . . . . . . . . . 1.7.5 Rational Functions . . . . . . . . . . . . . . . . . √ 1.7.6 The square root f (x) = x . . . . . . . . . . . 1.8 Derivative of sin x where x is in radians . . . . . . . . 1.9 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . 1.10 Proofs of some of the Rules . . . . . . . . . . . . . . . . 1.11 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . 1.11.1 Geometrical Interpretation of Second Derivative . 1.11.2 Kinematical Interpretation . . . . . . . . . . . . . 1.12 Limits & Continuity . . . . . . . . . . . . . . . . . . . . 1.12.1 Limits . . . . . . . . . . . . . . . . . . . . . . . . 1.12.1 Continuity . . . . . . . . . . . . . . . . . . . . . . 1.12.2 * Rolle’s Theorem and the Mean Value Theorem . iii 1 1 1 3 3 4 5 5 6 6 7 8 8 9 9 10 10 10 11 12 14 17 22 23 23 24 26 26 27 28

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iv 2 Further Diﬀerentiation 2.1 The Inverse Trigonometric Functions . . . . . . . . 2.1.1 Inverse Functions in general . . . . . . . . . 2.1.1 The Inverse Trigonometric Functions . . . . 2.1.2 Diﬀerentiating the Inverse Trig Functions . 2.1.3 Polar Coordinates . . . . . . . . . . . . . . . 2.2 Implicit Functions and their Diﬀerentiation . . . . 2.2.1 Diﬀerentiating Implicitly Deﬁned Functions 2.3 Parametric Representation . . . . . . . . . . . . . 2.3.1 Diﬀerentiation in Parametric Form . . . . . 3 The Exponential and Logarithm Functions 3.1 Exponential Population Growth . . . . . . . 3.2 The Exponential Function . . . . . . . . . . . 3.2.1 Deﬁnition . . . . . . . . . . . . . . . . 3.2.2 Properties of the Exponential Function 3.2.3 Summary . . . . . . . . . . . . . . . . 3.2.4 Diﬀerentiation Examples . . . . . . . . 3.3 The Natural Logarithm . . . . . . . . . . . . 3.3.1 Deﬁnition of the Logarithm . . . . . . 3.3.2 Derivative of ln x . . . . . . . . . . . . 3.4 Hyperbolic Functions . . . . . . . . . . . . . 3.4.1 More Diﬀerentiation Examples . . . . . 3.5 Powers . . . . . . . . . . . . . . . . . . . . . . 3.6 * Irrational Numbers . . . . . . . . . . . . . . A Some Useful Tables B Solutions to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

CONTENTS 31 31 31 32 34 35 38 39 42 43 47 47 48 49 51 52 53 54 55 56 56 57 59 60 63 65

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List of Figures
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 2.1 2.2 2.3 2.4 2.5 3.1 3.2 3.3 The tangent at a point . . . . . . . . . . . . . . . . . . . . The derivative of x2 from ﬁrst principles. . . . . . . . . . . The graph of y = |x|. . . . . . . . . . . . . . . . . . . . . . The derivative of sin x from ﬁrst principles. . . . . . . . . . Second derivative shape . . . . . . . . . . . . . . . . . . . Second derivative shape . . . . . . . . . . . . . . . . . . . The graph of a step function. . . . . . . . . . . . . . . . . A discontinuity. . . . . . . . . . . . . . . . . . . . . . . . . Rolle: somewhere between a and b, the graph must be ﬂat. Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . Graph of sin x. . . . . . . . . . . . Graph of arcsin(x). . . . . . . . . . Graph of arccos(x). . . . . . . . . . Polar co-ordinates (r, θ) of a point. The lemniscate of Example 2.6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3 6 15 24 24 28 28 29 30 33 34 34 36 41 48 52 55

Population growth behaviour. . . . . . . . . . . . . . . . . . . . . . . . . . Graph of ex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graph of ln(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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vi

LIST OF FIGURES

How to use these Notes
The notes are divided into • theory and explanation • worked examples • exercises Most of the theory will also be presented in lectures. The exceptions are the sections marked with a ∗. These are ‘extras’ and are not an examinable part of the course. They are not necessarily more diﬃcult than the oﬃcial bits—it’s just that they are not in the syllabus. Read them if you are interested. The more you read of everything the more you are likely to understand. Please study the worked examples carefully because that is usually the best way to grasp the theory. There is also a lot to be said for treating some of the worked examples as exercises in the ﬁrst instance, trying to solve them yourself and only then reading through my solution. It is usually rather silly to try reading mechanically through a worked example when you have not yet got into your head what the example is really about. The exercises come in two varieties—unstarred and starred. The starred exercises are either a bit more diﬃcult than the others or else are less directly relevant to the course. The more exercises you try the better, but do not get worried if you have not attempted many of the starred questions. The standard of the examination questions is based on the unstarred questions. There are some solutions and some answers at the back of the book. Use these sensibly. Make a serious attempt at a problem before looking up the answer or you will just be wasting your time.

vii

viii

LIST OF FIGURES

Chapter 1 The Diﬀerential Calculus
1.1 Introduction

The Calculus, comprising the Diﬀerential Calculus and the Integral Calculus, is one of the great achievements of Mathematics. Some aspects of it date back to the ancient Greeks but we normally regard it as the brain-child of Isaac Newton (1642–1727), with help from Gottfried Leibniz (1646–1716) and a push start from people like Pierre Fermat and Barrow (get used to the fact that Mathematics is resolutely international!). The basic problem that led up to the development of the Diﬀerential Calculus was that of constructing the tangent line to an arbitrary curve at a given point. But, for Newton, the much more serious problem was that of describing motion and making rigorous the concepts of velocity and acceleration. The Calculus solved this problem for him and with it Newton was able to construct his System of the World that has dominated our thought ever since. Over the last few centuries the Calculus has grown enormously to be a dominant inﬂuence on the development of Mathematics itself and an indispensable tool in the application of Mathematics to Science, Engineering, Biology, Economics etc. The Calculus is in the remarkable position of being intellectually profound, aesthetically beautiful and very practical all at the same time. There are very few other human achievements of which this can be said with the same force. This course does not pretend to give you more than a brief introduction to the Calculus, but it does contain most of the basic ideas that propel the subject.

1.2

The slope of a graph and the tangent

One of the problems that gave rise to the calculus was that of ﬁnding the equation of the tangent to a curve. The fundamental idea of the calculus is to obtain this information—the slope of the graph—by a limiting process. 1

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CHAPTER 1. THE DIFFERENTIAL CALCULUS
y=f(x) tangent P f(a+h) f(a) x a a+h

The argument goes as follows. Suppose we want to ﬁnd the slope of the graph y = f (x) at the point P(a, f (a)) on the graph. We take another point Q(a + h, f (a + h)) close to P on the graph (either side) and work out the slope of the chord PQ. This comes out as f (a + h) − f (a) f (a + h) − f (a) = (1.1) (a + h) − a h

Q

We call this the Newton Quotient. Figure 1.1: The tangent at a point Now for the critical step. We imagine Q moving closer and closer to P (h → 0). As it does so it seems clear that the slope of the line PQ will get closer and closer to the slope of the tangent line at P. So, we claim that the slope of the tangent line at P is given by the limiting value of (1.1) as h shrinks down to zero. This value, the slope of the tangent at P, is called the derivative of f (x) at x = a df f (a + h) − f (a) (a) = lim h→0 dx h (1.2)

Everything in the diﬀerential calculus comes from this. The clever, and diﬃcult, thing about this deﬁnition is its use of a limiting process. We shall have a little more to say about limits later on but for the moment note that taking the limit as h → 0 is not the same thing as putting h = 0 in the formula. If you put h = 0 in the Newton Quotient you just get 0/0—which does not mean anything (Newton’s critics had a lot to say about that!). Note that we frequently use the alternative notation f (a) for the derivative. Example 1.1. The derivative of f (x) = x2 Let me do an example at once so as to clarify matters. Let’s ﬁnd the slope of the graph y = x2 at the point (2, 4) on the graph. P is the point (2, 4) and Q is the point (2 + h, (2 + h)2 ) (NOT (2 + h, 4 + h)). The slope of the chord PQ is given by the Newton Quotient (2 + h)2 − 22 f (2 + h) − f (2) = h h We want to ﬁnd the limiting value of this slope as h → 0. Once more note that we cannot just put h = 0 in the formula. Now 4h + h2 (2 + h)2 − 4 = =4+h h h It is now clear that as we let h shrink to zero the value of the Newton Quotient tends to 4 as its ultimate limit. So the slope of the graph at (2, 4) is 4.

1.3. INCREASING AND DECREASING
y

3

y=x 2 Q P

x 2 2+h

Figure 1.2: The derivative of x2 from ﬁrst principles. More generally, if we take the point P(a, a2 ) on the graph the Newton Quotient becomes 2ah + h2 (a + h)2 − a2 = = 2a + h h h and as h → 0 this slope tends to the value 2a. So we have shown that the value of the derivative of f (x) = x2 at x = a is 2a, or f (x) = 2x

1.3

Increasing and Decreasing

The derivative f (x) of f (x) gives the slope of the graph y = f (x) at the point (x, f (x))—i.e. the slope of the tangent line to the graph at this point. You already know that a positive slope of a line means that the line is sloping upwards and that a negative slope of a line means that it is sloping downwards. This shows us the following important facts: if f (x) > 0 then the value of f (x) is increasing if f (x) < 0 then the value of f (x) is decreasing There is one other important point to be brought out here: The derivative of a constant function is zero and if f (x) is always zero then f (x) is constant.

1.4

Velocity

The other main impetus to the invention of the calculus was the study of dynamics. In particular, the clariﬁcation of the concept of speed (or velocity).

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CHAPTER 1. THE DIFFERENTIAL CALCULUS

If you ask a car driver how fast he is driving he will look at his speedometer and say something like ‘50 mph’. What does this mean? It obviously means something, or he wouldn’t have said it. It does not mean that he will travel 50 miles in the next hour. His speed is, presumably, changing all the time. What he is trying to say is that, at this particular moment, he is doing 50 mph. This is the idea that we have to clarify. The problem can be approached like this. Suppose, for simplicity, that he is moving along the x-axis and is in position x(t) at time t (x is a function of t). At a slightly later time t + h he is in position x(t + h). So, in time h he has covered distance x(t + h) − x(t) (assuming that he has not backed up in the meantime). So we can reasonably say that his average speed between times t and t + h is x(t + h) − x(t) distance = time h which is the Newton Quotient again! But we don’t want his average speed over the time from t to t + h, we want his speed at time t. The next step is predictable. We deﬁne his speed at time t to be the limiting value of this average speed as h → 0. speed at time t = lim x(t + h) − x(t) h→0 h

The right-hand side of this is just a standard Newton Quotient. This shows us that if a particle is moving along the x-axis and its x coordinate at time t is x(t) the the velocity of the particle at time t is given by the derivative of x with respect to t. dx velocity = dt Notes: 1. I use the technical term velocity rather than speed. Speed is the magnitude of velocity, without the sign. Velocities can be negative, speeds cannot. 2. For historical reasons it is common to use the notation x for a derivative with respect ˙ to time. 3. Try explaining all this to the driver.

1.4.1

Rate of Change

In general, if P (t) is a quantity that depends on time t then its rate of change is deﬁned to be its derivative with respect to time. You have to be a bit careful about signs here. If the volume of an object at time t is V (t) and I say that the volume is increasing at a rate α

1.5. SOME BASIC DERIVATIVES

5

˙ then I am saying that V (t) = α. If, on the other hand, I say that the volume is decreasing ˙ at a rate α then I am saying that V (t) = −α. Be careful to distinguish between rates of increase and decrease.

1.5

Some Basic Derivatives
f (x) = xn where n is a positive integer.

Let us now have a look at the derivatives of some other functions.

1.5.1

The Newton Quotient is (x + h)n − xn f (x + h) − f (x) = . h h How do we work out the limiting value of this as h → 0? As usual, there is nothing to be learned from just putting h = 0 in the RHS because, as usual, we just get 0/0. The complicated bit on the RHS is the term (x + h)n . Here are a few cases expanded out: (x + h)2 = x2 + 2hx + h2 , (x + h)3 = x3 + 3hx2 + 3h2 x + h3 , (x + h)4 = x4 + 4hx3 + 6h2 x2 + 4h3 x + h4 . In general we have n (x + h) factors multiplied together. If we think of what we are going to get when we multiply out these brackets we see that (1) if we take an x from each bracket that gives us a xn term, (2) if we take a h from one bracket and x’s from all the others that gives us a xn−1 h term. We could have chosen any of the n brackets to give the h so, all told, we get nxn−1 h. (3) any other term that we are going to get when multiplying out the product is going to contain at least a h2 . That is all that we need to know about the expansion. We have argued that (x + h)n = xn + nhxn−1 + terms involving at least h or h2 or ... or hn . So (x + h)n − xn = nhxn−1 + terms involving at least h or h2 or ... or hn . So the Newton Quotient can be written as (x + h)n − xn = nxn−1 + terms involving at least h. h As h → 0 all these extra terms tend to zero. So: (x + h)n − xn lim = nxn−1 . h→0 h So we have the result that d n x = nxn−1 dx for n = 1, 2, 3, 4, . . .

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CHAPTER 1. THE DIFFERENTIAL CALCULUS

1.5.2

f (x) =

1 x
1 f (x + h) − f (x) = h h 1 1 − . x+h x

Proceed as usual. The Newton Quotient is

Simplifying the RHS by algebra we get f (x + h) − f (x) −1 = . h x(x + h) Now we let h tend to zero. It is clear that, unless x = 0, the RHS is tending to the value −1/x2 , because (x + h) is tending to x. So we have the result that, if x = 0, d dx 1 x =− 1 . x2

1.5.3

The Absolute Value f (x) = |x|
x −x if if x ≥ 0, x < 0.

|x| is what is known as the absolute value of x. It is deﬁned by |x| =

You can describe this as: if x is negative then throw away the negative sign. So, for example, |3| = 3, |π| = π, | − 3| = 3, | − π| = π. You can easily check for yourself, by keeping track of positive and negative signs, that y |xy| = |x||y| |xn | = |x|n
y = |x|

Figure 1.3: The graph of y = |x|. The graph of y = |x| looks like Fig 1.3. For negative x it is the line y = −x and for positive x it is the line y = x. Note that it turns through a right-angle at the origin. The question “What is the slope of y = |x| at the origin?” does not seem to have a sensible answer. So we ought to ﬁnd ourselves in diﬃculties if we try to diﬀerentiate f (x) = |x| at x = 0. Let’s see what happens. The Newton Quotient is |x + h| − |x| f (x + h) − f (x) = h h

A less obvious property is that known as the Triangle Inequality |x + y| ≤ |x| + |y|

x

1.5. SOME BASIC DERIVATIVES We are interested in x = 0, so the Newton Quotient becomes |h| |h| − 0 = h h

7

What is the limiting value of this as h → 0? I said nothing in my original deﬁnition of the derivative to suggest that h had to be positive. We have to allow for the possibility of h tending to zero through negative values, or even a combination of positive and negative values. Now look at what happens in this case. If h is tending to zero through positive values we have, at all times, h |h| = =1 h h So the limiting value as h → 0 is 1. If h is tending to zero through negative values we have, at all times, −h |h| = = −1 h h So the limiting value as h → 0 is −1. So, in this case the Newton Quotient does NOT have a unique limiting value as h → 0. In such a case we say that the function f (x) does not have a derivative at this point, or that f (x) is a non-diﬀerentiable function (at this point). In conclusion, it is fairly easy to see that the function f (x) = |x| is diﬀerentiable for x = 0 with derivative −1 if x < 0 and derivative 1 if x > 0 and is not diﬀerentiable at x = 0. (Non-diﬀerentiability normally shows up as a kink or break in the graph, where we cannot make sense of the tangent.)

1.5.4

Derivatives of sin x and cos x

The derivatives of these two basic trig functions are d sin x = cos x dx d cos x = − sin x dx

It is very important to remember two things about these. First, remember the negative sign in the derivative of the cosine. Second, remember that the formulas only work if the angle x is measured in radians! We will come back to prove these results after doing the next section.

8

CHAPTER 1. THE DIFFERENTIAL CALCULUS (Hints and solutions start on page 65.)

Questions 1

Q 1.1. Go through the proof that the derivative of x3 is 3x2 , starting from the deﬁnition of the derivative. Q 1.2. Find the derivative of 1/(1 + x) from the deﬁnition. Q 1.3. The derivative of sin x is cos x. So the value of the derivative of sin x at x = 0 should be 1. Use your calculator to evaluate the Newton Quotient (sin(0 + h) − sin(0))/h for h = 0.1, 0.05, 0.01 and so on and compare with the limiting result. Remember to switch your calculator to radians! *Q 1.4. Draw a picture to illustrate the following statement. P is the point (x, f (x)) on the graph of y = f (x). R is the point (x + h, f (x + h)) and Q is the point (x − h, f (x − h)). The Newton Quotient is the slope of the chord PR. The limiting value of this slope as h → 0 gives the slope of the graph at P. On the other hand the slope of the chord QR seems to give a much better approximation to the slope of the graph at P and should give the value of the slope in the limit as h → 0. 1 (f (x + h) − f (x − h)) and try some examples Show that the slope of QR is given by 2h to compare this approximation to the derivative with that given by the Newton Quotient. Suppose I take the limiting value of this new expression as my new deﬁnition of f (x). Give an example of a function that is diﬀerentiable according to this new deﬁnition, but not diﬀerentiable according to the proper deﬁnition. Can you give a proof that if a function is diﬀerentiable at x according to the proper deﬁnition then it will also be diﬀerentiable at x according to the new deﬁnition.

1.6

Rules of Diﬀerentiation

So far we have had to work out all our derivatives from the deﬁnition. If this were the only way to do diﬀerentiation the Calculus would never have got oﬀ the ground. Luckily, it turns out that once we have a few derivatives under our belt the rest come more-or-less for free. This is because there are certain mechanical rules for diﬀerentiating functions that are combinations of functions that we know how to diﬀerentiate.

1.6.1

Rules

I will state six of the rules here. Then I will do some examples. Finally I will state the seventh rule.

1.7. EXAMPLES OF THE USE OF THE RULES Suppose that f and g are diﬀerentiable functions and that λ λ (λf ) (f + g) (f − g) (f g) f g = 0 derivative of constant is zero = λf can take out constants = f + g Sum Rule = f − g follows from [2] and [3] = f g + f g Product Rule = f g − fg g2 Quotient Rule
1

9 is a constant. Then [1] [2] [3] [4] [5] [6]

In the other notation we have, for example, df dg d (f g) = x.g + f. x. dx dx dx The rules are not independent of each other. You can easily check that [2] and [4] follow from the others. The rules tell us how to diﬀerentiate sums, diﬀerences, products and quotients of functions that we already know how to diﬀerentiate. The process can be used over and over again. For example, it follows immediately that (f + g + h) = (f + g) + h = f + g + h .

1.7
1.7.1

Examples of the Use of the Rules
Polynomials
p(x) = an xn + an−1 xn−1 + an−2 xn−2 + · · · + a1 x + a0

A polynomial of degree n is a function of the form

where the coeﬃcients an , an−1 , . . . are constants. For example, 2x − 3, x4 − 2x + 5 or πx10 − 1 x5 + 2.343 are polynomials. 2 Polynomials are easy to diﬀerentiate using the rule that we have produced for diﬀerentiating powers and rules [1], [2] and [3] above. Example 1.2. You will eventually diﬀerentiate polynomials in your head — they are that easy. But let me start by spelling the process out in great detail so as to show the way in which it uses the Rules. Consider the polynomial p(x) = x2 + 4x − 3. Using the Sum Rule (twice over) we get p (x) =
1

d d 2 d (x ) + (4x) + (−3). dx dx dx

For a discussion of greek letters, and a table of the alphabet, see the Appendix on page 64.

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CHAPTER 1. THE DIFFERENTIAL CALCULUS

d We already know how to get derivatives of powers, so we know that dx (x2 ) = 2x. The d d rule on taking out constants tells us that dx (4x) = 4 dx (x) = 4. Finally, the derivative of a constant is zero. So we end up with

p (x) = 2x + 4. As I said, you will get to do that mentally in time. Example 1.3. Now for a few more, without all the intervening details. d 2 d d 2 d (x + 3x − 4) = (x ) + 3 (x) − (4) = 2x + 3, dx dx dx dx d 5 d 5 d 4 d 4 (x − 4x + 2x − 3) = x − 4 x + 2 x = 5x4 − 16x3 + 2, dx dx dx dx d (2 − 3x8 + x16 ) = −24x7 + 16x15 . dx

1.7.2

The function f (x) = x−n

n = 1, 2, 3, 4 . . .

Using the quotient rule:
d 0.xn − 1. dx xn d 1 −nxn−1 d −n x = = = = −nx−n−1 dx dx xn (xn )2 x2n

So we now know that the rule d n x = nxn−1 dx works for all whole numbers n. In the same vein, if f (x) = 1/g(x) then f (x) = 0.g(x) − 1.g (x) g (x) =− . g(x)2 g(x)2

1.7.3

Long Products

This is an example of using the product rule repeatedly. Suppose that f , g and h are three diﬀerentiable functions and that p = f.g.h. Then, by the product rule: p = (f gh) = ((f g)h) = (f g) h + (f g)h = (f g + f g )h + f gh = f gh + f g h + f gh .

1.7.4

Trigonometric Functions
sin x , cos x cos x , sin x 1 , cos x 1 . sin x

We have already met sin and cos. The other standard trig functions are tan x = cot x = sec x = cosec x =

1.7. EXAMPLES OF THE USE OF THE RULES We can now use the rules (mainly the quotient rule) to diﬀerentiate these. d tan x = dx
d dx d sin x. cos x − sin x dx cos x cos2 x + sin2 x 1 = = . 2x 2x cos cos cos2 x

11

provided that x is measured in radians. So d tan x = sec2 x. dx Similarly
d 0. cos x − 1. dx cos x sin x tan x d sec x = = = . 2x 2x dx cos cos cos x

So sin x d sec x = = sec x tan x. dx cos2 x Similarly cos x d cosec x = − 2 = − cosec x cot x. dx sin x and d cot x = − cosec2 x. dx I suggest that you get to know the derivatives of sine, cosine and tan. The others can be worked out as you need them.

1.7.5

Rational Functions
p(x) q(x)

A Rational Function is a function of the form R(x) =

where p(x) and q(x) are polynomials. Here are two examples of rational functions: R(x) = x2 − 3x + 1 , x−3 R(x) = x10 3x − 5 . + x5 + 1

Rational functions can be diﬀerentiated by using the quotient rule, since we know how to diﬀerentiate polynomials. R(x) = For example R(x) = x2 + x + 1 x−3 R (x) = (2x + 1)(x − 3) − 1.(x2 + x + 1) x2 − 6x − 4 = . (x − 3)2 (x − 3)2 p(x) q(x) gives R (x) = p (x)q(x) − p(x)q (x) . q(x)2

12 Another example: R(x) = 1 − x2 x2 + x + 1

CHAPTER 1. THE DIFFERENTIAL CALCULUS

R (x) =

−2x(x2 + x + 1) − (1 − x2 )(2x + 1) , (x2 + x + 1)2

and I leave it to you to simplify the numerator.

1.7.6

The square root f (x) =

√

x

This can be done in a slightly indirect way2 as follows. Let g(x) = f (x).f (x) = x. Then, by the product rule, √ 1 = g (x) = f (x).f (x) + f (x).f (x) = 2f (x)f (x) = 2 xf (x). Therefore 1 d√ x= √ , dx 2 x 1 1 d 1 x 2 = x 2 −1 . dx 2 d a x = axa−1 . dx for all values of a. Example 1.4. Diﬀerentiate f (x) = 4x sin x + 3 cos x. Here we start to put together almost everything we know. In order, we use the sum rule, followed by the product rule and end up using the fact that we know how to diﬀerentiate the trig functions. f (x) = 4(1. sin x + x cos x) − 3 sin x = sin x + 4x cos x. Example 1.5. Diﬀerentiate f (x) = x sin x cos x. A function like this really needs the product rule twice over. f (x) = 1.(sin x cos x) + x. Example 1.6. Diﬀerentiate d (sin x cos x) = sin x cos x + x(cos x cos x − sin x sin x). dx f (x) =

or

We will see later that, in fact,

sin(x) + cos(x) . sin(x) − cos(x) This is, ﬁrst of all, a quotient. So we start by using the Quotient Rule: f (x) =
d (sin x dx d + cos x).(sin x − cos x) − (sin x + cos x) dx (sin x − cos x) . (sin x − cos x)2

2

A more direct way is given in Exercise 1.16

1.7. EXAMPLES OF THE USE OF THE RULES We then do the remaining derivatives in the numerator and tidy up f (x) = (cos x − sin x)(sin x − cos x) − (sin x + cos x)(cos x + sin x) . (sin x − cos x)2

13

Now multiply out the numerator and show that it simpliﬁes down to -2 and that the answer is therefore −2 f (x) = (sin x − cos x)2

Questions 2
f (x) = x4 ,

(Hints and solutions start on page 65.)

Q 1.5. Diﬀerentiate the following functions 1 g(x) = x5 + x − 3, 2 h(t) = 4t6 − 3t5 + 2t − 4, k(p) = p4 + 3p−3 + 2p−1 .

Q 1.6. Now diﬀerentiate these p(x) = 2 sin x, q(x) = tan x + sin x, r(θ) = 3 cos θ − 4 sin θ,
5 1

f (x) = x 2 − x− 2 ,
1

1

1

m(λ) = λ2 − 4 sin λ − 3 cos λ,

h(t) = 3t 4 − 6t 3 + t−2 ,

φ(x) = tan x + 3 cos x − x 2 .

Q 1.7. Diﬀerentiate the following functions using the product rule. f (x) = x sin x, p(x) = x(1 + 2 sin x), u(x) = (tan x − 1)(tan x + 1), g(x) = cos x sin x, r(x) = cos2 x.

q(x) = (sin x + cos x)(sin x − cos x). v(x) = tan2 x, w(t) = (1 + t 4 )(t2 − t3 ).
5

Q 1.8. Diﬀerentiate the following, starting with the quotient rule. p(x) = 1 , x r(x) = x2 x , +1 s(x) = 3x + 1 , 2x + 3 t(x) = x2 2x + 1 . −x+1 1 1. 1+ x

x2 + 1 u(x) = 2 , x −1 sin x , 1 + cos x

t2 + t + 1 e(t) = 2 , t −t+1 cos x , 1 − sin x

√ t−t m(t) = √ , t+t 1 , sin x

l(x) =

v(x) =

w(x) =

z(x) =

f (x) =

1 . cos x sin x

14

CHAPTER 1. THE DIFFERENTIAL CALCULUS

Q 1.9. Find the equation of the tangent line to the curve y = x3 at the point (1, 1). Where does the tangent meet the x-axis? Q 1.10. Find the equation of the tangent to the graph of y = sin(x) at the point where x = π. Find the points where this tangent meets the x and y-axes.
1 1 Q 1.11. Let P be the point (a, a ) (a = 0) on the graph y = x where a > 0. Find the equation of the tangent line at P. Find where this line meets the x- and y-axes. What is the area of the triangle formed by the tangent and the two axes? What is surprising about the answer?

Q 1.12. The Normal at a point P on a curve is the line through P that is perpendicular to the tangent line. What is the slope of the normal in terms of the slope of the tangent? Find the equation of the normal at the point (a, a2 ) on the curve y = x2 . Where does it meet the y-axis? Q 1.13. For what values of x is the function f (x) = x(1 − x)(2 − x) increasing and for what values is it decreasing? Draw a picture. Q 1.14. Consider the function f (x) = (x2 +1)/(x2 −1). For what values of x is this function deﬁned? For what values of x is it increasing and for what values of x is it decreasing? Try to draw a picture in this case too. *Q 1.15. Sketch the graph of y = x2 . Find the equation of the tangent to this graph at the point P with coordinates (a, a2 ). Find the point Q on the graph at which the tangent is perpendicular to the tangent at P (a = 0). Show that the tangents at P and Q meet on the horizontal line y = − 1 . What is the corresponding result for a circle? 4 *Q 1.16. Show that √ √ a−b √ . a− b= √ a+ b √ Use this to ﬁnd the derivative of f (x) = x from the deﬁnition of the derivative using Newton Quotients.

1.8

Derivative of sin x where x is in radians

As promised (or threatened) earlier, we are now going to ﬁnd out the derivative of sin(x) from ﬁrst principles. Please remember that x is measured in radians. The Newton Quotient is sin(x + h) − sin x f (x + h) − f (x) = h h

1.8.

DERIVATIVE OF SIN X WHERE X IS IN RADIANS

15

What happens to this as h → 0? This turns out to be slightly diﬃcult to answer. Once more we get nowhere by putting h = 0. We have to be cleverer than that. There is a trig formula that says sin A − sin B = 2 cos A+B 2 sin A−B 2

(you should prove this by expanding out the RHS) So sin(x + h) − sin x = 2 cos(x + h ) sin( h ) and hence 2 2 Newton Quotient = sin(x + h) − sin x h sin(h/2) = cos(x + ) · h 2 h/2

How does this help us? Let me start with the easy bit. As h → 0 the value of cos(x + h/2) tends to cos x. It’s the other bit that gives problems. What happens to the ratio sin(h/2) as h/2 h → 0? Let us try plugging in some values to see what we get h 0.2 0.1 0.01 sin(h/2) h/2 0.99831 0.99958 0.9999958

This rather suggests that the ratio is tending to 1. That is indeed true and we will now prove it. Theorem 1.17. As θ → 0 then sin(θ)/θ → 1 (θ in radians). Proof. Consider Fig. 1.4, showing a sector of a circle of radius R and having angle θ (where θ is positive and less than a right-angle).

C

D

O

θ A B

Figure 1.4: The derivative of sin x from ﬁrst principles. By trigonometry, OA = R cos θ, AC = R sin θ and BD = R tan θ. It is clear from the diagram that area OAC < area of sector OBC < area OBD

16 Or

CHAPTER 1. THE DIFFERENTIAL CALCULUS

1 1 1 .OA.AC < R2 θ < .OB.BD 2 2 2 Putting in the values that we have obtained (with OB = OC = R) we get 1 2 1 1 R sin θ cos θ < R2 θ < R2 tan θ 2 2 2 1 2 Now divide through by 2 R and get sin θ cos θ < θ < tan θ =

sin θ cos θ We are thinking of θ as a small positive angle (!) so sin θ > 0. So we can divide through by sin θ and get θ 1 cos θ < < sin θ cos θ Now let θ → 0. The outer terms in this inequality are both tending to the value cos 0=1. The term in the middle is sandwiched between them at all times, so it must be tending to the limit 1 as well. So θ → 1 as θ → 0 sin θ and hence sin θ → 1 as θ → 0 θ Strictly speaking, we have only proved this for positive values of θ. But changing the sign of θ does not aﬀect the value of sin(θ)/θ, so our argument works for negative values of θ as well. Now we go back to our original Newton Quotient h sin(h/2) sin(x + h) − sin(x) = lim cos(x + ) · = cos(x).1 = cos x lim h→0 h→0 h 2 h/2 So, so long as x is measured in radians d sin x = cos x dx In a similar way it can be shown that d cos x = − sin x dx
Footnote: Look at the following argument. We know how to diﬀerentiate sin(x). We also know that sin2 x + cos2 x = 1. So, diﬀerentiating both sides of this equation wrt x, using the product rule on sin2 x = sin(x) sin(x) etc., we get 2 sin(x) cos(x) + 2 cos(x) Solving this for the derivative gives us d cos x = − sin x. dx The answer is right, but what do you think of the logic? Is it acceptable? d (cos(x)) = 0. dx

1.9. THE CHAIN RULE

17

1.9

The Chain Rule

This is the ﬁnal rule of diﬀerentiation. It is a bit more complicated than the others. Let me start by saying something about functions. Up to now we have been treating functions as being, basically, formulas “y = x2 + 1”. A better way to think of a function like f (x) is as a machine, a piece of electronic wizardry, which takes in x as its input and gives out f (x) as its output. x → f → f (x) A formula, like f (x) = x2 + 1, is then just describing what the machine does. If x goes in x2 + 1 comes out. If α goes in α2 + 1 comes out. If 2 goes in 5 comes out. If x + 1 goes in (x + 1)2 + 1 comes out. All that we ask of a function is that it be ‘well-deﬁned’, i.e. that the output is totally determined by the input. There is no necessity that the function should be given by a ‘formula’ in any straightforward sense. For later reference I should also point out that we sometimes need to put conditions on the type of input that is allowed — the domain of deﬁnition of the function. A function that is supposed to work out the smallest factor of a whole number is obviously not going to take too kindly to being given an input like 23.2342 because a number like this does not have ‘factors’. Similarly, a function that is supposed to be working out the square root of a number will not approve of a negative input. If we have two functions, f and g—two machines, we can think of forming a single machine by plugging the output of one into the input of the other (connecting them in series). g → f → f (g(x)) x → This new machine represents a function called the composition of f and g. Note that the order matters, the following is probably a diﬀerent machine x → f → √ g → g(f (x))

So, for example, the function h(x) = 1 + x2 can be thought of as the result of com√ posing two functions f (x) = 1 + x2 and g(x) = x √ f → 1 + x2 → g → 1 + x2 x → √ More elaborately, the function k(x) = sin 1 + x2 can be thought of as the result of composing three functions in series √ √ x → f → 1 + x2 → g → 1 + x2 → h → sin 1 + x2 √ f (x) = 1 + x2 g(x) = x h(x) = sin x The Chain Rule is a rule for diﬀerentiating functions that are built up in this way. Let me start with the simple case of two functions. Rule 7: Chain Rule If f and g are diﬀerentiable functions and h(x) = f (g(x)) then h (x) = f (g(x))g (x)

18

CHAPTER 1. THE DIFFERENTIAL CALCULUS Another way to say this is as follows: if y = h(x) then y = f (u) where u = g(x) and dy du dy = dx du dx

This is an easy formula to remember because of the ‘cancellation’. More generally, if we have a whole string of functions composed together, e.g. y = f (u) where u = g(v), where v = h(w), where w = k(x) (so y = f (g(h(k(x))))) then dy du dv dw dy = dx du dv dw dx —you can see the pattern. Example 1.7. f (x) = sin(x2 ) Think of this as y = sin u where u = x2 . Then dy = cos u and du So the Chain Rule says that dy du dy = = cos(u).2x = 2x cos(x2 ) dx du dx (A small note in passing: sin xn always means sin(xn ), whereas (sin x)n is usually written as sinn x.) √ Example 1.8. f (x) = 1 + x2 √ Think of this as y = u, where u = 1 + x2 . Then 1 1 dy = √ du 2 u So the Chain Rule says that dy du 1 1 x dy = = √ .2x = √ dx du dx 2 u 1 + x2 Example 1.9. f (x) = sin(1 + cos x) Think of this as y = sin u, where u = 1 + cos x. Then dy = cos u and du So the Chain Rule says that dy du dy = = cos(u).(− sin x) = − sin(x) cos(1 + cos x) dx du dx du = − sin x dx and du = 2x dx du = 2x dx

1.9. THE CHAIN RULE Example 1.10. f (x) = 1 + sin2 (1 + x2 ) This goes deeper. Think of it as y= Then 1 1 dy = √ du 2 u So the Chain Rule says that dy dy du dv dw 1 1 2x sin(1 + x2 ) cos(1 + x2 ) = = √ .2v. cos(w).2x = dx du dv dw dx 2 u 1 + sin2 (1 + x2 ) du = 2v dv dv = cos w dw dw = 2x dx √ u, u = 1 + v2, v = sin w, w = 1 + x2

19

Note that there is often nothing clear cut about the way in which you choose to break down a function. The usual idea is that you should try to break the function up into pieces that are small enough to diﬀerentiate easily. The more experienced you are, the bigger the pieces can become. Example 1.11. f (x) = x2 + tan(x2 ) This requires a combination of rules. First treat it as a sum: d df = 2x + tan(x2 ) dx dx Now do the remaining derivative by chain rule: y = tan u, u = x2 dy du dy = = sec2 (u).2x = 2x sec2 (x2 ) dx du dx So we ﬁnally get f (x) = 2x + 2x sec2 x2 Example 1.12. x 1 + x2 Here we have to start by using the quotient rule: f (x) = √ √ √ d 1. 1 + x2 − x. dx ( 1 + x2 ) f (x) = 1 + x2 Now we complete the job by using the chain rule to show that x d √ ( 1 + x2 ) = √ dx 1 + x2 I leave it as an exercise to show that the ﬁnal result simpliﬁes down to f (x) = 1/(1+x2)3/2 .

20

CHAPTER 1. THE DIFFERENTIAL CALCULUS (Hints and solutions start on page 66.)

Questions 3

Q 1.18. If f (x) = x2 + 1 and g(x) = 1/x what are f (g(x)), g(f (x)) and g(g(x))? Q 1.19. If f (x) = x + 1, g(x) = 1/x and h(x) = 1 − x what are the functions f (g(h(x))), f (h(g(x))), g(h(f (x))) and so on through all six possibilities? Q 1.20. Diﬀerentiate the following, using the chain rule. √ 1 1 + x2 , (2x + 1)5 , (3x − 2) 3 , sin3 x, cos10 x, tan(1 + x2 ), (x3 + 2x − 6) 3 ,
2

sin 5x,

cos(3x + 1), √

sin(cos x), √

1 cos(t + ), t sin √ sin t,

tan x,
1

2 sin3 x − 3 cos3 x,

sin x − cos x,

√ sin2 ( t),

sin 3 x2 .

Q 1.21. If f (x) = x2 + c what are f (f (x)) and f (f (f (x)))? Q 1.22. Now for a general pile of functions using all the rules. 1+x , 1−x sin x + cos x,
2 2

1 , sin(x2 )

√

x , 1 + x2 √ u √ , u− u u+

1 , 1 + sin4 t 1+x 1−x
2 3

x+

x2 + 1 , x2 − 1

tan(tan x),

,

√ sin( 1 + x2 + x),

sin(2x) sin(3x),

3 sin2 x + 2 cos2 x.

Q 1.23. You should know from school that, if x = 1, 1 + x + x2 + x3 + x4 + · · · + xn = xn+1 − 1 . x−1

Diﬀerentiate both sides of this equation and put x = −1. Show that you get 1−2+3−4+···±n = Check that this is correct for n = 4 and n = 5. *Q 1.24. If f (x) = sin(sin(sin(sin x))) what is f (x)? 1 − (2n + 1)(−1)n . 4

1.9. THE CHAIN RULE

21

*Q 1.25. [Only attempt this if you already know about the natural logarithm function ln(x) and that its derivative is 1/x.] Consider the two functions f (x) = ln(x) and g(x) = sin(x). What is the function p(x) = f (f (g(x)))? What is p (x)? For what values of x is p(x) deﬁned? For what values of x is p (x) deﬁned ?!!? *Q 1.26. s(x) and c(x) are two diﬀerentiable functions of x which are deﬁned for all values of x. Suppose that s(0) = 0, c(0) = 1 and that, for all values of x, s (x) = c(x) and c (x) = −s(x). Let A(x) = s(x)2 + c(x)2 . Show that A (x) = 0 and deduce that s(x)2 + c(x)2 = 1 for all values of x. Let B(x) = (s(x) − sin(x))2 + (c(x) − cos(x))2 . Show that B (x) = 0 and deduce that s(x) = sin(x) for all values of x. *Q 1.27. You know that the functions sin x and cos x take values between −1 and 1. You also know that a function with a positive derivative is increasing in value. Consider the function f (x) = x − sin x. Obviously, f (0) = 0. Now diﬀerentiate f (x) and show that for 0 < x < π this derivative is positive. Deduce that for this range of values x > sin x (why did I stop at π, and does it really matter?) Use a similar argument to show that, on this range of values, 1 cos x > 1 − x2 2 and and, if you still have the energy, 1 1 1 1 − x2 < cos x < 1 − x2 + x4 2 2 24 How accurately does this tell you the value of cos 0.2 and cos 0.1 ? If you are feeling totally masochistic you can try to show that 1 1 1 6 1 1 x < cos x < 1 − x2 + x4 1 − x2 + x4 − 2 24 720 2 24 How accurately do you know cos 0.2 now? What about cos 0.4? 1 x − x3 < sin x < x 6 c(x) = cos(x)

22

CHAPTER 1. THE DIFFERENTIAL CALCULUS

1.10

Proofs of some of the Rules

I cannot give you strict proofs of the rules because they are based on the manipulation of limits and we have not done the theory of limits precisely—I have taken them as being ‘obvious’. But I can indicate the ideas behind the proofs. I will just do two of them, the Sum Rule and the Product Rule. First the Sum Rule. Let f (x) and g(x) be diﬀerentiable functions. Let p(x) = f (x) + g(x). By deﬁnition, p(x + h) − p(x) p (x) = lim h→0 h (f (x + h) + g(x + h)) − (f (x) + g(x)) = lim h→0 h f (x + h) − f (x) g(x + h) − g(x) + = lim h→0 h h f (x + h) − f (x) g(x + h) − g(x) = lim + lim h→0 h→0 h h = f (x) + g (x) as required. You can now easily supply a similar proof for the subtraction rule, it just involves changing a few signs. The Product Rule is a lot more tricky (in fact it managed to fox Newton himself and the rule was ﬁrst found by Leibniz). The method is similar to the one we have just used, but we have to be rather more canny in how we do the manipulation. Let f and g be as before and let p(x) = f (x)g(x). Now write down the Newton Quotient for p: f (x + h)g(x + h) − f (x)g(x) p(x + h) − p(x) = h h At ﬁrst sight the numerator of the RHS does not seem to break apart very easily into an f bit and a g bit. But then we remember that the product rule itself (f g) = f g + f g is rather complicated. The trick is to write the RHS of the above equation as 1 (f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)) h All we have done here is add in a term and then subtract it oﬀ again. That, as it turns out, is the required sleight of hand. We can now re-bracket this expression as g(x + h) − g(x) f (x + h) − f (x) · g(x + h) + f (x) · h h (check that I have not changed anything!). Now take the limit as h → 0. The two ‘newton quotients’ in the expression tend to the corresponding derivatives in the limit and the only other thing that is changing is g(x + h) which just tends to g(x). So, as h → 0 we get the limit p (x) = f (x).g(x) + f (x).g (x) and we have proved the Product Rule.

1.11. HIGHER DERIVATIVES

23

1.11

Higher Derivatives

The derivative of a function is a function and it may be possible to diﬀerentiate it again— and again and again. The derivative of the derivative is called the Second Derivative d d2 f = 2 dx dx also denoted by f (x). The derivative of this is called the Third Derivative by f (x). And so on. The nth derivative is denoted by f, f , df dx d d3 f = 3 dx dx d2 f dx2

also denoted

dn f . dxn

In ‘dash’ notation the derivatives are called f iv , f v, . . . f (n)

f ,

So, for example, the tenth derivative of f is written f (10) . Just where the Roman numerals stop and the brackets take over is a matter of personal preference. Example 1.13. If f (x) = x3 + 2x2 + 3x + 1 then f (x) = 3x2 + 4x + 3 f (x) = 6x + 4 f (x) = 6 f iv (x) = 0 f v (x) = 0 and so on. If f (x) = sin x then f (x) = cos x f (x) = − sin x f (x) = − cos x f iv (x) = sin x f v (x) = cos x and so on.

1.11.1

Geometrical Interpretation of Second Derivative

f (x) gives the slope of the function—the ‘rate’ at which f is changing. f (x) gives the rate at which the slope is changing and gives, roughly, the rate at which the graph is bending.

24

CHAPTER 1. THE DIFFERENTIAL CALCULUS

f’’(x) > 0

f’’(x) < 0

Figure 1.5: The sign of the second derivative shows the shape of the curve. You will often hear politicians, in desperation, talking about the rate of decrease of the rate of increase of something unpopular. Look at the graphs in Fig 1.5 In the ﬁrst the slope is increasing as x increases, so the derivative of the slope is positive. f (x) > 0 (Remember that slope is a signed quantity. Most people would say that the ﬁrst graph slopes more at the left-hand end than it does in the middle and that therefore the slope is initially decreasing rather than increasing. Actually, it is going from very negative to less negative and that is an increase—(-2) is bigger than (-3).) In the second the slope is decreasing as x increases, so the derivative of the slope is negative. f (x) < 0 Example 1.14. Fig 1.6 indicates the various possibilities

f’(x) > 0

f’(x) < 0

f’(x) > 0

f"(x) < 0

f"(x) > 0

Figure 1.6: First and second derivatives geometrically.

1.11.2

Kinematical Interpretation

If a particle is moving along the x-axis so that its position at time t is given by x(t), then we have seen that x(t) is the velocity of the particle at time t. ˙ The second derivative, denoted by x(t), gives the rate of change of velocity—the accel¨ eration.

1.11. HIGHER DERIVATIVES

25

This is a very important concept because Newton’s Laws of motion relate the force acting on a particle to its acceleration. Example 1.15. Suppose that a particle has position x(t) = A sin(ωt) at time t, where A and ω are constants. Find its velocity and acceleration. Its velocity is x(t) = Aω cos(ωt) and its acceleration is x(t) = −Aω 2 sin(ωt). ˙ ¨

Questions 4

(Hints and solutions start on page 68.)

Q 1.28. Work out the ﬁrst ten derivatives of f (x) = sin x. What is the 300th derivative of f (x)? Now work out the ﬁrst ten derivatives of f (x) = x sin x. Can you guess the value of f (300) (x)? √ Q 1.29. Work out the ﬁrst 5 derivatives of f (x) = x. Can you work out an expression for the nth derivative of f (x) using factorials? (Last bit is quite hard.) Q 1.30. Show that the function x(t) = A sin(ωt + α), where A, ω and α are constants, satisﬁes the so-called Harmonic Oscillator equation x = −ω 2 x ¨ Q 1.31. Show that y = (cos(2x) + sin(2x))/x satisﬁes the equation 2 y + y + 4y = 0 x *Q 1.32. I have a function f (x). I diﬀerentiate it repeatedly and eventually end up getting zero. What kind of function must f have been? *Q 1.33. The product rule gives you a formula for (f g) . Work out similar formulas for the second, third and fourth derivatives of f g. If you know about such things as Binomial Coeﬃcients, or the Pascal triangle, you can guess at the formula for the nth derivative (Leibnitz’s Theorem). *Q 1.34. Let T stand for tan(x) and let Tk stand for the k th derivative of tan(x). Why is T1 = 1 + T 2? Use the chain rule to show that T2 = 2T + 2T 3 and T3 = 2 + 8T 2 + 6T 4. Also work out T4 , . . . , T7 . Now look at the pattern of the coeﬃcients of all these polynomials (make up a table). Conﬁrm the following claims in the cases that you have calculated: (1) Tn is a polynomial of degree n + 1 in T ; (2) if n is odd then Tn just involves even powers of T , and vice versa if n is even; (3) the coeﬃcient of T n+1 in Tn is n!. Can you spot a formula for the second highest coeﬃcient of Tn ? Can you now have a shot at justifying these claims in the general case — not just in the examples that you have worked out.

26 *Q 1.35. If y = x + y = z n .) √ 1 + x2
n

CHAPTER 1. THE DIFFERENTIAL CALCULUS show that (1 + x2 )y + xy − n2 y = 0. (Could try writing

*Q 1.36. Let fn (x), where n is a positive integer, be given by f (x) = xn if x ≥ 0 and f (x) = 0 if x < 0. Sketch the graphs of f1 , f2 and f3 . Show that f1 (x) is not diﬀerentiable at x = 0. What is the derivative of f2 (x)? How many times can we diﬀerentiate f2 (x)? How many times can we diﬀerentiate fn (x)?

1.12

Limits & Continuity

This is a brief section meant to give you the basic idea about these things. A proper treatment will have to wait until second year.

1.12.1

Limits

I have been using the idea of a ‘limiting value’ of a function, limx→a f (x), and I have been taking the meaning as being ‘obvious’—though it really ought to be deﬁned properly (see later year). 1 Of course, limits may not exist. For example, we would say that x does not have a limit as x → 0. In this kind of case we often write something like 1 → ∞ as x x→0

1 More complicatedly the function f (x) = sin x does not have any kind of limiting value as x → 0. It is reasonably easy to calculate with limits, if we know that they exist and behave themselves. We have the following simple rules for some simple situations. If f (x) and g(x) are functions that have limiting values α and β as x → a:

x→a

lim f (x) = α

x→a

lim g(x) = β

then f (x) ± g(x) → α ± β f (x)g(x) → αβ f (x)/g(x) → α/β if

β=0

Questions 5

(Hints and solutions start on page 69.)

Q 1.37. What is the limiting value of f (x) = 1/x as x → ∞? What is the limiting value of f (x) = cos(1/x) as x → ∞?

1.12. LIMITS & CONTINUITY Q 1.38. Work out the values of the following limits 1 , x→∞ 1 + x2 lim x2 + 1 , x→∞ x2 − 1 lim
x→∞

27

lim

√

x+1−

√

x .

I suggest that you try dividing top and bottom by x2 in the second one. For the last one try a − b = (a2 − b2 )/(a + b). Q 1.39. Consider the function f (x) = (x2 − 1)/(x − 1). If we put x = 1 in the formula we get f (0) = 0/0, which makes no kind of sense. What is the limiting value of f (x) as x approaches 1? (This is where you tend to hit a doctrinal diﬀerence between Pure and Applied mathematicians. Pure mathematicians tend to say that the function f (x) is not deﬁned for x = 1 though, as you should have seen, it has a perfectly decent limiting value as x → 1. Applied mathematicians tend to argue that f (x) = ((x − 1)(x + 1))/(x − 1) = x + 1 and is therefore a perfectly well deﬁned function everywhere.)

1.12.1

Continuity

Funny business with quotients apart, you would normally expect to be able to work out limx→a f (x) simply by putting x = a in f (x). In other words, you would expect that as x→a f (x) → f (a)

This may not be true. For example, consider the function given by f (x) = 1 0 x≥0 x<0

The graph of the function is sketched in Fig. 1.7. As x → 0 through negative values f (x) is always zero. So, approaching 0 in this direction: f (x) → 0 as x→0

But the value of f at x = 0 is not zero, it is 1. This happens because of the ‘break’ in the function at x = 0. This leads to the following important deﬁnition: a function f (x) is said to be Continuous if, for every value a, f (x) → f (a) as x → a.

Fig. 1.8 shows an example when this fails - in this case, the function has a jump discontinuity. It is reasonably easy to prove, by looking at the Newton Quotient, that a function that can be diﬀerentiated everywhere must be continuous. Since the bottom line of the Newton

28

CHAPTER 1. THE DIFFERENTIAL CALCULUS

y

x
Figure 1.7: The graph of a step function.

discontinuity Figure 1.8: A discontinuity.

Quotient is tending to zero the top line must as well, for the limit to exist. This means that f must be continuous. (We actually used this property when ‘proving’ the product rule of diﬀerentiation— recall that I had to use the fact that g(x + h) → g(x) as h → 0. This was o.k. because we were assuming that g was diﬀerentiable.) The converse is certainly not true. The function f (x) = |x| is certainly continuous, but is not diﬀerentiable at x = 0.
It is easy enough to produce examples of functions that are not continuous at any value of x, though it is not easy to draw their graphs! For example, deﬁne f (x) to have the value 1 if x happens to be a number of the form n/2m , where n and m are whole numbers, and value 0 everywhere else. (N.B. functions do not have to be given by nice algebraic formulas — it is suﬃcient that we know what the value of the function is going to be for each value of x.) Functions that are not continuous anywhere certainly cannot be diﬀerentiable anywhere either. What about functions that are continuous but not diﬀerentiable? We have seen examples of this (e.g. f (x) = |x|) but all our examples have just involved functions that have failed to be diﬀerentiable at a few isolated points. Can we conceive of a function that is continuous but which does not have a derivative anywhere? Put another way, can we conceive of a graph that has no breaks in it (continuous) but which does not have a slope anywhere (the derivative would give the slope). Such functions were in fact discovered about 100 years ago by Weierstrass. Until recently they were regarded as belonging to one of the more esoteric branches of pure mathematics—the type of kinky example that we have to think up to use as a counterexample for over-hasty conjectures. In the last few years there has been a remarkable turn around. I suspect that most of the readers of these notes have seen pictures of such functions. Nowadays they go under the name of Fractals. Fractals, by the way in which they are constructed, almost always turn out to be continuous but never diﬀerentiable because they are usually made by an inﬁnite process of building kinks on kinks.

1.12.2

* Rolle’s Theorem and the Mean Value Theorem

Rolle’s theorem is an important basic result about diﬀerentiable functions. Like many basic results in the calculus it seems very obvious. It just says that between any two points where the graph of the diﬀerentiable function f (x) cuts the x-axis there must be a point where f (x) = 0. The following picture illustrates the theorem. Like most important theorems, Rolle’s theorem has to be stated rather carefully in order to make sure that it is true. I will now state a simple, but precise, version of the theorem: Theorem 1.40 (Rolle’s Theorem). Suppose that f (x) is a diﬀerentiable function whose derivative is a continuous function. Suppose that f (a) = 0 and f (b) = 0 (with a < b, let’s

1.12. LIMITS & CONTINUITY
f’(c) = 0

29

x a c b

Figure 1.9: Rolle: somewhere between a and b, the graph must be ﬂat. say). Then there must be at least one point c between a and b (a < c < b) at which f (c) = 0. (The precise behaviour of f (x) outside the interval a ≤ x ≤ b is not really relevant and the theorem can be stated in a more general form.) Suppose we accept this result as being intuitively obvious (it can be proved). Now consider the following extension. As before, let f (x) be a function that is diﬀerentiable and whose derivative is continuous. Choose any points a < b and deﬁne a new function F (x) by the formula F (x) = f (x) − f (a) − x−a (f (b) − f (a)) b−a

F is also a diﬀerentiable function with a continuous derivative. The derivative is easy to calculate—it is f (b) − f (a) F (x) = f (x) − . b−a Now notice that F (a) = F (b) = 0 (check this). So we can apply Rolle’s theorem to F (x) and say that there must be a point c between a and b at which F (c) = 0. But this says that there is a point c between a and b at which 0 = f (c) − f (b) − f (a) b−a

Rearranging this equation we get the following important theorem Theorem 1.41 (Mean Value Theorem). If f (x) is a diﬀerentiable function with a continuous derivative and if a < b are any two points then there is a point c between a and b at which f (c) = f (b) − f (a) b−a

This result is also ‘geometrically obvious’ once you draw the right picture. The expression (f (b) − f (a))/(b − a) is just the slope of the chord AB. So the theorem is saying that there must be a point c between a and b at which the slope of the graph is equal to the slope of the chord AB. This may seem very simple—it is very simple—but it is also very important. In most of its applications we think of the result in the following way, which is easily derived. f (b) = f (a) + (b − a)f (c)

30

CHAPTER 1. THE DIFFERENTIAL CALCULUS

B

A

a

c

b

Figure 1.10: Mean Value Theorem: somewhere between a and b, the graph must be parallel to the chord AB. (Note that the theorem does not tell us what c is—just that it must be there.) Here’s a simple application of this theorem. Suppose that f (x) > 0 between a and b (a < b). Then (b − a)f (c) must be positive (whatever c is). So f (b) must be bigger than f (a). So, in an obvious sense, the function must be increasing. Here’s another typical application. Let f (x) = tan(x). Then f (x) = 1/ cos2 (x) > 1. So we must have tan(b) − tan(a) > b − a This result is rubbish. For example, tan(π) = tan(0) = 0. But 0 − 0 is not bigger than π − 0. What’s gone wrong? The problem here is that tan(x) is not a diﬀerentiable function between 0 and π—it blows up at π/2. When using these theorems, no matter how ‘obvious’ they are, you really must check that the conditions of the theorem are precisely satisﬁed.

Chapter 2 Further Diﬀerentiation
2.1
2.1.1

The Inverse Trigonometric Functions
Inverse Functions in general

If y = 3x − 1 then x = (y + 1)/3. The ﬁrst formula deﬁnes a function f (x) = 3x − 1 x → f → 3x − 1 The second formula gives the inverse (reverse) of this function, which we usually denote by the very unfortunate notation f −1 (x). x → f −1 → x+1 3

All that we are doing is switching round the input and output of the machine—putting in at the ‘out’ end and getting out at the ‘in’ end. The inverse function undoes the eﬀect of the original function: x → f → f −1 → x (Note the notational problem. Does f −1 mean the inverse function of f or does it mean 1/f ? I’m afraid that the answer has to be ‘it depends on the context’. You hope that it is obvious which interpretation the author has in mind. If it is not, then ﬁnd another author! We really should have a better notation but once these things settle down it is very diﬃcult to change them.) There are problems. We cannot always deﬁne an inverse in a straightforward way. Consider the function f (x) = x2 . What can we make of f −1 ? If we put x = 1 into the f machine we get out 1. If we put −1 into the machine we also get out 1. So what do we expect to get out of the inverse machine if we put in 1? You see the problem. This machine is not uniquely reversible. 1, −1 → f → 1 31 1 → f −1 →????

32

CHAPTER 2. FURTHER DIFFERENTIATION

The problem that we face here is described by the technical term one-to-one. A function f is one-to-one if you never have f (x) = f (y) when x = y. The function f (x) = x2 is NOT one-to-one because, for example, f (1) = f (−1). √ Nevertheless, we can say that, in some sense, g(x) = x is an inverse for f . Certainly, most people would say that g undoes the eﬀect of f x → f → x2 → g → x but this is not quite correct as it stands. It becomes correct if we put on the extra condition that x is not allowed to be negative and insist that the result of the square root is not negative.

Questions 6

(Hints and solutions start on page 69.)

Q 2.1. What are the inverses of the functions f (x) = 2x + 4 and g(x) = 2 − 5x? Q 2.2. What is the inverse of the function f (x) = 1/x for x > 0? Q 2.3. The function f (x) is deﬁned as follows: if x ≥ 0 then f (x) = 1 + x and if x < 0 then f (x) = 1 + x/2. What is the inverse of f ? *Q 2.4. If the inverse of the function f (x) for some range of values of x is also f (x) what does that tell you about the graph of f (x) on that range? If f (x) is a diﬀerentiable function and f (f (x)) = x for all values of x can f be anything other than ±x? (This is a problem for thinking about rather than answering precisely.) *Q 2.5. Suppose that a, b, c, d are constants. Consider the following function (known as a M¨bius Transformation) o ax + b f (x) = . cx + d Suppose that ad = bc. Show that if c = 0 there is one value of x at which f (x) is not deﬁned and that there is one number that is not a possible value of f . Show that if you forget about the point where f is not deﬁned then f is a one-to-one function. What is the inverse of f (if you exclude the bad points)? Show that any M¨bius Transformation can be o built up as a composition of some number of the functions p(x) = x + u, r(x) = vx (v = 0) and q(x) = 1/x (note that these are all M¨bius Transformations themselves). o

2.1.1

The Inverse Trigonometric Functions

The Inverse Trigonometric Functions are meant to be the ‘inverses’ of the standard trigonometric functions. There are obviously going to be diﬃculties in trying to deﬁne them because none of the standard trig functions are one-to-one.

2.1. THE INVERSE TRIGONOMETRIC FUNCTIONS

33

We have to start somewhere so let me start by giving the ‘obvious’ (and obviously wrong) deﬁnitions of the inverse trig functions. I will then try to put them right. The inverse functions of sin, cos and tan are called arcsin, arccos and arctan. The notations sin−1 (x), cos−1 (x) and tan−1 (x) are also common, though they tend to be confusing to beginners. arcsin(x) is the angle whose sine is x arccos(x) is the angle whose cosine is x arctan(x) is the angle whose tan is x if y = sin x then x = arcsin y if y = cos x then x = arccos y if y = tan x then x = arctan y That’s what we want to mean by the inverse trig functions, but the deﬁnitions do not make sense as written, simply because the trigonometric functions are not one-to-one. For example, sin 0 = 0 and sin π = 0 and so on. So what is arcsin 0? Is it 0 or π or what? We get round this diﬃculty by carefully restricting the values that the inverse trigonometric functions can take. Let me start with sin x. This function is not one-to-one, but it becomes one-to-one if we re1 strict attention to the range −π/2 ≤ x ≤ π/2. It also takes all its possible values in this range. If we make the range any wider then the function ceases to be one-to-one. If we make the range any 0 π narrower then the function does not take all its −π 0 2 2 possible values. Deﬁnition 2.6. arcsin(x) is the angle between − π and π whose sine is x. (−1 ≤ x ≤ 1) 2 2 For the cosine we have to pick a diﬀerent range because cosine is not one-to-one on the range −π/2 ≤ x ≤ π/2. Instead, we choose the range 0 ≤ x ≤ π. This works.

-1

Figure 2.1: Graph of sin x.

Deﬁnition 2.7. arccos(x) is the angle between 0 and π whose cosine is x. (−1 ≤ x ≤ 1) The graphs of the inverse trigonometric functions are given in Figs. 2.2 and Fig. 2.3.
These show the general pattern for the graphs of inverse functions. All we are doing is reversing the original x and y axes — which is the same thing as ﬂipping the graph over the line y = x. The resulting graph is a genuine graph (doesn’t give more than one y value for any given x value) provided that the original function has an inverse.

For the tan we can use more or less the same range as we did for sine, except that we have to leave out the end points −π/2 and π/2 because tan is not deﬁned at these values (it blows up).

34
π 2

CHAPTER 2. FURTHER DIFFERENTIATION y π y arccos(x) x
1

arcsin(x)
-1

π 2

−π 2 Figure 2.2: Graph of arcsin(x). Deﬁnition 2.8. arctan(x) is the angle in the range − whose tan is x. (x can have any value.) π π <x< 2 2

x
-1 1

Figure 2.3: Graph of arccos(x).

Other Notations: As commented earlier we also use the notations sin −1 (x), cos−1 (x) and tan−1 (x) for the inverse trig functions. This is traditional but dangerous, because of the possible confusion with 1/ sin(x) etc. I actually have a copy of a computer programming manual that gives a program for calculating arctan and uses that ‘formula’ arctan(x) = cos(x)/ sin(x)! Your pocket calculator will almost certainly be able to calculate inverse trig functions for you. It is usually a matter of pressing some kind of INV button before pressing the required trig function button.

2.1.2

Diﬀerentiating the Inverse Trig Functions

We use a standard method for inverse functions. Consider arcsin ﬁrst. Let y = arcsin x. Then x = sin y. Diﬀerentiate this equation wrt x using the chain rule: d d dy d x= sin y = sin y. dx dx dy dx Hence 1 dy = dx cos y so 1 = cos y dy dx

That’s the answer, but we would prefer to have the right-hand side as a function of x. Now cos2 y = 1 − sin2 y = 1 − x2 . So √ cos y = ± 1 − x2

2.1. THE INVERSE TRIGONOMETRIC FUNCTIONS

35

But, by our deﬁnition of arcsin, y lies between −π/2 and π/2. So cos y ≥ 0. So we must choose the positive sign. 1 d arcsin x = √ dx 1 − x2 In exactly the same way you can show that −1 d arccos x = √ dx 1 − x2 The story for arctan is very similar. Let y = arctan x. Then x = tan y. Diﬀerentiate wrt x and use the Chain Rule: dy 1 = sec2 (y) dx So 1 1 1 dy = = = 2 2 (y) dx sec 1 + tan (y) 1 + x2 So 1 d arctan x = dx 1 + x2 Example 2.1. Diﬀerentiate f (x) = x arctan(1 + x). By product rule d f (x) = arctan(1 + x) + x arctan(1 + x) dx By Chain rule and the formula for diﬀerentiating arctan: 1 d arctan(1 + x) = .1 dx 1 + (1 + x)2 Therefore x f (x) = arctan(1 + x) + 1 + (1 + x)2 Example 2.2. Diﬀerentiate f (x) = arcsin x + arccos x. By the formulas, 1 −1 +√ =0 f (x) = √ 2 1−x 1 − x2 So f (x) is always 0! So f (x) must be a constant. If we put x = 0 we get π π f (0) = arcsin 0 + arccos 0 = 0 + = 2 2 So we now have a signiﬁcant result, which is actually obvious if you think about it: π arcsin x + arccos x = 2

2.1.3

Polar Coordinates

36

CHAPTER 2. FURTHER DIFFERENTIATION

We have always used Cartesian Coordinates to describe y P points in a plane. There are lots of other coordinate (x, y) systems in use as well, particularly in specialised applications. One well-known coordinate system is that known as Polar Coordinates. It is especially useful in situations where information is most conveniently expressed in terms of distance from the origin. x θ Let OX and OY be cartesian axes in the plane. Let P be a point in the plane, other than the origin. O The Polar Coordinates of P are (r, θ) where r = OP > 0 and θ is the angle from OX round to OP antiFigure 2.4: Polar co-ordinates clockwise. By convention, we take the angle range to be −π < (r, θ) of a point. θ ≤ π. (some people take 0 ≤ θ < 2π). Note that we do not give polar coordinates for the origin, because the angle does not make sense there. Now consider the problem of converting between cartesian and polar coordinates. One way round is easy: x = r cos θ y = r sin θ (these are really just the deﬁnitions of sin θ and cos θ.) The other way round is more tricky and needs some understanding of inverse trig functions. First of all, without any diﬃculty, r= x2 + y 2

The real problem is to get θ in terms of x and y. Dividing the above equations gives tan θ = So we are tempted to write y x

y θ = arctan( ) x

This is wrong. First of all, we have to be careful about the case x = 0 (which corresponds to the y-axis) because we don’t want to divide by zero. Inspection of the picture tells us that if x = 0 then if y > 0 then θ = π/2 if y < 0 then θ = −π/2 Our problems don’t end there. Recall that arctan is deﬁned as taking values in the range π π − <θ< 2 2

2.1. THE INVERSE TRIGONOMETRIC FUNCTIONS

37

So it is not good enough in itself to solve our problem. The obvious trouble is that, for example, (1, 1) and (−1, −1) give the same value for y/x even though they have diﬀerent angles (π/4 and −3π/4). We have to take account of the quadrant in which the point lies. The correct conversion rules are these—which you should check through as an exercise. if x > 0 then θ = arctan(y/x) if x < 0 then if y ≥ 0 then θ = arctan(y/x) + π if y < 0 then θ = arctan(y/x) − π

Questions 7

(Hints and solutions start on page 69.)

Q 2.9. Get used to the inverse trig buttons on your calculator by checking the following arcsin 0.7 ≈ 0.7754 arccos(0.1) + arcsin(0.1) ≈ 1.5708 tan−1 0.7 ≈ 0.6107

Q 2.10. Give the values of the following without using your calculator (check afterwards). Remember what you know about values of the trig functions! arcsin(1), arcsin(−1), cos−1 (1), arccos(−1), arccos(0) √ sin−1 (1/ 2), arccos(1/2)

arctan(0),

arctan(1),

Q 2.11. Without using a calculator, ﬁnd the polar coordinates of the following points: (1, 0), (0, 1), (−1, 0), (0, −1), (1, 1), (−1, 1), (−1, −1). Now use your calculator to ﬁnd the polar coordinates of the following points: (2, 1), (−3, 4), (5, −2). Q 2.12. Diﬀerentiate the following functions. 2 arcsin x − 3 arccos x, arccos(3x + 1), sin−1 (2x), x. tan−1 x, √ √ tan−1 (1 + x2 ), arcsin( 1 − x2 ), cos−1 x

Q 2.13. Diﬀerentiate the function f (x) = arctan(x) + arctan(1/x). Explain the answer that you get (think in terms of a right-angled triangle). *Q 2.14. Sketch the graph of the function f (x) = arcsin(sin x) for 0 ≤ x ≤ 3π. Do the same for g(x) = arccos(cos x) and h(x) = arcsin(cos x). *Q 2.15. Show that if a and b are any constants then y = (arcsin x)2 + a arcsin(x) + b satisﬁes the equation (1 − x2 )y − xy = 2.

38

CHAPTER 2. FURTHER DIFFERENTIATION

*Q 2.16. This is nothing to do with Inverse Trig functions. You know that tan 2x = 2 tan x 1 − tan2 x

It can be shown that the approximation tan x ≈ x+ 1 x3 gives an error of less than 1.4×10−11 3 if |x| < 0.01. Use both of these results to obtain an approximate value for tan 0.1 by a process of repeated halving and doubling (I mean you to use a calculator). Show that if you know the value of tan x, with 0 ≤ x ≤ π/2 then you can easily work out the values of sin x and cos x, so long as you can take square roots. Get an approximation to sin 0.1. By inspecting the graphs of the trig functions show that you can work out the sin and cos of any angle once you know the values of sin and cos for angles in the range 0 ≤ x ≤ π/2 (at least, in principle). Put all this together to calculate sin 6.4 to an accuracy of 5 decimal places. You can check your answer with your calculator. (If you know anything about computer programming you could now write a decent program to evaluate the trig functions.) Why did I say “in principle” a couple of paragraphs ago?

2.2

Implicit Functions and their Diﬀerentiation

Suppose that the variable y is a function of the variable x. Up to now the deﬁnition of y in terms of x has been given in the form y = expression in x This is called an Explicit Formula for y in terms of x. You get the value of y by plugging the value of x into the RHS. It is also possible, with a bit of care, to deﬁne y in terms of x by a equation of more general form involving x and y. For example, the equation of a straight line: ax + by + c = 0 or the equation of a circle: or, more elaborately, x2 + y 2 = r 2 x2 y 2 + y 4 = sin(xy)

Such an equation is called an Implicit Formula for y in terms of x. In some cases it is easy to turn an Implicit Equation into an Explicit Equation, simply by ‘solving for y’. For example, if b = 0 in the equation of the straight line, then the equation can be changed to c a y =− x− b b

2.2. IMPLICIT FUNCTIONS AND THEIR DIFFERENTIATION

39

which is now explicit. As another example, the Implicit Equation 3y + x2 = 2x2 y can be rearranged to give the Explicit Equation x2 y= 2 2x − 3 Things are not always so simple. There are two problems that we face. The ﬁrst and most obvious is that we may not know how to solve the equation for y — see the third example at the start, you cannot do it and nor can I. The second problem is more theoretical. An implicit equation on its own may not deﬁne y as a function of x in a straightforward way. Consider the example of the circle equation x2 + y 2 = r 2 . If we ‘solve for y’ we actually get two explicit equations: √ √ y = r 2 − x2 andy = − r 2 − x2 Since, within the range −r < x < r there are always two y-values for each x-value we cannot hope to express y as a single function of x. So when we talk of the function deﬁned by an Implicit Equation we should always be aware that there may be more than one and that we may need extra information to decide on which to choose.

2.2.1

Diﬀerentiating Implicitly Deﬁned Functions
x2 + y 2 = sin(x + y)

Suppose that y is a function of x that satisﬁes the Implicit Equation

Can we work out the derivative dy/dx? This would be no problem if we could rearrange our equation into an Explicit equation y = . . . , but in this case we cannot. Even if we cannot convert the equation into explicit form it is still quite easy to work out the derivative by using the chain rule. Just diﬀerentiate both sides of the equation, remembering that y is a function of x and that we will have to use the Chain Rule on it. One or two examples will show you how easy this is. Example 2.3. x2 + y 2 = 1 Diﬀerentiating both sides wrt x we get 2x + By the chain rule d dy d 2 (y ) = (y 2). dx dy dx So the equation becomes x dy =− dx y Note the almost inevitable snag: the expression for the derivative contains both x and 2x + 2yy = 0 or y. d 2 (y ) = 0 dx

40

CHAPTER 2. FURTHER DIFFERENTIATION

Example 2.4. Find the equation of the tangent line to the curve x2 + y 3 = 2 at the point (1, 1) on the curve. To ﬁnd the equation of the tangent we need to ﬁnd the slope of the curve at (1, 1). This means working out the derivative. We diﬀerentiate both sides of the equation with respect to x. The derivative of x2 is 2x and, by the chain rule, the derivative of y 3 is 3y 2 y . So we get 2x 2x + 3y 2y = 0 hence y = − 2 3y So the slope of the curve at the required point is y = −2/3. So the equation of the tangent 2 is y − 1 = − (x − 1). 3 Example 2.5. x3 + sin(xy) = xy 2 Diﬀerentiating both sides wrt x we get 3x2 + d d sin(xy) = 1.y 2 + x y 2 dx dx

This still leaves us with some work to do: d 2 y = 2yy dx as before. The sin term involves a double application of the chain rule. We have, let’s say, z = sin u where u = xy. So dz du d dz = = cos(u) (xy) dx du dx dx Now d d (xy) = 1.y + x. y = y + xy dx dx Finally we get 3x2 + (y + xy ) cos(xy) = y 2 + 2xyy We can now rearrange this to get all the y terms onto the LHS: (x cos(xy) − 2xy)y = y 2 − 3x2 − y cos(xy) So y 2 − 3x2 − y cos(xy) dy = dx x cos(xy) − 2xy

—a mess, but a successful mess. Example 2.6. Find the tangent at (0, 0) to the curve (x2 + y 2)2 = x2 − y 2. Diﬀerentiate implicitly to get 2(x2 + y 2)(2x + 2yy ) = 2x − 2yy ory = x(1 − 2x2 − 2y 2) y(1 + 2x2 + 2y 2 )

2.2. IMPLICIT FUNCTIONS AND THEIR DIFFERENTIATION

41

Now put x = 0 and y = 0 — and get problems: 0/0. It doesn’t work. What’s happening? We are interested in the point (0, 0). Suppose that we are very close to it, so that x and y are very small. Then x2 and y 2 are extremely small and (x2 + y 2 )2 is absolutely tiny. So, rather approximately, we can say that, near (0, 0), the equation is x2 − y 2 ≈ 0 or y 2 ≈ x2 or y ≈ ±x. This gives two straight lines through the origin at right angles—which explains why we had diﬃculty in ﬁnding the slope at that point. In fact the curve, a Lemniscate, is shown in Fig. 2.5

Figure 2.5: The lemniscate of Example 2.6. Example 2.7. Here is another kind of example of the use of implicit diﬀerentiation. dy We are told that the function y(x) satisﬁes the equation + xy 2 = x + 1 for all values dx of x. We are also told that y(0) = 1. What, in that case, are y (0), y (0) and y (0)? Put x = 0 and y = 1 into the equation and get y (0) + 0 = 0 + 1. So y (0) = 1. Diﬀerentiate the equation wrt x and get y + y 2 + 2xyy = 1 Put x = 0 in this, together with y = 1 and y = 1 and get y (0) = 0. Diﬀerentiate again wrt x and get y + 4yy + 2xy 2 + 2xyy = 0 Put x = 0 in this together with y = 1, y = 1, y = 0 and get y (0) = −4.

Questions 8

(Hints and solutions start on page 70.)
dy dx

Q 2.17. For each of the following implicit equations ﬁnd out the equation of the tangent line at this point

at the given point. Also work

x2 + x + y 2 = 1at(0, 1) x3 + y 3 = 9at(2, 1) x3 + xy + y 3 = 1at(1, 0) π π sin(x + y) + cos(x − y) = 0at( , − ) 4 4 (x2 + y 2 )2 + (x2 − y 2)2 = 4at(1, 1)

42

CHAPTER 2. FURTHER DIFFERENTIATION

*Q 2.18. Show that the equation of the tangent to the curve x2 y 2 + 2 =1 a2 b at the point (x0 , y0) on the curve can be written as xx0 yy0 + 2 =1 a2 b *Q 2.19. The problem in this question is to work out the shape of the curve produced by all the points P such that the product of the distances from P to two given points A and B is a constant: P A.P B = α. Suppose we choose axes so that A = (−p, 0) and B = (p, 0). Show that the equation of the curve can be written as (x2 + y 2 + p2 )2 = α2 + 4p2 x2 . Note that the curve must be symmetrical about both the x and the y axes (why?). Find where, if anywhere, the curve cuts the axes. Now try to ﬁnd the rough shape of the curve. The shape depends on the value of α. What is special about the case α = p2 ?

2.3

Parametric Representation

Suppose we have a point moving around in the (x, y) plane. At each time t the particle will be at some point whose coordinates we can write as (x(t), y(t)). i.e. the x and y coordinates of the point are given as functions of the parameter t. Many curves can be most conveniently expressed in the form x = x(t) y = y(t)

where t is some parameter. t need not be time, though it often helps to think about it in that way. This is called a parametric representation of the curve. It may in some cases be possible to eliminate t between the two equations and get an Implicit Equation just involving x and y. Even when possible, this is not always desirable. Example 2.8. The Circle Any point on the circle x2 + y 2 = r 2 can be written in the form x(θ) = r cos θ y(θ) = r sin θ — just using polar coordinates. This is now a parametric representation of this circle. As θ goes from 0 to 2π the point (x(θ), y(θ)) goes once round the circle.

2.3. PARAMETRIC REPRESENTATION Similarly, the circle (x − a)2 + (y − b)2 = r 2 can be represented parametrically as x(θ) = a + r cos θ y(θ) = b + r sin θ

43

There are always lots of diﬀerent ways of parameterizing a given curve, some more helpful than others. As a trivial example, x(t) = sin(t), y(t) = cos(t) still gives a circle, as does x(t) = cos(t2 ), y(t) = sin(t2 ). Notice that a parametric representation can represent a whole curve in situations where is is not possible to do this by a formula of the form y = f (x), because there may be more than one value of y corresponding to a given value of x. This representation can even handle cases where a curve crosses over itself, as in the case x(t) = cos(t), which produces a shape like the ∞ sign. y(t) = sin(2t)

2.3.1

Diﬀerentiation in Parametric Form

Suppose that we are given a curve in the parametric form x = x(t) y = y(t)

where t is a parameter. Can we ﬁnd the value of dy/dx (in other words, the slope of the curve) without actually having to eliminate t between the equations? The answer is yes and the process is simple and based on the following simple consequence of the chain rule: dy dx dy = · dt dx dt hence, y ˙ dy = dx x ˙ where the dot denotes diﬀerentiation wrt t. Note something that might be a problem in some cases: the answer is given in terms of the parameter t rather than in terms of x and y. Example 2.9. Consider the circle in the parametric form x = r cos t Then x = −r sin t and ˙ so y = r cos t ˙ y = r sin t

y ˙ r cos t dy = = = − cot t dx x ˙ −r sin t

44

CHAPTER 2. FURTHER DIFFERENTIATION

Example 2.10. The parabola y 2 = 4ax can be parametrised as x(t) = at2 y(t) = 2at

where t takes all values. Why is this true? Firstly, y 2 = 4a2 t2 and 4ax = 4a2 t2 so the point (x(t), y(t)) certainly satisﬁes the equation and lies on the parabola. That’s half the story. The second question is: do we get the whole parabola? i.e. is there a value of t corresponding to each point on the curve? This question is easy to answer in this case. Let (x, y) be a point on the parabola, so y 2 = 4ax. Let t = y/2a. Then y = 2at and x = y 2/4a = at2 . Final slight worry: do diﬀerent values of t give us the same point on the parabola? No they do not — as the above argument actually shows. Put another way: if (at2 , 2at) = (as2 , 2as) then we must have s = t, as can be seen by comparing the y-coordinates. Having settled all that, let’s diﬀerentiate: x = 2at ˙ y = 2a so ˙ y ˙ 1 dy = = dx x ˙ t

To expand on the points raised in the previous example, consider the curve given parametrically by x(t) = at4 y(t) = 2at2

where t takes all values. What curve is this? Well, it is easy to see that x(t) and y(t) satisfy y 2 = 4ax. So we seem to have the parabola again. No we don’t. We get half the parabola. In this parametric representation, assuming that a > 0, the y value can never be negative (2at2 ). Furthermore, we get half the parabola twice over: (at4 , 2at2 ) = (as4 , 2as2 ) if t = ±s. If we think of (x(t), y(t)) as the path of a particle parametrised by time then the particle comes in from inﬁnity along the top half of the parabola, gets to the origin, stops and then reverses back the way it came.

Questions 9

(Hints and solutions start on page 70.)
dy dx

Q 2.20. For each of the following parametric representations ﬁnd the parameter. x(t) = t2 ,

at the given value of

y(t) = t3 att = 1 π x(t) = cos(t), y(t) = sin(t)att = 4 2 2 x(λ) = λ + 1, y(λ) = λ − 1 at λ = 1 1 y(t) = 2t/(1 + t2 ) at t = x(t) = (1 − t2 )/(1 + t2 ), 2

2.3. PARAMETRIC REPRESENTATION Q 2.21. Consider the curve given parametrically by x(t) = t2 y(t) = t3

45

Find the slope of the curve at the point with parameter value t. Now try to sketch the curve. Hints: what’s the relation between the points with parameter t and −t?. What exactly is happening at the origin? *Q 2.22. Show that if you put t = tan(θ/2) in the following formulas x(t) = 1 − t2 1 + t2 y(t) = 2t . 1 + t2 (1)

then you get x(θ) = cos θ, y(θ) = sin θ. So (1) seems to be a parametrisation of the circle x2 + y 2 = 1. This is rather interesting because (1) does not use trig functions and does not use square roots either. Is it actually true that you get every point on this circle from (1)? *Q 2.23. Sketch the curve given parametrically by the equations x(t) = 1 1 cos t − cos 2t, 2 4 y(t) = 1 1 sin t − sin 2t 2 4

Be sure to give decent arguments for what you draw. (For those of you who know about such things, this is the main blob of the Mandelbrot Set.) *Q 2.24. L1 and L2 are two straight lines. Initially L1 lies along the y-axis and L2 lies along the line y = 1. The two lines now start to move. L1 turns clockwise about the origin and L2 moves downwards. At time t the line L1 makes angle πt/2 with the y-axis and L2 is the line y = 1 − t. Show that the point of intersection of the two lines moves on the curve ((1 − t) tan(πt/2), 1 − t), called the Quadratrix of Hippias. Try to give a rough sketch of the curve (in the ﬁrst quadrant). *Q 2.25. This is similar to the previous example. This time the two lines are initially parallel to the y-axis. L1 passes through P = (−1, 0) and L2 passes through Q = (1, 0). L1 rotates around P and L2 rotates around Q, both clockwise. L1 is turning at twice the rate that L2 is turning. Sketch the path followed by the point of intersection of the two lines.

46

CHAPTER 2. FURTHER DIFFERENTIATION

Chapter 3 The Exponential and Logarithm Functions
You are already familiar with some of the ‘standard functions’ of mathematics — polynomials, rational functions, square roots, trigonometric functions and their inverses. We are now going to complete the list of standard functions by introducing two more basic functions. These are the exponential function and the logarithm.

3.1

Exponential Population Growth

Let’s start by looking at a problem in biology concerning the growth of animal populations. The theory that I am going to present here is usually associated with Malthus—whose work had an enormous inﬂuence in the last century, particularly on Charles Darwin. The theory is not taken very seriously nowadays, but it still makes for a nice illustration. Suppose a certain group of animals or plants has population P (t) at time t. We want to ﬁnd out something about how this population changes in time. We cannot deduce anything without making some assumptions. I am going to study a very simple ‘growth model’ which starts from the following assumptions. Each year each individual ‘produces’ on average b new individuals. Each year, for each individual in the population, d individuals die. We assume that b and d are constants. In a short time interval from time t to time t + δt we can say, to the ﬁrst approximation, that b.δt.P (t)individuals are produced d.δt.P (t)individuals die So, with this approximation, P (t + δt) = P (t) + b δt P (t) − d δt P (t) 47

48

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Rearrange this equation as P (t + δt) − P (t) = (b − d)P (t) δt You will notice that the left-hand side of this equation is just the usual Newton Quotient for P(t). Now let δt → 0 and get dP = κP dt Exponential Growth Law

where κ = b − d is what is called the growth rate. We have deduced from our assumptions a mathematical relationship based on the time derivative of P . Suppose we start at time t = 0 with population P (0). ˙ If κ > 0 then P > 0 so the population is increasing. As it increases the RHS of the equation gets bigger, so the rate of increase also increases—the growth feeds upon itself. The population grows more and more rapidly. ˙ If κ = 0 then P = 0 always, so the population remains constant. ˙ If κ < 0 then P < 0 at ﬁrst, so the population is decreasing. As it decreases (so long as it does y not die out) the RHS gets smaller, so the rate of decrease slows down. We will see later that, in fact, the population → 0 as t → ∞. In real life κ>0 this just means that the population dies out. We can give a rough sketch of the three cases κ=0 as shown in Fig. 3.1. In the next section we are goP (0) ing to take a more careful look at the exponential κ<0 x growth equation. The main weakness in this model is that it has to assume that b and d are constants. This is Figure 3.1: The behaviour of the solurarely true in reality. We will consider this point tion depends crucially on the sign of κ. further later on in the course.

3.2 The Exponential Function
This is one section of the course where I am going to do some fairly careful and detailed mathematics. If you ﬁnd this too diﬃcult then just concentrate on learning the summary of properties of the exponential function at the end.

3.2. THE EXPONENTIAL FUNCTION

49

3.2.1

Deﬁnition
dy =y dx

Consider the following equation for a function y(x): y(0) = 1 (3.1)

This is just a simpliﬁed version of the growth law that we derived in the previous section (the growth rate is unity). It can be proved that this equation has a solution that is deﬁned and diﬀerentiable for all values of x. Please accept this. I am now going to try to use the equation to discover properties of its solution. The work from here on will consist of a number of ‘theorems’. Some of these are important in their own right and others are just stepping stones along the way. This is one of the problems with studying mathematics. Sometimes you have to take the text on trust. The argument is getting to a punch-line eventually and when you get there you should be able to look back over it and see why the various results along the way had to be proved. The mathematician who originates the argument often has to think backwards. He decides that he wants to prove that something is true. In trying to prove it he realises that he needs to prove that something more basic is true, so he sets out to prove that and so on until he ﬁnishes. The results then get presented in the ‘logical order’, which is actually the reverse of what he did. The ﬁrst theorem looks rather odd and inconsequential. Proposition 3.1. If y(x) is a solution to (3.1) then y(x)y(−x) = 1 for all values of x. Proof. This is easy to prove once you get the idea, and the idea is going to be used frequently in this section. Consider the function A(x) = y(x)y(−x) where y(x) is a solution to (3.1). Diﬀerentiating both sides of this equation we get, using (3.1): A (x) = y (x)y(−x) − y(x)y (−x) by product and chain rules = y(x)y(−x) − y(x)y(−x) by (3.1) =0 So the derivative of A(x) is always zero. So A(x) is a constant. But when x = 0 we have A(0) = y(0)y(0) = 1. So A(x) = 1 for all values of x. That proves the proposition. Proposition 3.2. If y(x) is a solution to (3.1) then y(x) is never zero. Proof. This is obvious: if y(x) = 0 then we can hardly have y(x)y(−x) = 1. Now we come to a very important result. Up until now I have talked loosely about ‘the’ solution to equation (3.1). Could it actually have more than one solution (like quadratic equations can have more than one solution)? The common-sense of our population model strongly suggests that it cannot. If I know how large the population is at the start and also know the law for its growth then that really ought to ﬁx the population thereafter. But common sense is not mathematical proof. So we need a theorem.

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CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Proposition 3.3. The equation (3.1) has precisely one solution, deﬁned for all x. Proof. Thanks to the results that we have already proved this theorem is quite easy to prove. If you are trying to prove that at most one of a certain kind of thing exists then a standard approach is to use ‘proof by contradiction’. You assume the opposite—that there is more than one—and then try to show that this leads to a logical contradiction. If it does then you must have been right in thinking that there was at most one. In our case we already know that there is at least one solution, so we will have proved that there is exactly one. So suppose that y1 (x) and y2 (x) are two diﬀerent solutions to (3.1). Consider the function y1 (x) z(x) = y2 (x) There is no danger of this function being undeﬁned for any value of x because we have just proved that the denominator can never be zero. Diﬀerentiate z(x) and get z (x) = y1 (x)y2 (x) − y1 (x)y2 (x) 2 y2 (x) y1 (x)y2 (x) − y1 (x)y2 (x) = y2 (x)2 =0

by (3.1)

So the derivative of z(x) is always zero, so z(x) is constant. When x = 0 we have z(0) = 1/1 = 1. So z(x) = 1 for all values of x and therefore y1 (x) = y2 (x) for all values of x. This contradicts our original assumption that y1 and y2 were diﬀerent. So (3.1) can only have one solution. Now we are getting places. Equation (3.1) deﬁnitely deﬁnes a single function of x. The single solution to equation (3.1) is called the Exponential Function and is written as exp(x) or ex . The second notation makes the exponential function look like a power. It is, but we have not proved that yet. So think of it as a peculiar notation for the time being. The number exp(1) is denoted by e and has the value e = 2.71828182845904523536 . . . There is almost certainly a button on your pocket calculator for working out exp(x). It will be called Exp or ex . Try using it to check some of the claims that have been made. We have already proved the following facts about the exponential function:

3.2. THE EXPONENTIAL FUNCTION

51

d exp(x) = exp(x) dx

or

d x e = ex dx

i.e. it has the remarkable property that it does not change when you diﬀerentiate it. exp(0) = e0 = 1 and exp(1) = e1 = e

ex is never zero. exp(−x) = 1 exp(x) or e−x = 1 ex

The ﬁrst one is just saying that y(x) = exp(x) satisﬁes the equation in (3.1), which it does by deﬁnition. The second one is saying that exp(x) satisﬁes the condition y(0) = 1 in (3.1), which is also true by deﬁnition. The third and fourth are consequences of our theorems.

3.2.2

Properties of the Exponential Function

Now we push ever onwards. The next theorem is the really juicy one. It shows that the exponential function has a remarkably nice property that you would hardly guess at on the basis of equation (1). It goes a long way towards showing that exp(x) is ‘really’ a power, as I said above. Proposition 3.4. For all values of x and y exp(x + y) = exp(x) exp(y) or ex+y = ex ey

Proof. Same trick as usual. Pick any value of y. Now let A(x) = exp(x + y) exp x

Diﬀerentiate this with respect to x (we are thinking of y as a constant): A (x) = exp(x + y) exp(x) − exp(x + y) exp(x) exp(x)2 =0 using chain rule and (3.1)

So A(x) has a constant value as x varies. Put x = 0 and get A(0) = exp y. So, for all values of x and y, we have exp(x + y) = exp y exp(x) and hence exp(x + y) = exp(x) exp(y)

This is the crucial ‘law of indices’ property of the exponential function.

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CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Proposition 3.5. exp(x) is positive for all values of x. Proof. Using Prop. 3.4 we have exp(x) = exp(x/2) exp(x/2) = exp(x/2)2 . So exp(x) cannot be negative and we have already shown that it cannot be zero either. So it is always positive. Proposition 3.6. The exponential function exp(x) is always increasing and has the following limiting behaviour: as x → ∞ exp(x) → ∞ as x → −∞ exp(x) → 0

Proof. We know that exp(x) is always positive, so equation (3.1) tells us that the derivative of exp(x) is always positive. So exp(x) is always increasing. We can prove that it tends to ∞ as x → ∞ if we can prove that exp(x) takes arbitrarily large values. This is easy, thanks to Prop 3.3. We know that exp(0) = 1, so exp(2) > 1. Now exp(4) = exp(2) exp(2) and exp(8) = exp(4) exp(4) and so on up. In general, exp(2n ) = exp(2)2
n−1

Since exp(2) > 1 its powers get bigger and bigger without limit. That’s what we wanted. On the other side: exp(−x) = 1/ exp(x) so exp(−x) → 0 as x → ∞ and that proves the second point. That’s the end of my basic work on the exponential function.

3.2.3

Summary

You can open your eyes now. Here is a summary of the information that we have obtained about the exponential function. Even if you did not follow the arguments of the previous section you should still make sure that you know the following information. The exponential function exp(x) (or ex ) is the single solution of dy =y dx It has the properties d x e = ex dx e0 = 1 e−x = 1 ex ex+y = ex ey y(0) = 1

The number e1 = exp(1) is called e and has the value e = 2.71828 . . . . ex is always positive and increasing. ex → ∞ as x → ∞ ex → 0 as x → −∞ Part of the graph of ex is shown in Fig 3.2.

4

3

y = ex

2

1

3.2. THE EXPONENTIAL FUNCTION

53

3.2.4

Diﬀerentiation Examples

These are some more examples of diﬀerentiation, now that we have got a new function to diﬀerentiate. There is nothing new in the methods. Example 3.1. y = e4x This is a simple, but important, example of the chain rule. We can express the function as y = eu where u = 4x. So, by the chain rule, dy du dy = = 4eu = 4e4x dx du dx In general, if a is a constant, d ax e = aeax dx Example 3.2. y = e3x (sin x + 2 cos x) Think of this ﬁrstly as a product d 3x dy = (e )(sin x + 2 cos x) + e3x (cos x − 2 sin x) dx dx = 3e3x (sin x + 2 cos x) + e3x (cos x − 2 sin x) = e3x (sin x + 7 cos x) Example 3.3. f (x) = ex Chain rule again. Let y = eu where u = x2 . Then dy du dy 2 = = eu .2x = 2xex dx du dx Example 3.4. Show that the function y = ex (sin(x) + cos(x)) satisﬁes the equation y − 2y + 2y = 0 for all values of x. The basic plan is straightforward. Work out y and y , substitute them into the equation and hope for the best. This is a case where the diﬀerentiation is quite easy but is also quite messy. It is important to keep workings clear and well-organized or you are bound to make a silly mistake somewhere. y = ex (sin(x) + cos(x)) + ex (cos(x) − sin(x)) It is best to tidy this up at once to get y = 2ex cos(x) Now diﬀerentiate again. y = 2ex cos(x) − 2ex sin(x)
2

54 So

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

y − 2y + 2y = 2ex cos(x) − 2ex sin(x) − 4ex cos(x) + 2ex (sin(x) + cos(x)) If you now simplify down the RHS of this equation you will ﬁnd that you get the answer 0, regardless of the value of x. That solves the problem. Example 3.5. Find the values of α for which y = eαx sin(2x) satisﬁes the equation y − 4y + 8y = 0 This is much the same as the previous example. We calculate y and y and then put the results into the equation. We then hope that the fact that the equation has to be satisﬁed for all values of x will tell us the possibilities for α. Again, good organization is important—simplify as you go along, or things get out of hand. y = αeαx sin(2x) + 2eαx cos(2x) = eαx (α sin(2x) + 2 cos(2x)) y = αeαx (α sin(2x) + 2 cos(2x)) + eαx (2α cos(2x) − 4 sin(2x)) = eαx ((α2 − 4) sin(2x) + 4α cos(2x)) So y − 4y + 8y = eαx ((α2 − 4) sin(2x) + 4α cos(2x) − 4α sin(2x) − 8 cos(2x) + 8 sin(2x)) = eαx ((α2 − 4α + 4) sin(2x) + (4α − 8) cos(2x)) and we want to know what values of α make this identically zero. We can forget about the exponential at the start (that will never be zero). The expression will only vanish identically if the coeﬃcients of the sine and the cosine both vanish (see exercises for the justiﬁcation for this). The cosine term tells us that the only possibility for α is 2 and, if you check, you will see that α = 2 kills oﬀ the sine term as well. So the answer to the question is that α has to be 2.

3.3

The Natural Logarithm

The Logarithm is an older function than the Exponential. It is also seen as being more elementary—in the sense that you are likely to meet logarithms at a fairly early stage of your mathematical education (at least, you did before the days of pocket calculators). The main elementary use of the logarithm is that it can be used to turn multiplication into addition: log(ab) = log(a) + log(b), thus making calculations easier to do. That aspect of the logarithm has rather faded away with the advent of pocket calculators (life was hard when I was young). Most calculus books still tend to take the logarithm as basic and derive the exponential function from it. I am adopting the opposite approach of starting with the exponential

3.3. THE NATURAL LOGARITHM

55

function (which really is more important) and then deriving the logarithm function from it. You may have studied logarithms ‘to base 10’ in school. You can now forget about them.

3.3.1

Deﬁnition of the Logarithm

We have seen in the previous section that the exponential function is positive and always increasing. Consequently, given any positive value y there is one and only one value x for which exp x = y. This means that we can construct an inverse function for exp. The inverse of the exponential function is called the Natural Logarithm and is denoted by ln x. As immediate consequences of the deﬁnition we have: ln(exp(x)) = ln(ex ) = xfor all x exp(ln(x)) = eln x = xfor x > 0 Reading some of the properties of the exponential ‘backwards’ we get: ln(x) is deﬁned for x > 0 but not for x ≤ 0. ln(1) = 0, ln(e) = 1, ln(x) < 0 for 0 < x < 1 and ln(x) > 0 for x > 1. ln(x) is an increasing function and ln(x) → ∞ as x → ∞, ln(x) → −∞ as x → 0. The other important properties of the logarithm function are: ln(xy) = ln x + ln y 1 ln( ) = − ln x x for for x, y > 0 Figure 3.3: Graph of ln(x). x>0 y = ln(x)

1

0 1 2 3 4

-1

These can be proved as follows. Let x and y be two positive values. We know that there are values a and b such that x = exp a and y = exp b. So ln(xy) = ln(exp a exp b) = ln(exp(a + b)) = a + b = ln x + ln y Secondly, exp(−a) = 1/ exp(a) = 1/x. So 1 ln( ) = ln(exp(−a)) = −a = − ln x x We can get the graph of ln x simply by turning round the graph of ex . It is given in Fig 3.3.

56

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

3.3.2

Derivative of ln x

We do this in the same way as we did the inverse trig functions. Let y = ln x. Then, by deﬁnition, x = ey . Diﬀerentiate this wrt x, using the Chain Rule: 1 = ey So dy dx

1 1 dy = y = dx e x 1 d ln(x) = dx x

Therefore

This is important.

3.4

Hyperbolic Functions

There are some simple combinations of exponential functions that have been given special names. There is really nothing much to them at this level, but they should be known. The functions sinh x, cosh x and tanh x are deﬁned as follows cosh x = 1 (ex + e−x ) 2 sinh x = 1 (ex − e−x ) 2 sinh x tanh x = cosh x They are pronounced as ‘cosh’, ‘shine’ and ‘tansh’ (or ‘than’) and are called the Hyperbolic Trigonometric Functions. You may have buttons for them on your calculator. Their properties can be deduced easily from those of ex . You should check that cosh(−x) = cosh(x) sinh(−x) = − sinh(x) cosh(0) = 1 x → ±∞ tanh(−x) = − tanh(x)

as

sinh(0) = 0   cosh x → ∞  sinh x → ±∞   tanh x → ±1 d sinh x = cosh x dx

d cosh x = sinh x dx

cosh2 x − sinh2 x = 1 There are also formulas corresponding to all the usual trigonometric formulas, e.g. for sinh(x + y) and sinh 2x. They are the same as the trig formulas except for some of the signs.

3.4. HYPERBOLIC FUNCTIONS

57

3.4.1

More Diﬀerentiation Examples

Now we can do derivatives with logs in, and with hyperbolic functions. Example 3.6. f (x) = x ln x By the product rule f (x) = 1. ln x + x. 1 = ln x + 1 x Example 3.7. f (x) = ln2 x This is chain rule. Let y = u2 where u = ln x. Then dy du 1 2 dy = = 2u. = ln x dx du dx x x Example 3.8. f (x) = ln(1 + x2 ) Chain Rule. Let y = ln u where u = 1 + x2 . Then dy du 1 2x dy = = .2x = dx du dx u 1 + x2 Example 3.9. f (x) = sinh3 (1 + x2 ) Chain Rule. Let y = u3 where u = sinh v, where v = 1 + x2 . Then dy du dv dy = = 3u2 . cosh v.2x = 6x sinh2 (1 + x2 ) cosh(1 + x2 ) dx du dv dx

Questions 10
e2.3 ≈ 9.974,

(Hints and solutions start on page 71.)

Q 3.1. To get used to using the right buttons on your calculator, check the following values: e−1 ≈ 0.3679, ln(2) ≈ 0.6931, ln(1000000000) ≈ 20.7233

(If you got ln 2 = 0.3010 then you are using the wrong logarithm button. You are using ‘common logarithms’ ln10 2.). What is the value of ln 101000 ? Q 3.2. Diﬀerentiate the following functions f (x) = 2ex + x, k(x) = ln x , x2
2x

g(x) = 2 ln x − ex , 1 l(t) = p(− t2 ), 2

h(x) =

ex − 1 ex + 1

m(s) = ln(1 + s2 )
x

n(x) = 2e r(x) =

−e ,

x2

p(x) = ln(e ), χ2 + 2χ + 1 ), χ2 − 2χ + 1

ex q(x) = x e

2

ex − e−x , ex + e−x

s(χ) = ln(

t(x) = ln(ln x)

58

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Q 3.3. Diﬀerentiate the following functions f (x) = cosh(2x), g(x) = 2 cosh x − 3 sinh x, h(x) = 1 , cosh x k(x) = sinh x cosh x

l(x) = sinh(sin x),

m(x) = cosh(ln x),

n(x) = sinh2 (x2 )

Q 3.4. Work out the equation of the tangent at the point (a, ea ) on the graph y = ex . What condition does a have to satisfy if this tangent is to pass through the origin (0, 0)? Draw a picture to illustrate your answer. Now ﬁnd the values of m for which the straight line y = mx fails to meet the graph y = ex . Q 3.5. Show that x(t) = ept , where p is a constant, only satisﬁes the equation a¨ +bx+cx = x ˙ 2 0 if ap + bp + c = 0. Q 3.6. Show that x(t) = e2t (sin(t) − cos(t)) satisﬁes the equation x − 4x + 5x = 0 for all ¨ ˙ values of t. Q 3.7. For what values of the constant µ does the function x(t) = et cos(µt) satisfy the equation x − 2x + 10x = 0 for all values of t? ¨ ˙ Q 3.8. Use the properties of the logarithm, in particular ln ab = ln a + ln b, ln a/b = ln a − ln b and ln xa = a ln x, to simplify the following expressions as much as you can — you can then diﬀerentiate them. y = ln((1 + x)(2 + x)), y = ln((1 + x)2 (1 + 2x)3 ), 1−x 1+x
2

y=

ln((1 + x)2 ) ln(1 + x)

y = ln

,

y = ln

1−x 1+x

Q 3.9. This question is about the inverse functions of the hyperbolic functions. Show that √ √ cosh−1 (x) = ln(x + x2 − 1) if x ≥ 1 and sinh−1 (x) = ln(x + x2 + 1) (which is a genuine inverse). *Q 3.10. This goes back to a point raised in the text. I claimed, at one point, that if we have constants A and B such that A sin x+B cos x = 0 for every value of x then A = B = 0. Prove this (remembering that ‘every’ automatically includes ‘any’). Now prove the same kind of result for A sin x + B cos x + C sin 2x + D cos 2x = 0 and for A + Bex + Ce2x = 0.

3.5. POWERS *Q 3.11. Consider the equation dy = ayandy(0) = 1 dx where a = 0 is a constant. Deﬁne a new variable z by z = ax and show that dy = yandy(0) = 1 dz

59

(1)

Show that the single solution to (1) is y = eax . Did I really need to put on the restriction a = 0? *Q 3.12. This is an alternative view of the exponential function. Suppose I invest a sum P for one year at an interest rate of I%. Explain why, at the end of the year, I will have sum (1 + k)P where k = I/100. Now suppose that instead of getting I% per annum I get I/2% every six months. What sum do I now have at the end of one year? Now suppose that I get I/N% interest added N times during the year. Give a formula for the total sum at the end of one year. Now use your calculator to work out some numbers. Take a capital P = 1 an interest rate of 10% and work out the sum at the end of one year for N = 2, 10, 100, 1000, 10000. Do you notice anything? This all boils down to the following theorem about the exponential function (which I am not asking you to prove!). x N lim 1 + = ex N →∞ N

3.5

Powers

Let n be a positive integer. Then xn = x.x.x . . . x n times. So ln(xn ) = ln(x.x.x . . . x) = ln(x) + ln(x) + · · · + ln(x) = n ln(x) (for example, ln(x3 ) = ln(x.x.x) = ln(x) + ln(x) + ln(x) = 3 ln(x).) Now consider x1/n . (x1/n )n = x so Therefore ln(x1/n ) = Finally, by combining the two results, we get ln(xn/m ) = n ln x m ln((x1/n )n ) = ln x so 1 ln x n n ln(x1/n ) = ln x

60 Consequently,

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS n ln x) m

xn/m = exp( because exp(

n ln x) = exp(ln(xn/m )) = xn/m m This looks ridiculously complicated, but is extremely useful. Firstly, it the the way that your calculators evaluate non-integral powers. If you ask your calculator to evaluate 3.451.23 what it actually evaluates is exp(1.23 ln 3.45)
√ Secondly, 2 π

2

this gives us a simple way to say what we actually mean by expressions like or e . We are just going to say that for any x and a, with x > 0, xa = ea ln x

This is our deﬁnition of a power. We have seen that it is consistent with our normal ideas of powers. As an exercise you can now prove all the usual ‘rules of indices’ by using this deﬁnition. Theorem 3.13. For x > 0 and for any value of a d a x = axa−1 dx Proof. We claimed this earlier on, now having a deﬁnition we can prove it. It is a simple use of the chain rule. xa = exp(a ln x) so Now we go through the chain rule motions: y = exp u where u = av where v = ln x d a d x = exp(a ln x) dx dx

dy dy du dv 1 xa = = exp(u).a. = a = axa−1 dx du dv dx x x

3.6

* Irrational Numbers

I commented in the last section that there is no great diﬃculty in understanding what is meant by a power like x3/7 but that we needed the exponential and logarithm functions to make easy sense of a power like √ √ Why? Can we not just write 2 as a fraction and use the old deﬁnition? No we can’t. It turns out x 2. √ that 2 is one of a class of numbers that cannot be written as fractions (ratio of two whole numbers). They are called the Irrational Numbers and the ones that can be written as fractions are called the Rational Numbers.

3.6. * IRRATIONAL NUMBERS
The following numbers are certainly Rational Numbers. 7 , 2 −13 , 23 56, 123.456 (= 123456 ) 1000

61

It is not√ always easy to decide whether a number is Irrational. Let me show you the argument that proves that 2 must be Irrational. This proof is in Euclid’s Elements (published circa 300 B.C.) and is beautifully simple. √ Theorem 3.14. 2 is an Irrational Number. Or, equivalently, we cannot ﬁnd whole numbers p and q such √ that 2 = p/q. Proof. Once more we use ‘proof by contradiction’ (we used it when developing the properties of the √ exponential function). We assume that it is possible to ﬁnd whole numbers p and q such that 2 = p/q and then try to get ourselves into a logical tangle. We are only interested in the ratio of p and q so any common factors that the numbers have can be cancelled out. Suppose that this has been done. Then p and q cannot both be even numbers (or there is a common factor of 2). Squaring up and rearranging, we get 2q 2 = p2 . It is at once clear that p2 must be an even number (it’s twice something). In that case p must be an even number as well because the square of an odd number is still odd. So we can choose to write p = 2r, where r is another whole number. Put this back into the earlier equation and get 2q 2 = 4r2 or q 2 = 2r2 . So, as before, we see that q must be an even number. So both p and q are even numbers. But we started with the assumption that they were not both even! √ This gives us a contradiction and so we can conclude that 2 cannot be written as a fraction and hence is an Irrational Number. It is not recorded who it was ﬁrst realised that not all numbers are fractions. We do know that, whoever he was, he probably lived in the fourth century BC. There is an ancient tradition that says that he was murdered for his pains. Why murder somebody for making a mathematical discovery? The answer is that he was probably a member of (or associated with) the religious cult started by Pythagoras (of the triangle). The Pythagoreans seem to have had a theory of the world that was based on whole numbers and consequently on their ratios. The discovery that whole number ratios were not everything seems to have put the cat amongst the pigeons. Eliminating the discoverer was a predictable reaction, though not nowadays regarded as an acceptable form of mathematical proof. Here are some Irrational Numbers √ 2, √

3,

21/3 ,

π,

e,

ln 2

√ (in fact the square root of a whole number is either another whole number, like 4, or else irrational). √ As an exercise, try adapting my earlier proof to prove that 3 is irrational. Then check that your √ proof does not also prove that 4 is irrational, because it ain’t. The diﬀerence between rational and irrational numbers shows up nicely in their decimal expansions. The decimal expansion of a rational number is always eventually periodic (repeats itself). 1 3 2 7 1 6 1 2 = 0.333333333333333333 . . . = 0.285714285714285714 . . . = 0.166666666666666666 . . . = 0.500000000000000000 . . .

62

CHAPTER 3. THE EXPONENTIAL AND LOGARITHM FUNCTIONS

On the other √ hand the decimal expansion of an Irrational number can never become periodic.The decimal expansion of 2 starts oﬀ as 1.4142135623730950488016887242096980785696718753769480731766797 379907324784621070388503875343276415727350138462309122970249248 360558507372126441214970999358314132226659275055927557999505011 527820605714701095599716059702745345968620147285174186408891986 095523292304843087143214508397626036279952514079896872533965463 —not that that proves anything, the period might have been big.

Questions 11

(Hints and solutions start on page 72.)
√ 3 is an irrational number. Now prove that the cube root of 2 is irrational.

*Q 3.15. As I said, prove that

*Q 3.16. How would you use geometric series to work out the value of the following periodic ‘decimal’ as an exact fraction? 0.123123123123123 . . . What’s the answer? What is the value of 0.11112121212121 . . .? Now go on and give a reasonably coherent argument to prove that any periodic (or eventually periodic) decimal must represent a rational number? *Q 3.17. X is any point on the x-axis and is any positive quantity (as small as you like). Prove that there must be a rational number within of X. Prove that there must be an irrational number within of X as well. Deduce that between any two diﬀerent numbers there are inﬁnitely many rational numbers and inﬁnitely many irrational numbers.

Appendix A Some Useful Tables
Table of Derivatives
Here, for reference, is a table of the basic derivatives that you have met so far. f (x) Powers x xa sin(x) cos(x) tan(x) cosec(x) sec(x) cot(x) 1 axa−1 cos(x) − sin(x) sec2 (x) − cos(x)/ sin2 (x) sin(x)/ cos2 (x) − cosec2 (x) a any constant (x in radians) f (x)

Trigonometric Functions

Inverse Trig Functions √ arcsin(x) 1/ 1 − x2 √ arccos(x) −1/ 1 − x2 arctan(x) Exp & Log eax ln(x) sinh(x) cosh(x) aeax 1/x cosh(x) sinh(x) 63 a any constant x>0 1/(1 + x2 )

64

APPENDIX A. SOME USEFUL TABLES

Greek Letters
Mathematicians have a habit of using Greek letters in their work. This is partly to show that they are educated and partly because there are not enough Roman letters to go round (Mathematicians don’t really like the Computing habit of using words for quantities — because of the problem with multiplication: is CAT a thing with a tail or C.A.T?). This can be a nuisance if you are not familiar with the Greek alphabet. So here it is as a reference, together with the pronunciations of the names of the letters. Capital Name Pronunciation of name α β γ δ ζ η θ ι κ λ µ ν ξ o π ρ σ τ υ φ χ ψ ω A B Γ ∆ E Z H Θ I K Λ M N Ξ O Π P Σ T Υ Φ χ Ψ Ω alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu nu xi omicron pi rho sigma tau upsilon phi chi psi omega alfa beeta gamma like gammy as in river epsuylon or epsi-lon zeeta eat a (cake) th as in thing not the I owe Ta cap a(long) lamda mew like cat new like old zi short ‘o’ pie rho(dodendron) sigma ta(lk) oopsilon ﬁ(ne) (UK) fee (US) ki(nd) (shar)p si(ght) alpha and omega

Appendix B Solutions to Exercises
Solutions for Questions 1
(page 8).

Solution 1.1: The Newton Quotient is ((x + h)3 − x3 )/h. Now (x + h)3 = x3 + 3hx2 + 3h2 x + h3 , as you can check by multiplying out. So, subtracting x3 and dividing by h we get 3x2 + 3hx + h2 . The limiting value of this as h → 0 is 3x2 , so this is the required derivative. Solution 1.2: The Newton Quotient is (1/(1 + x+ h) −1/(1 + x))/h = −1/(x+ 1)(x+ 1 +h). As h → 0 we have (x + 1 + h) → (x + 1). So the derivative is −1/(x + 1)2 . Solution 1.3: For example, with h = 0.05, (sin(0.05) − 0)/0.05 = 0.99958. Solution 1.4: Consider the function |x| at x = 0, draw a picture.

Solutions for Questions 2

(page 13).

Solution 1.5: 4x3 , 5x4 + 1 , 24t5 − 15t4 + 2, 4p3 − 9p−4 − 2p−2 . 2 Solution 1.6: p (x) = 2 cos x, q (x) = sec2 x + cos x, r (θ) = −3 sin θ − 4 cos θ, m (λ) = ˙ 2λ − 4 cos λ + 3 sin λ, f (x) = 1 x−1/2 + 1 x−3/2 , h(t) = 15 t1/4 − 2t−2/3 − 2t−3 , φ (x) = 2 2 4 1 −1/2 2 . sec x − 3 sin x − 2 x Solution 1.7: f (x) = sin x + x cos x, g (x) = − sin2 x + cos2 x, r (x) = −2 cos x sin x, p (x) = 1 + 2 sin x + 2x cos x, q (x) = 4 sin x cos x, u (x) = v (x) = 2 sin(x)/ cos3 (x), w(t) = 2t − 3t2 + 13 t9/4 − 17 t13/4 . ˙ 4 4 Solution 1.8: p (x) = − 1 , x2 r (x) = 1 − x2 , (1 + x2 )2 s (x) = 7 , (2x + 3)2 t (x) = 3 − 2x − 2x2 . (x2 − x + 1)2 1 . (x + 1)2

u (x) =

−4x , 2 − 1)2 (x

e(t) = ˙

2 − 2t2 , (t2 − t + 1)2 65

m(t) = √ ˙

−1 √ , t(1 + t)2

l (x) =

66 v (x) = 1 , (1 + cos x) w (x) =

APPENDIX B. SOLUTIONS TO EXERCISES 1 , (1 − sin x) z (x) = − cos x , sin2 x f (x) = sec2 x − cosec2 x.

Solution 1.9: y = 3x2 . So the slope at (1, 1) is 3. So y − 1 = 3(x − 1) or y = 3x − 2. This meets the x-axis when y = 0, so x = 2/3. Solution 1.10: y = cos(x). You know that cos(π) = −1 and sin(π) = 0. So tangent is y − 0 = −1.(x − π) or x + y = π. This meets the axes at (π, 0) and (0, π). Solution 1.11: Derivative is −1/x2 . So the slope at P is −1/a2 . The equation of the tangent is therefore y − 1/a = −1/a2 (x − a). This meets the x-axis at x = 2a and the y-axis at y = 2/a. These give the base and height of a right-angled triangle. The area is therefore A = 1 (2a)(2/a) = 2. Note that the area does not depend on the value of a, i.e. on the 2 choice of the point P. Solution 1.12: The product of the slope of the tangent and the slope of the normal must be −1. In this case the slope of the tangent is 2a and so the slope of the normal must be −1/(2a). So the equation of the normal is y − a2 = −1/(2a)(x − a). It meets the y-axis at y = a2 + 1/2. Solution 1.13: f (x) = 2x − 3x2 + x3 so f (x) = 2 − 6x + 3x2 . We just want to know when this is positive and when it is negative. It is a quadratic with roots 1 ± 3−1/2 . Between these the derivative is negative, so the value of the function is decreasing. Outside, the derivative is positive so the function is increasing. Solution 1.14: The function fails to be deﬁned only at x = ±1 (where the denominator becomes zero). The derivative works out as −4x/(x2 − 1)2 and this has the opposite sign to x. So the graph is always increasing when x < 0 (and not = −1) and is always decreasing when x > 0 (and not = 1). Solution 1.15: The slope at (a, a2 ) is 2a, so the slope at Q is −1/(2a) and hence Q is the point (b, b2 ) where b = −1/(4a). Now work out the equations of the tangents and solve them simultaneously. You should get y = −1/4 regardless of the value of a. √ √ Solution 1.16: Multiply ‘top & bottom’ by a + b to get the formula. For the derivative: √ 1 1 (x + h) − x 1 √ ( x + h − x) = √ √ =√ √ . h h x+h+ x x+h+ x √ The ﬁnal denominator tends to 2 x as h → 0.

Solutions for Questions 3

(page 20).

Solution 1.18: f (g(x)) = 1 + (1/x2 ), g(f (x)) = 1/(1 + x2 ), g(g(x)) = x. Solution 1.19: f (g(h(x))) = 1 + 1/(1 − x), f (h(g(x))) = 2 − 1/x, g(h(f (x))) = −1/x, etc. Solution 1.20: 10(2x + 1)4 , (3x − 2)−2/3 , 3 sin2 x cos x, −10 cos9 x sin x, √ x/ 1 + x2 ,

67 5 cos 5x, −3 sin(3x + 1), 2x sec2 (1 + x2 ), − sin(x) cos(cos x)), √ 1 sec2 (x)/ tan x, 2 √ √ t−1/2 sin( t) cos( t),

2 3 (x + 2x − 6)−1/3 (3x2 + 2), 3 6 sin2 (x) cos x + 9 cos2 x sin x,

−(1 − 1/t2 ) sin(t + 1/t),

√ 1 (cos x + sin x)/ sin x − cos x, 2

√ √ 1 cos( sin t) cos t/ sin t, 2

2 x cos(x2 ) sin−2/3 (x2 ). 3

Solution 1.21: f (f (x)) = (x2 + c)2 + c = x4 + 2cx2 + c2 + c, and so on Solution 1.22: 1 1−x · , 1 + x (1 − x)2 −2x cos(x2 )/ sin2 (x2 ), 1 − 4x/(x2 − 1)2 , 1/(1 + x2 )3/2 , 0, and so on . . . .

−4 cos(x) sin3 (x)/(1 + sin4 x)2 ,

Solution 1.23: Diﬀerentiating the LHS gives 1+2x+3x2 +4x3 +· · ·+nxn−1 . Diﬀerentiating the RHS gives nxn+1 − (n + 1)xn + 1 . (x − 1)2 Putting x = −1 gives 1 − 2 + 3 − 4 + · · · ± n on the LHS and the RHS becomes 1 − (2n + 1)(−1)n n(−1)n+1 − (n + 1)(−1)n + 1 = . 4 4 Solution 1.24: c(s(s(s(x)))).c(s(s(x))).c(s(x)).c(x), where c, s stand for cos and sin. Solution 1.25: p(x) = ln ln sin(x). sin x takes values between −1 and 1. ln sin(x) is only deﬁned if sin(x) > 0 and then takes values ≤ 0. But that means that the ﬁnal logarithm is not deﬁned at all! So p(x) is not deﬁned for any value of x. If we ignore this and diﬀerentiate like a machine we get the derivative cot(x)/ ln(sin(x)). This is certainly deﬁned for quite a lot of values of x! Solution 1.26: No solution given. Solution 1.27: No solution given.

68

APPENDIX B. SOLUTIONS TO EXERCISES (page 25).

Solutions for Questions 4
sin(x), cos(x),

Solution 1.28: The derivatives, starting from f (x) are − sin(x), − cos(x), sin(x)

and now they start to repeat. In fact the sequence of derivatives goes through a cycle of four functions. To ﬁnd the 300th derivative we just have to decide where 300 comes in the cycle. Now 300 is a multiple of 4 and therefore the required derivative is sin(x). The next problem is harder. Let me put down a number of derivatives x sin x, sin x + x cos x, 2 cos x − x sin x 5 sin x + x cos x

−3 sin x − x cos x,

−4 cos x + x sin x,

Now you have to try to spot the pattern. There is still a kind of cycle of 4, except that the numerical coeﬃcient of the ﬁrst term is steadily increasing. The 300th derivative is −300 cos x + x sin x. Solution 1.29: 1 x−1/2 , − 1 x−3/2 , 3 x−5/2 , − 15 x−7/2 , 105 x−9/2 . It should by now be easy to 2 4 8 16 32 see what is happening. The problem is to turn it into a formula. The power of x in the nth derivative is −(2n − 1)/2. If n is even the sign is negative and if n is odd the sign is positive. The denominator of the coeﬃcient is 2n and the numerator is 1.3.5.7.9...(2n − 3). This actually works out to give f (n) (x) = (−1)n+1 — so there! Solution 1.30: We just have to ﬁnd out the second derivative of x(t). x(t) = Aω cos(ωt + α), ˙ x(t) = −Aω 2 sin(ωt + α) = −ω 2 x(t) ¨ 2(2n − 2)! −(2n−1)/2 x 4n (n − 1)!

Solution 1.31: This is just hard work. Work out the values of y and y , plug them into the equation together with y and hope that the resulting mess simpliﬁes down to 0. It probably easiest if you write y as a product, rather than a quotient. Here are clues to help you check. cos 2x + sin 2x = x−1 cos 2x + x−1 sin 2x x y = −x−2 cos 2x − 2x−1 sin 2x − x−2 sin 2x + 2x−1 cos 2x y = 2x−3 cos 2x + 2x−3 sin 2x4x−2 sin 2x − 4x−2 cos 2x − 4x−1 cos 2x − 4x−1 sin 2x. y= Now compute y + 2x−1 y + 4y. Solution 1.32: It must have been a polynomial.

69 Solution 1.33: (f g) = f g + 2f g + f g , (f g) = f g + 3f g + 3f g + g The layout is the same as for (x + y)n . Solution 1.34: No solution given. √ Solution 1.35: y = ny/ 1 + x2 , y = n2 y/(1 + x2 ) − nxy/(1 + x2 )3/2 Solution 1.36:b fn (x) diﬀerentiates to give nfn−1 (x) if n > 1. So fn can be diﬀerentiated n − 1 times. and so on.

Solutions for Questions 5

(page 26).

Solution 1.37: 1/x → 0 as x → ∞, so cos(1/x) → 1. Solution 1.38: Use common sense on the ﬁrst one. The numerator is constant and the denominator is getting bigger and bigger. In fact the denominator is tending to inﬁnity. So the fraction tends to 0. In the second limit both the numerator and the denominator are tending to inﬁnity, so the result is not quite so obvious. If you plug in a few values you will soon convince yourself that the limiting value is 1. The easy way to see this is to write the function as (1 + 1/x2 )/(1 − 1/x2 ). In this form both numerator and denominator tend √ √ to 1 in the limit, so we were right. For the last one, following the hint, x + 1 − x = √ √ 1/( x + 1 + x) which makes it obvious that the limit is 0. Solution 1.39: f (x) = x + 1 for any value of x other than 1. So the limiting value is clearly 2.

Solutions for Questions 6

(page 32).

Solution 2.1: f −1 (x) = (x − 4)/2, g −1 (x) = (2 − x)/5. Solution 2.2: The inverse is also f (x) = 1/x. Solution 2.3: The inverse g(x) is given by: if x ≥ 1 then g(x) = x − 1 and if x < 1 then g(x) = 2(x − 1). Solution 2.4: The graph must be symmetrical about the line y = x — i.e. if (x, y) is on the graph then so is (y, x). Solution 2.5: f is not deﬁned at x = −d/c. It cannot take the value a/c. Suppose we have f (x) = f (y) (with neither x nor y making the denominator zero). Then (ax + b)(cy + d) = (ay + b)(cx + d) or (ad − bc)(x − y) = 0. So since ad = bc we must have x = y. The inverse is f −1 (x) = (b − dx)/(cx − a).

Solutions for Questions 7
Solution 2.9: No solution given.

(page 37).

Solution 2.10: π/2, −π/2, 0, π, π/2, 0, π/4, π/4, π/3.

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APPENDIX B. SOLUTIONS TO EXERCISES

√ √ Solution 2.11: (1, 0), (1, π/2), (1, π), (1, −π/2), ( 2, π/4), ( 2, 3π/4) and √ ( 2, −3π/4). √ √ , Solution 2.12: 5/ 1 − x2√ arctan(x) + x/(1 + x2 ), 2/ 1 − 4x2 , −3/ 1 − (3x + 1)2 , 2x/(2 √ 2x2 + x4 ), ±1/ 1 − x2 (+ if x < 0 and − if x > 0, not diﬀ at x = 0), + −1/(2 x − x2 ). Solution 2.13: You should get 0 because f (x) = π/2. Solution 2.14: You should get a ‘sawtooth’ graph in both cases. √ Solution 2.15: y = (a + 2 arcsin(x))/ 1 − x2 , y = 2/(1 − x2 ) + (ax + 2x arcsin(x))/(1 − x2 )3/2 . Now plug in. Solution 2.16: No solution given.

Solutions for Questions 8

(page 41).

Solution 2.17: (1) 2x + 1 + 2yy = 0, Put x = 0, y = 1 and get y = −1/2. So equation of tangent is y − 1 = −x/2. (2)3x2 + 3y 2y = 0. Put x = 2, y = 1 and get y = −4. So equation of tangent is y − 1 = −4(x − 2). (3) 3x2 + y + xy + 3y 2y = 0. Put x = 1 and y = 0 and get y = −3. Tangent is y = −3(x − 1). (4) (1 + y ) cos(x + y) − (1 − y ) sin(x − y) = 0. x + y = 0, x − y = π/2. So y = 0 and tangent is y = −π/4. (5) After simplifying we get x4 + y 4 = 2. Diﬀerentiate: x3 + y 3y = 0. So y = −1 and tangent is y − 1 = 1 − x or y = 2 − x. Solution 2.18: No solution given. Solution 2.19: No solution given.

Solutions for Questions 9

(page 44).

Solution 2.20: (1) x = 2t, y = 3t2 , so dy/dx = 3t/2 = 3/2. (2) x = − sin t, y = cos t, ˙ ˙ ˙ ˙ so dy/dx = − cot t = −1 (3) x = 2λ, y = 2λ, so dy/dx = 1. (4) x(1/2) = −32/25, ˙ ˙ ˙ y(1/2) = 24/25 so dy/dx = −3/4. ˙ Solution 2.21: x = 2t, y = 3t2 , dy/dx = 3t/2. x(−t) = x(t) ≥ 0, y(−t) = −y(t). Curve is ˙ ˙ symmetrical about x-axis and has slope 0 at origin, where it has a cusp. Solution 2.22: The only point on the circle that you do not get is (−1, 0), which is what corresponds to t = ±∞. Solution 2.23: No solution given. Solution 2.24: No solution given. Solution 2.25: No solution given.

71

Solutions for Questions 10
Solution 3.1: No solution given.

(page 57).

Solution 3.2: f (x) = 2ex + 1, g (x) = 2/x − ex , h (x) = 2ex /(ex + 1)2 , k (x) = (1 − 2 2 2 ln x)/x3 , l (x) = −te−t /2 , m (x) = 2s/(1 + s2 ) n (x) = 4e2x − 2xex , p (x) = 1, 2 −x q (x) = (2x − 1)ex , r (x) = 4e2x /(1 + e2x )2 , s (x) = −4/(χ2 − 1), t (x) = 1/(x ln x). Solution 3.3: f (x) = 2 sinh(2x), k (x) = 1/ cosh2 x, g (x) = 2 sinh x − 3 cosh x, l (x) = cos(x) cosh(sin x), h (x) = − sinh(x)/ cosh2 (x) m (x) = (1 − (1/x2 ))/2

n (x) = 4x sinh(x2 ) cosh(x2 ) Solution 3.4: Slope is ea . Tangent is y − ea = ea (x − a). This is satisﬁed by (0, 0) if −ea = −aea or a = 1. Point is (1, e) and slope is e. Your picture should make it clear that y = mx fails to meet the graph if 0 ≤ m < e. ¨ x ˙ Solution 3.5: x = pept and x = p2 ept . So a¨ + bx + cx = (ap2 + bp + c)ept Since the ˙ exponential function is never zero this can only be zero if ap2 + bp + c = 0. Solution 3.6: Hard pounding! x = 2e2t (sin t − cos t) + e2t (cos t + sin t) = e2t (3 sin t − cos t) ˙ x = 2e2t (3 sin t − cos t) + e2t (3 cos t + sin t) = e2t (7 sin t + cos t) ¨ So, putting it all together, x − 4x + 5x = e2t ((7 − 12 + 5) sin t + (1 + 4 − 5) cos t) = 0 ¨ ˙ So the equation is satisﬁed for all t. Solution 3.7: Same process x = et cos µt − µet sin µt = et (cos µt − µ sin µt) ˙ x = et (cos µt − µ sin µt) + et (−µ sin µt − µ2 cos µt) = et ((1 − µ2 ) cos µt − 2µ sin µt) ¨ Putting it all together x − 2x + 10x = et ((1 − µ2 − 2 + 10) cos µt + (−2µ + 2µ) sin µt) = et (9 − µ2 ) cos µt ¨ ˙ This will certainly be identically zero if µ = ±3. et is never zero and cos µt can only be zero at certain points. So the equation can only be satisﬁed for all values of t if µ = ±3. Solution 3.8: (1) y = ln(1 + x) + ln(2 + x), so y = 1/(1 + x) + 1/(2 + x). (2) y = 2 ln(1 + x) + 3 ln(1 + 2x), so y = 2/(1 + x) + 6/(1 + 2x). (3) y = 2 ln(1 + x)/ ln(1 + x) = 2,

72

APPENDIX B. SOLUTIONS TO EXERCISES

so y = 0. (4) y = 2 ln(1 − x) − 2 ln(1 + x), so y = −2/(1 − x) − 2/(1 + x). (5) y = 1 (ln(1 − x) − ln(1 + x)), so y = 1 (−1/(1 − x) − 1(1 + x)). 2 2 Solution 3.9: We follow the usual procedure, but have to be careful about signs. I will just do cosh−1 . The other goes the same way. If y = cosh−1 (x) and x ≥ 1 then x = cosh(y) = (ey + e−y )/2. Multiply through by ey to get e2y − 2xey + 1 = 0 which is a quadratic for ey . √ Its solutions are ey = x ± x2 − 1. We choose to have y ≥ 0, so ey ≥ 1, so we must pick the + sign. Solution 3.10: Put x = 0 and get B = 0; put x = π/2 and get A = 0. The next one can be done by looking at x = 0, π/2, π. For the last one, it helps to diﬀerentiate once or twice before you start putting in values. Solution 3.11: dy/dx = dy/dz.dz/dx = a.dy/dz. So the original equation becomes dy/dz = y with condition y(0) = 1. We know that the solution to this is y = ez , so the equation in its original form has solution y = eax . Solution 3.12: No solution given.

Solutions for Questions 11

(page 62).

√ Solution 3.15: Suppose that 3 = p/q as a fraction in its lowest terms. Then p2 = 3q 2 . This means that 3 divides p2 . Convince yourself that this means that 3 must divide p as well, say p = 3k. Then√ 2 = 3q 2 so q 2 = 3k 2 and therefore 3 divides q. Contradiction. 9k (This does not work for 4 because knowing that 4 divides p2 does not imply that 4 divides p (e.g. p = 6). The cube root argument goes along in much the same way as the square root one. Solution 3.16: The number is 1/10 + 2/102 + 3/103 + 1/104 + 2/105 etc. Because of the periodicity we can arrange this as (1/10 + 2/102 + 3/103) + (1/10 + 2/102 + 3/103 )/103 + (1/10 + 2/102 + 3/103 )/106 + · · · a(1 + b + b2 + b3 + b4 + · · · ) where a = 0.123 and b = 1/103. The Geometric series sums to give 1/(1 − b) = 1000/999. So the number is 123/999. The number 0.121212 . . . works out to give 4/33, so the number in the question is 3667/33000. The general argument just follows the ideas of the examples. Gather a single periodic block together and get it as the coeﬃcient of a geometric series in powers of 10. Then add on the header bit, which certainly gives a fraction. Solution 3.17: The fractions 1/2n can be made as small as we like by making the whole number n big enough. Choose n so that 1/2n is less than the gap between x − and x + . Then one of the numbers 1/2n , 2/2n , 3/2n , . . . √ must land in the gap, and these are all rational numbers. For the second bit, note that k 2/2n is irrational if k = 0 and then argue as before. So the number is

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