# CHAPTER THREE APPLICATIONS OF THE DERIVATIVE by dsp14791

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```									       Hare Krsna Hare Krsna Krsna Krsna Hare Hare
Hare Rama Hare Rama Rama Rama Hare Hare
Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of
Math and Science)
KRSNA CALCULUS™ PRESENTS:

CHAPTER THREE:
APPLICATIONS OF THE
DERIVATIVE
Edited on October 7, 2003
HARE KRISHNA!

   Welcome back, dear students! I pray that you are all
well.
   In order to understand this chapter, you will have to
know the following:
   Definition of the derivative (at least the concept of it
difference quotient.), derivatives of polynomials, trig
functions, implicit differentiation, chain rule, and the
rectilinear motion relationships (position, velcoity, and
acceleration.)
   If you do not understand this, please review the last
chapter (Chapter 2) and keep practicing that chapter
until you understand those topics above.
WHERE WE HAVE GONE?

   As one mathematician said, “Calculus is a
journey. A good tourist will think about the
past here and use it to him or her in the
future.”
   Before we move onto the next chapter, let’s
take a step back and see what we have
covered so far, just to help us remember
important things.
SUMMARY OF PREVIOUSLY
STUDIED MATERIAL
   In chapter 1, we had studied limits. An informal and
most commonly used definition of the limit is simply
the question, “When I make x closer and closer and
closer to some number c, what f(x) am I
approaching to?”
   Asymptotes were explored using limits. Limits help
determine end-behavior.
   Continuity was briefly mentioned. Continuity is
present at x=c is f(c) exists, if the limit as x
approaches c from both sides exist and are the
same value, and if the limit as x approaches c and
f(c) are equal.
SUMMARY

   With the help of limits, we were able to
calculate instantaneous rates, and the slope
of the line tangent to the curve at a certain
point. By letting the change in x be
infinitesimally small in the slope formula
(difference quotient), we are able to get
derivatives.
   A differentiable function is continuous
function, but a continuous function is not
always differentiable.
DERIVATIVES.

   Practice of derivative rules and formulas
must be emphasized. Derivatives and
Integrals (next chapter) will be the backbone
to calculus.
   Definitely study the differentiation rules from
the last chapter!!!!!!
APPLICATIONS OF THE
DERIVATIVE
   This chapter will underscore the uses of
derivatives in the following cases.
   How to graph functions by knowledge of
critical points and derivatives.
   How to differentiate inverse, exponential and
logarithmic functions.
   Differentials
   Related rates and optimization problems
CURVE SKETCHING

   You can easily graph any function by
knowing three things.
   1) ZEROS AND UNDEFINED SPOTS
   2) MAXIMUM AND MINIMUM POINTS
   3) CONCAVITY AND INFLECTION POINTS.
CRITICAL POINTS
   Let’s find the points where the function is zero or undefined.
   We want to graph f(x)=x3+2x2+x.
   Obviously, there are no undefined spots, due to the fact that this
is a polynomial function. All polynomial functions are continuous,
thus differentiable.
   To find the zeros, set the entire function to 0.
   0= x3+2x2+x  Original function
   0=x(x2+2x+1)  Factored
   0=x 0=(x+1)(x+1)  Factored the trinomial and set both factors
equal to zero.
   x=0, -1  x is solved for.
   Thus x=0 and x=-1
MAXIMUM OR MINIMUM POINTS

   Here is the graph of the
f(x).
MAXIMUM AND MINIMUM
Notice  that if you draw
tangent lines at the
maximum and the
minimum, you will see that
you will get horizontal
lines. That means the
slope is zero. Thus, the
derivative is zero. In other
words, at a maximum or
minimum, the derivative is
zero.
MAXIMUM AND MINIMUM POINTS

   Since the f’(x)=0, we can solve for the maximum and minimum
points.
   First, find the derivative of f(x)=x3+2x2+x.
   f’(x)= 3x2+4x+1  Differentiation using power rule and
sum/difference rule.
   0 = 3x2+4x+1  Max/Min always have f’(x)=0.
   0=(3x+1)(x+1)Factoring to solve for x.
   0=3x+1 0=x+1  Setting factors equal to 0.
   x=-1/3 and x=-1  Solve for two answers of x.
   We know that x= -1/3 and x=-1 are the possible points, however,
without a graph, we cannot determine which is the max and
which one is the min.
SIGN ANALYSIS TEST FOR
MAX/MIN.
   You can do a sign chart and determine the sign between possible
zeros. Look how it is set up.
   What the sign of the two factors to the right of x=-1/3?
   What is the sign of the two factors between -1 and -1/3?
   What is the sign of the two factors to the left of -1?
   Since they are factors, you multiply them. Account for overall sign.
DERIVATIVE OF SIGNS

   If the derivative goes
from negative slope to
zero to positive slope,
then x is where the       zero
minimum is.
   If the derivative goes
from positive slope to    zero
zero to minimum slope,
then x is where the
maximum is.
WHAT WE KNOW

   We know that this is a cubic function, with x=-
1 and x=0 as the two zeros. No undefined
points
   We know that the maximum is at x = -1 and
the minimum is at x= -1/3.
HORIZONTAL TANGENTS
   Be careful when solving for x
when f’(x) is 0. Perhaps you will
have such function where the
slope is zero at a certain point,
but that point is neither
maximum nor minimum. y(x)=x3
for example. Once you
differentiate that, you get
y’(x)=3x2. y’(x)=0. Therefore,
you equate 0 and 3x2. You will
see that x = 0. However, by
doing a sign analysis chart, you
will see that before and after 0,
there is no sign change. The
slope continues to be positive.
CONCAVITY

   Concavity is very difficult to define.
   A good way to define concavity is using a
spoon. When you hold a spoon the right way
(where the milk is in the spoon), the spoon is
said to be concave up.
   When the spoon is down (when the milk is
not in the spoon… the spoon is not in the
right way), then the spoon is said to be
concave down.
CONCAVITY

   An upward parabola is concave up. A downward
parabola is concave down.
   The f(x) is above the tangent line when it is concave
up. The f(x) graph is under the tangent line when it
is concave down.
   The concavity is the rate of slope change. Hint hint!!
RATE OF CHANGE!! The rate of change of the
slope, means the derivative of the slope… The
slope is the derivative of the function. So the
concavity is the derivative of the derivative. The
first derivative of slope and the second
derivative of the function.
CONCAVITY GRAPHS
Both function
Both function                         have negative
have positive                         slopes. f’(x)<0.
slopes. f’(x)>0.                      The blue
However, the                          function is
red function is                       concave up,
concave up,                           since it is
since it is                           above the
above the                             tangent line.
tangent line.                         The red
The blue                              function is
function is                           concave down,
concave down,                         since it is
since it is      RED       BLUE                          RED        BLUE
below its
below its        f’(x)>0   f’(x)>0                       f’(x)>0    f’(x)>0
respective
tangent lines. f’’(x)>0    f’’(x)<0
tangent line.
f’’(x)<0   f’’(x)>0
ZERO CONCAVITY

   Remember when the function, f(x), was at a
maximum or minimum, the derivative, the
slope, f’(x) was 0.
   When the slope is at its maximum or
minimum, the derivative of the slope is 0.
   In order words, when f’(x) is at its maximum
or minimum, then the derivative, f’’(x), the
concavity will be 0.
   Concavity is the slope of the slope. !!
Side note: Vertical tangents.
   Let’s examine for a moment, the vertical
line graph.
   There is no such function that when
graphed has a vertical line.
   If you look at the slope of a line (Dy/Dx).
You will see that the difference in y is some
number. However, since there is no change
in x, that will be equal to zero. n/0 is
infinitely big that is undefined.
   However, there are functions with vertical
tangents. The cube root of x at x=0 has a
vertical tangent. Therefore, the derivative of
the cube root function will be defined
everywhere except x=0.
   Just be aware of that!
CONCAVITY AND INFLECTION
POINTS
   The point where the change in derivative sign
(positive vs. negative slope) is called extrema (i.e.
max/min points) f(max) and f(min), their derivatives
at those points will both be 0.
   The point where the change in concavity occurs is
called the inflection point.
   Since this is where the slope, f’(x), is the maximum,
the steepest or the flattest, the derivative of slope,
f’(x), will be 0.
   After solving for x and doing a sign analysis, you will
find the inflection points.
FINDING INFLECTION POINTS
   The original function was x3+2x2+x.
   Since we want to find the inflection point, we must obtain the second
derivative.
   f(x)= x3+2x2+x.  GIVEN
   f’(x)= 3x2+4x+1.  FIRST DERIVATIVE
   f’’(x)=6x+4  SECOND DERIVATIVE
   Since the inflection points are found by solving for x when f’’(x) = 0.
   0=6x+4 f’’(x) =0
   x= -2/3  SOLVED FOR X.
   Now we must do sign analysis.
   Everything right of -2/3 for the factor 6x+4 is positive. Everything left
of it is negative. Thus, x=-2/3 is an inflection point for x3+2x2+x.
   Everything right of x=-2/3 is concave up. Everything the to left of x=-
2/3 is concave down.
When sketching this curve…

   Make sure your zeros are at the x values.
   Make sure your max, mis, and horizontal
tangents are drawn at the proper spots.
   Make sure your inflection point is accurately
drawn.
   A good way of thinking about inflection points
is when the tangent lines switch sides with
respect to the graph.

f’(x)>0 f’’(x)>0
f(x)=0
f’(x)=0
f’’(x)<0
MAXIMUM                     ZERO
ZERO
f’(x)<0               f(x)=0
f’’(x)>0
f’(x)=0
f’(x)<0                    f’’(x)>0
f’’(x)<0
f’(x)>0 f’’(x)<0                                              MINIMUM
INFLECTION
POINT
f’(x)<0
f’’(x)=0
ROLLE’S THEOREM

   Some people hate calculus a lot. For example, they
might like to find instantaneous velocity by saying
distance/time. But we know that is not true.
   Rolle was a mathematician who tried to disprove
calculus. He disliked it very much. He wanted to a
lot of geometry and algebra. Well, you know, if you
hate something a lot, it is just as much as loving it a
lot. Similarly, Rolle hates calculus so much, that he
   This theorem will not be tested very frequently in
exams, but its best to know it anyway.
ROLLE’S THEOREM (MEAN VALUE
THEOREM)
   There will always be an
average rate line that will be
parallel to the tangent line     f (b)  f (a)
of a certain point. This                        f ' (c )
means they have the same
slope.
ba
   You can see that the
average rate line between
x=-2 and x=2 is parallel to
the tangent line of x=0.
Thus, you can conclude,
that they have the same
slope.
DIFFERENTIATION OF INVERSE
FUNCTIONS
   An inverse of a function is found by reflecting the
graph over the line y=x.
   The function is found simply by switching the y with
the x and vice verse.. For example: If y = 3x+2 with
coordinates (0,2) as one possible point, then its
inverse is x=3y+2 with coordinates (2,0). I just
switched the x and y around. That’s all.
   Sometimes, a function may have an inverse that is
not a function. y=x2 is function and x=y2 is not.
Remember, a function is a function if for any x, there
is one and only one y value. The vertical line test
determines that. (Course III Info for N.Y.S students).
DIFFERENTIATION OF INVERSE
FUNCTIONS
   From pre-calculus, you know that if f(x) has its inverse
g(x), then the composition f(g(x)) is x.
   If f(x) has g(x) as its inverse function, then the formula
for finding the derivative of a function is….

dy       1

dx f ' ( g ( x))
DERIVATIVE OF THE INVERSE OF
y=x2.
   f(x)                       f ( x)  x   2

   g(x)                       g ( x)  x
 
   Composition of f and g                        2
   dy/dx of f(x)              f ( g ( x))  x
   Composition of f’ and g.   f ' ( x)  2 x
f ' ( g ( x))  2 x
   Inv. Deriv. Formula        dy            1

dx f ' ( g ( x))
dy         1
   Substitution.                    
dx 2 x
CHECK
   You can check it also using the power rule.

1
y xx                  2

1
dy 1                           1
 x               2

dx 2                        2 x
DERIVATIVE OF EXPONENTIAL
FUNCTIONS
   The natural exponential function, f(x)=ex, has the base e.
e is a transcendental number (just like 16,108, p, f...). It
is named after its founder, Leonard Euler. e is derived
and found many ways. e is a very special number in
calculus. You will see why in a moment. e is
approximately 2.718… If you define it using limits, e = …

x
 1
e  lim 1  
x 
 x
DERIVATIVE OF THE
EXPONENTIAL FUNCTION
f x  h   f x 
   Derivative definition     lim
h 0            h
   f(x)=ex                         e xh  e x
   Law of exponents,         lim
h 0       h
ex+h=(ex)(eh)
e xeh  e x
   Factor out ex.            lim
h 0        h
e x (e h  1)
lim
h 0        h
   Previously proven limit         eh 1
lim            1
h 0     h
   Product Rule of limits    e x (1)
   Derivative                ex
THE DERIVATIVE OF THE
EXPONENTIAL FUNCTION
   This is one of the two functions whose derivative is
its own function! d(ex)=ex! Amazing how
transcendental functions work! Jaya! Jaya!

d x
dx
e e  
x
THE NATURAL LOGARITHM
FUNCTION y=ln x
   The natural logarithm is NOT taking a piece of log
naturally from a tree, play rhythms on it! 
   The natural logarithm is the inverse of the natural
exponential function. It has e as its base. There are
many ways y=ln x is derived by.
   y=ln x is defined for all x>0.
   We can find the derivative using the inverse
differentiation rule since ln x is the inverse of ex.
DIFFERENTIATING y=ln x.
   f(x)                                         f ( x)  e   x

   g(x)  inverse of f(x)                       g ( x)  ln x
Composition of f(x) and g(x)
f ( g ( x))  e ln x


   dy/dx of f(x)
   Composition of f’(x) and g(x)                f ' ( x)  e x
   If f’(g(x)) = f(g(x)), and if f(g(x)) = x,   f ' ( g ( x))  e   ln x
x
then f’(g(x))=x.
dy       1
   Inverse Differentiation Rule                    
dx f ' ( g ( x))
   Replacing f’(g(x)) with x.                   dy 1

dx x
THEREFORE…

   I find it pretty cool that an algebraic function like y
= 1/x is the derivative of transcendental function
y = ln x. Transcendental functions are functions
that cannot be derived simply by algebra.

d
ln x   1
dx           x
INVERSE TRIGONOMETRIC
FUNCTIONS
   The inverse of the function y = sin x is y= arcsin x.
   NOTE: arcsin x is NOT a function. For example, arcsin (½) = p/3, 5p/6,
-7p/6, -11p/6, so on… since one x value produced an infinite number of
y values, you know that this is not a function
   However, Arcsin x IS a function (notice capital A, compared to previous
lower-case a). Arcsin x is restricted from [-p/2, p/2], just as Arccos x is
restricted from [0,p] and Arctan x is restricted from (-p/2, p/2). Note that
the () parenthesis represent the open interval excluding endpoints.
Remember than tan (p/2) = ∞.
   If you look at the inverse cosine function, you will notice that it is just the
inverse sine function reflected over the x axis. Therefore, the slope, the
derivative, of the inverse cosine function will be negative the slope,
derivative, of the inverse sine function. Same thing will occur with
tangent and cotangent, secant and cosecant functions.
   d(arcsin x)=-d(arccos x)
   d(arctan x)=-d(arccot x)
   d(arcsec x)=-d(arccsc x)
DIFFERENTIATION OF THE
ARCSINE FUNCTION
   f(x)                       f ( x)  sin x
   g(x)  inverse function    g ( x)  arcsin x
   Composition of f and g.    f ( g ( x))  sin(arcsin x)
   dy/dx of f
f ' ( x)  cos x
   Composition of f’ and g.
   Derivative formula for     f ' ( g ( x))  cos(arcsinx)
finding inverses of        dy       1
functions                     
dx f ' ( g ( x))
   Applying f’(g(x)).
dy          1
   What is cos(arcsin x)?        
dx cos(arcsinx)
The denominator

   To simplify the denominator to more practical terms,
we know that y=arcsin x is the same as sin(arcsin
x)= sin y = x. Then to solve for cos y.

Pythagorean Trig Identity   sin y  cos y  1
2       2

sin y = x                   x  cos y  1
2       2

Solved for cos y.           cos y  1  x    2
AFTER APPLYING cos y IN THE
DENOMINATOR…
   We see that…

d arcsin x     1

dx         1 x 2

d arccosx        1

dx          1 x  2
USING THE SAME PROCESS…

   If you use the derivative of an inverse rule for the
arctangent and the arcsecant functions, you will see
that

d
arctan x     1          d
arc sec x         1
dx               1 x 2       dx                x     x 1
2

d
arc cot x    2 1       d
arc csc x            1
dx                 1 x       dx                  x       x2 1
MISCELLANEOUS THINGS TO
KNOW
   Some textbooks will make great emphasis on this.
Other textbooks won’t talk about it too much. Since
the AP and many college calculus I courses do not
test this topic greatly, I won’t discuss it in great
depth.
   You can treat dy/dx as a fraction of the
infinitesimally small change in y over the
infinitesimally small change in x.
   You can consider y = x2,  dy = 2x dx since dy/dx
= 2x. dy is said, “a differential amount of y.”
Differential means an infinitesimal change in a
specific direction.
EXAMPLE DIFFERENTIAL
PROBLEM
   If you take a cube with each side having
length 8, what is the differential volume if you
have a cube that has each side having length
7.99?
PROBLEM WORKED OUT

   The volume of a cube is
given

V x
   dV/dx is the differential             3
amount of volume over the
differential amount of
change in length. The
change in length is simply     dV  3 x dx2
8-7.99=.01=dx. Multiply
both sides by it so you
isolate dV. Then plug in       dV  3(7.99) (.01)
2

7.99 for x, and plug in 0.01
for dx and calculate the
differential amount of
dV  1.92
volume.
APPLICATION

   At the Sri Sri Gaura
Nitai temple, they want
to make the rectangular
temple room such that
the perimeter is 1000
feet. Find the maximum
length and width so that
everyone can have the
fortune to the Sri Sri
find the area.
OPTIMIZATION

   Optima- what???
   Optimization problems is a fancy way of saying
maximum minimum problems.
   The best way to start out is to draw a picture to help
yourself visualize the problem.
THEN..

   Since we have two variables, ℓ and w, we
have to solve everything in terms of one
variable.
   We know that A= ℓ w and Prect = 2 ℓ +2w.
   So we can solve for ℓ in either equation. Most
preferably use the P equation.
Prect  2w

2
And finally…

   Substite the expression for l into the area
formula.        Prect  2w

2
A  w
 Prect  2w 
A             w
      2     
1
A  Prectw  w2
2
According to the problem…

   P = 1000 ft so…

A  500w  w   2

   But to find maximum w,      A  500w  w2
the derivative will have    dA
to equal 0.                      500  2w  0
dw
   Then solve for w.
500  2w
250 ft  w
Then…

   Plug w in for the              1000  2w
expression for length.     
2
   Coincidentially, the
width and length are         500  w
the same. Therefore,
the temple room will be      500  250
square so that
everyone could see the       250 ft
Lord Sri Sri Gaura Nitai
 Jaya!
FIND THE AREA…

   Simple.. The length is
250 ft. The width is 250
ft. Multiply them
together..
   A= 625,000 sq ft.

   Nice big temple ain’t
it?? HARI HARI BOL!!!!
RELATED RATES

   Remember how I constantly emphasized,
RATE = DERIVATIVE
   Rate specifically deals with time.
   Here is an example:
   The diameter and height of a paper cup in the
shape of a cone are both 4 in. and water is
leaking out at the rate of 0.5 in3/ sec. Find the
rate at which the water level is dropping when
the diameter of the surface is 2 in.
VARIABLES SOUND NICE.. (laughs)

   V is volume in cubic
inches                    1 2
   h is height.           V  πr h
   t is time                 3
2
   Volume formula            1 h
(known)                V  π  h
   2r = h, since             3  2
diameter=h.
πh3
V
12
Differentiate with respect to time
πh 3
   Remember now… we are                 V
differentiating with time!!! So,        12
we must differentiate implicitly!!
Remember chapter two??               dV  3πh 2  dh  πh 2  dh
   Since we know that the rate of            12  dt   4  dt
                
leak is ½ cu in/ sec, dV/dt is –     dt                  
½ since we are decreasing.
dV      1
   Solve for the rate of change for
the height.

dt      2
   Since there is no variable for
time, the rate of change for the     dh  1  4 
height is constant for any value            2 
of t.                                dt  2  πh 
   Replace the h with 2 in and the
rate of the water level is           dh  1  4         1

         
dt  2  π22 
dropping with the height and
time is 1/2p in/sec.                                    2π
SUMMARY

   We covered a good amount of material in this
chapter. Let’s go over what we learned.
   You can find the maximum or minimum of f(x)
by setting f’(x) = 0. Solve for x and do a sign
analysis.
   You can find concavity of f(x) by setting
f’’(x)=0. Solve for x and do a sign analysis.
DERIVATIVES                           
d x
dx
e  ex

d
ln x   1
dx            x
   We learned how to              d arcsin x      1

differentiate more functions         dx         1 x2
using the inverse rule.        d arccosx         1

   COMMIT THESE TO                      dx          1 x2
MEMORY AS WELL AS
THE FUNCTIONS IN
d
arctan x   1 2
dx               1 x
CHAPTER TWO!!! IF YOU
DO NOT REMEMBER ANY
d
arc sec x   12
FUNCTION’S DERIVATIVE,         dx                x x 1
DO NOT GO ON TO
CHAPTER FOUR!!
d
arc csc x    12
dx                 x x 1
COMMIT EVERYTHING TO
MEMORY!!!!                     d
arc cot x    21
dx                 1 x
OPTIMIZATION/RELATED RATES

   For optimization problems, relate everything to one
variable. For example, area and perimeter with sides
a and b. Solve for a, for example, and plug the a
expression in for the area formula so you have an
area function A(b). Then differentiate that and set it
equal to 0. DON’T FORGET SIGN ANALYSIS!!!
   For related rate problems, gather all of your
variables together and use any geometry formulas
you need. Differentiate with respect to TIME. You
most likely will need to use implicit differentiation. It
   It helps in both scenarios to draw a picture of the
problem to get a better understanding.
TEST

   Please visit the following website to take a
practice exam and please look over the