# Brief History of Calculus by sir17308

VIEWS: 25 PAGES: 57

• pg 1
```									                   Brief History of Calculus
• In the first half of the XVII century, mathematicians were
interested in finding horizontal tangents to the graph of a
function ( in order to find the maximum and minimum values of
the function) and physicists were interested in finding the
velocity and acceleration of a particle whose position function
was given.
• By 1666 Isaac Newton (English), introduced the 'method of
fluxions' which allowed him to find the derivative of a function.
The derivative can be interpreted as the slope of the tangent to
the graph of y=f(x) and also as the instantaneous velocity of an
object whose position is ruled by s=f(t). The derivative of a
function f(x) is given by the limit of the quotient
[f(x+h) – f(x)] / h when h0.
• A few years later, Gottfried Leibniz (German), working
independently introduced the idea of infinitesimals (a quantity
whose square is zero) which allowed him to calculate the
derivative without limit. He define the derivative of y=f(x) as the
quotient dy/dx, where the differentials dy & dx are the
respective values of f(x+x) - f(x) and x when x  0.
• Leibniz’s approach was taught until XIX century. It was the great
mathematician Euler who did a gigantic work in applying
Leibnitz’s method to both math & science.
• But in late 1700 some mathematicians (like Berkeley, Cauchy,
Weierstrass) began to criticize the lack of rigor of Calculus and in
particular the Leibniz’s approach. In 1826 Cauchy taught an
improved version of Newton’s approach in Paris.
• “Leibniz’s differentials approach” didn’t pass the validation rigor
test and the improved Newton’s approach was preferentially
taught.
• In 1965 A. Robinson (American), constructed an extension of
real number and he called “Hipper-real numbers” that contains
the real number and “the Leibniz’s infinitesimals elements”. This
fact gave full validation to the Leibniz’s approach to Calculus.
Calculus over the “Hipper-Real” is called Non-Standard Analysis
and requires more theoretical knowledge of math.
• In this project we begin with Newton’s approach and then we’ll
introduce the differentials (Leibnitz’s way) as soon as possible
and it will then make the work simpler.
Assuming that x=f(t) is the function that relates the position x of a particle
with time t, the derivative allows us to define instantaneous velocity at
t=a as the limit of the average velocity from t =a to t=a+h when h  0.

So, instantaneous velocity when t=a is defined as lim f(a+h) – f(a)       .
h 0      h
Numerical approximation:
Example1: Let f(t) = -16t2 +64t +312 be the function that relates the
position (in feet) of a ball thrown vertically up versus with the time t in
sec). Find instantaneous velocity of the ball when t=1 sec.
Since the average velocity = f(1.1) – f (1)
.                                  1.1 - 1    = 243.04 - 240 = 30.4 ft/sec.
. 0.1
We guess that the instantaneous velocity is approximately 30.4 ft/sec
Now, taking the values of t even closer to 1, we’ll get more accurate
values of v(t). See the table below.
t     1.01    1.001    1.0001 1.00001    …....     1.0000001
v      31.84 31.984 31.984 31.9984       ……..      31.99998
We see (and it can be proved) that lim v(t) = 32 .
t1
It seems natural to say that the instantaneous velocity when t=1 is 32 ft/sec.
Example 2: The position of an object is given by s(t) = t2 +10 with s
in feet and t in seconds. Find the speed of the object when t= 5sec
Speed = lim s(5+h) – s(5) = lim (5+h)2+10- (52+10)
.        h0      h             h0   h
= lim 25+10h+h +10 – 25 -10 = lim 10h = 10
2

h0          h               .h0 h

So speed when t=5 is v=10 ft /sec
Example 3: The position of an object is given by s(t) = at2 + bt+ c,
with s in feet, t in seconds and a, b & c are constants. Find the speed
of the object at any time t.
s(t+h) – s(t) a(t+h)2 +b(t+h)+c – (at2 +bt+c)
=
.     h          .              h
=. at +2aht +h +bt+bh+c- at –bt-c
2        2            2

.              h
= 2aht+h +bh.
2

.    h
Speed at any time t = lim h(2at+b) / h = lim 2at+b = 2at +b
.                     h0              . h0
GEOMETRIC INTERPRETATION OF THE DERIVATIVE
Let’s assume that a particle is moving along a line and it’s position
is given by the function s = f(t). The graph f is shown below.
To visualize the (instantaneous) velocity when t=a we have to
identify the expression [f(a+h)-f(a)] / h] on the graph.
Let call A and B the points on the graph corresponding
to t=a and t=a+h. See orange triangle.
s
So [f(a+h)-f(a)] / h] = AB slope = mAB.
B      s= f(t)
Taking an h1 smaller than h we get point                           B1

B1 (more close to A than B) . The Average               Bn                  f(a+h) – f(a) = f
Velocity is represented by mAB1 (the slope              A
of AB1). See blue triangle.                                        h = x
Taking the h smaller every time ( h 0),                                        f(a+h)
the secant’s line  tangent line at A                       f(a)
and msec  mtan                                         aa+hn a+h1
t
a+h

So lim [ f(a+h)-f(a)] / h] = f ’(a) = mtan = slope of the tangent line
.   h 0                                      to the curve at x = a
GLOSARY OF TERMS

I)     f(x+ x)-f(x) or f(x) represents:
.    x          x
1) the Rate of Change of f(x) = change of f(x) / change of x
2) Average Speed when x represents time.
3) The slope of the secant line passing though the
points A= (x, (f(x)) & B=(x+ x, f(x+ x)).

II)     The derivative of f(x), f ’(x), is used for the value of
lim    f(x+ x) - f(x) = lim f
x0          x           x0 x
By “differentiating f(x)” we means “get the derivative of y=f(x)” .
The notation for the derivative of y=f(x) is f ’(x) or df /dx or dy /dx.

III)   f ’(c) , the derivative of f(x) evaluating at x=c, represents the:
1) Instantaneous rate of change of f(x) at x=c
2) Instantaneous speed at x=c when x represents time.
3) Slope of the tangent line to the graph y=f(x) at x=c
4) Marginal value of f(x) in Business.
Example1: Find the derivative of f(x) = 2x+1
Since y=2x+1 is a line and the tangent line coincides with it, the
slope of the tangent line is the slope of y =2x+1, e.g. m=2. So f ’(x)=2
Example 2: Find the value of x where the tangent line to the graph
of f(x)= 5x2+x+1 has slope 0
Since f ’(x)= lim 5(x+h)2 +(x+h) +1– ( 5x2+x+1) .
.             h0            h
= lim 10xh +5h2+h . = lim h(10x+5h+1) = 10x+1
. h0     h          . h0    h
So m = 0 when 10x +1=0  x= - 0.1
Example 2: Find the points where the tangent line to f(x)= x3-3x2-9x+1
is parallel to the X axis.
f ’(x)=3x2-6x-9 = mT= 0
But 3x2-6x-9 = 3(x+1)(x-3)=0  x= -1 or x= 3
Since f(-1)= 6 & f(3)= -26
The points are (-1,6) and (3,-26)
Theorem : f is differentiable on the interval (a,b).  f is continuous
on the interval (a,b).
Proof: Assume that f ’(c) exists for any c in (a,b).
Then lim [ f(c+h)- f(c)] = lim f(c+h) - f(c) • h
.      h0
. h 0         h
= f ’(c) • lim h = f ’(c) • 0 = 0
.        h 0
So lim [ f(c+h) - f(c)] = 0 , and from here we get lim f(c+h) = f(c) .
.     h0                    .                      h0
So f is continuous at c for every c in (a,b).
Example: Since the derivative of f(x)= 5x2+x+1 is f ’(x) =
10x+1, which exists for every real number x. So f(x)= 5x2+x+1
is continuous everywhere.
Remark. The reverse of this theorem is not true.
Counter example: We know that f(x) = |x| is continuous on R ,
but at x=0 it’s not differentiable since:
lim    l0+hl –l0l = lim lhl                           +1 if h 0
, which approaches to     –1 if h0
h 0      h       . h 0 h
Let f be a continuous function. A point P=(c,f(c)) is called:
1) a corner at P , when the graph of f has two different tangent lines
at p (one for x c- & the other for x c+ ).
2) a cusp when the tangent line at (c,f(c)) is vertical
Y
We’ll usually say that f is smooth on (a,b) if f
is differentiable on the interval (a,b).           CORNER

The graph on the right shows the two
types of points where f’(x) doesn’t exist.
X
CUSP

Example 1: Where is a parabola not smooth?
The general equation of a parabola is f(x)= ax2+bx+c .
Differentiating we get f ’(x) = 2ax+b; therefore f ’(x) exists for every
real number x and it is smooth everywhere. In other words,
parabolas have no corner points or cusps. All parabolas are smooth.

Remark: The same result is valid for any polynomial functions.
Ex 2: Find the value of x that f(x) = (2x-6) 2/3 is not smooth.
f ’(x) = 4/3(2x-6)-1/3 , exists for every x except at x=3.
f ’(3)  + ∞ or -∞ when x 3. But f is continuous at x=3, this means
that the tangent line at (3,0) is a vertical line. So f has a cusp at x=3 .
Y
Since f(x) is never negative, the shape of                        y= (2x-6) 2/3
the graph of f like the figure on the right.
X
3
DIFFERENTIALS
Let f(x) be a differentiable function on an interval (a,b). Then
lim f - f ’(x) x = lim f – f ’(x) = 0
x         x0 x                                                y=f(x)
x0                                                   Y
Meaning that [f - f ’(x) x] 0 when x0
Q
See figure on the right
f = f(x+ x) – f(x)
f
(f is represented by the blue color curve. )
To help us to identify the expression f ’(x) x        P         df
we draw the tangent line in red color to the                  dx
graph of f at P and check the red right triangle.           . ||
x                      X
Now we’ll rename x as dx and we’ll call df
the other leg of the red right triangle (which          x         x+x
measures the difference from the tangent line).
On the red triangle we have: f ’(x) = m tan at P =f(x+dx)-f(x) = df
. dx           dx
which implies that df = f ’(x) dx .
(Note that this is an independent selection of x ).
In summary, when using df = f ’(x) dx           Y                      y=f(x)
you can find derivatives without limits.
Let’s review that df = f ’(x) dx one more
Tangent
time and that m=dy/dx                                                  y=mx+b
The graph on the right shows the tangent                     df   f
to y=f(x) at the point P(x,y) and dx, df &         P
f at x+x as well.                                     dx
The difference f – dy is represented in
green color                                                                     X
x        x+ x
On the graph see how      f – dy  0 as x  0.

Remark: This could explain why Leibniz used dy/dx notation for the
derivative instead of the f ’(x) notation.
THE DIFFERENTIAL TO FIND DIFFERENTIATION RULES.
First we’ll prove that (dx)2 = 0 (as mentioned before) .
Let f(x)= x2 and find df in two ways:
1) f ’(x)= lim (x+x)2 – x2 = lim x2 +2x x + (x)2 – x2 = lim (2x+x) = 2x
.          x0    x           x0      x               x0

So df = f ’(x) dx = 2xdx
2) Using differential definition
df =(x+dx)2 - x2 = x2+ 2xdx +(dx)2 - x2 = 2xdx +(dx)2
From 1) & 2) we get: 2x dx = 2xdx+(dx)2 . So (dx)2 = 0.
REMARK: Using df = f(x+ dx) – f(x) plus the fact that dx2 =0, or better
(dx) n = 0 for n ≥ 2 , we can perform the differentiation without using
limit process.
Example 1: Find f ’(x) for f(x)= x3
df = (x+ dx)3 – x3        0      0
df = x3 + 3x2dx + 3x(dx)2 +(dx)3 – x3= 3x2 dx
So dividing both sides by dx , we get d x = 3x2
3
.dx
Example 2. Differentiate f(x) = 5x3 – 2x2 –10
df(x) = 5(x+ dx )3 – 2(x+ dx)2 –10 –(5x3 – 2x2 –10)
= 5(x3 +3x2 dx +3xdx2+ dx3 )–2(x2 + 2xdx +dx2 ) -10 –(5x3 – 2x2 –10)
Replacing dx2 = 0 and collecting …
d f(x) = 5x3 +15x2 dx - 4x dx - 2x2 - 10– 5x3+2x2+10
Factoring out dx … df(x) = (15x2 +4x)dx
Dividing by dx … f ’(x)= 15x2 +4x
Example 3: Find dy/dx if y=xn , for any positive whole number n
0
Since (x+dx) n - xn = xn +nxn-1dx + (sum of terms containing (dx)n with n≥2) - xn

So d((xn)= nxn-1dx and dividing by dx we get dy/dx = nxn-1
Example 4. Find d( x) / dx
Let’s call y =  x . Squaring both side we get y2 = x
So d(y2) = dx
2ydy = dx
dy = dx / (2x )  dy/dx = ½ x
BASIC FORMULAS TO DIFFERENTIATE A FUNCTION

Let u & v be functions of x, then
1) d(u ± v) = du ± d v
Proof: d(u ± v) = [u(x+ dx) ± v(x+ dx)] – [u(x) ± v(x) ]
=[ u(x+ dx) – u(x) ] ± [ v(x+ dx) - v(x) ] = d(u) ± d( v)
2) d( ku ) = k du where k is a constant
Proof: d(ku) = (ku)(x+dx) – (ku)(x) = k[u(x+dx)] – k[u(x)]
= k[u(x+dx) – u(x)] = kdu
3) d(u v) = (du)v + u(dv)
Proof: d(u·v) = [ u(x+dx) · v(x+ dx) ] – u(x)·v(x)
= [u(x) + u’(x) dx ]·[v(x) + v’(x) dx ] – u(x)· v(x)
Expanding …
= u(x)v(x)+ u(x)v’(x) dx + u’(x) dx v(x)+ u’(x)dx v’(x) dx – u(x)v(x)
= u(x)v’(x) dx + u’(x) dx v(x) + u’(x) v’(x) dx2
0

So, d(u v) = du·v + u·dv
4) d( 1 ) = - du
u      . u2
Proof: Notice that d ( u · 1 ) = d(1) = 0
u
But d(u ·  1 ) = du·( 1 ) + u·d( 1) = 0  du + u·d( 1) = 0
u           u         u               .u        u
Isolating d(    1) we get d( 1 ) = - du
u              u      . u2
5) d( u ) = (du) v – 2 (dv)
u
v      .       v
Proof: d(   u ) = d( u· 1 ) = du ( 1 ) + u ( -dv ) = (du)v–u(dv)
v            v         v         . v2    .    v2
FORMULAS TO FIND DERIVATIVES OF BASIC FUNCTIONS

Let u & v function of x. Dividing by dx above formulas we get.
1) d (u  v) = d u  d v           2) d (k·u) = k d u. , k is a constant
.     dx       dx    dx            .    dx        dx
3) d (u·v) = du v + u dv                d( u )    du v - u dv
v      dx
4) ______ ___________  dx
.     dx     dx       dx                  dx =          v2
dxn
Besides was proved that dx = nxn-1 for any integer n.
Ex 1. Find [√(x) - 2x3 +1]’ = [√ x]’- [2x3 ]’+[1]’
= ½(x-1/2) – 2 ·3x2 + 0
= 1/(2√ x ) – 6x2
Ex 2. For y= (5x2– 3x + 6) / (x2– 5) , find y’(3)
y’ = (10x – 3)( x2-5) – (5x2 – 3x+6) (2x) .
(x2– 5)2
y’(3) = 27(4)-42(6) = -9 .
16
Ex 3: Find dy/dx when x=1, for y = (x2+3x-1)(5x3-x2-2x+5)

y’ = (x2+3x-1)’(5x3-x2-2x+5)+ (x2+3x-1)(5x3-x2-2x+5)’
= (2x+3) (5x3-x2-2x+5) + (x2+3x-1)(15x2-2x-2)
So, y’(1) = (5)(7)+(3)(11)= 35+33 = 68

NON CONTINUITY v/s CONTINUITY v/s DIFFERENTIABILITY
A) A function f is discontinuous at x=c when lim f(x) f(c)
.                                                       x c
Two different situations can occur.                                            1)
there exists an interval containing x=c , where the function is
continuous except only at x=c. Graphically this happens if:
a) f(x) has a jump.                          b) A point is missing or has been
OR
See figures below                            relocated. See figure below.
Finite jump           Infinite jump       Point missing or relocated
Y                     Y                     Y                        Y

X                    X                     X                    X
c                    c                      c                       c

2) It f doesn’t exist on the interval as defined in part A) then the
there exists an interval containing x=c where the function is
discontinuous everywhere. See next example.
1 if x is rational
Example: Let f(x) =
0 if x is irrational
This function is discontinuous everywhere. Its graph cannot be drawn.
B) A continuous function f is not differentiable at x=c , if and only if,
lim f(c+x) – f( c) does not exist.
x0       x
At x=c the graph of f can either:
2) has a corner ( there are two
1) has a cusp ( the tangent line          tangent lines instead of just
at x=c is vertical)               OR      one tangent line at x=c.)
See graph below                          See graph below
Y                                         Y

X                                            X
c                                           c
CHAIN RULE

Theorem: Let y = f(u) and u= g(x) be differentiable functions
(Notice that y = fg(x) = f(g(x)) depends only on x ). Then
dy = dy du                                                       y
(See figure on the right).
dx     du dx                                                         dy/du
Using Newton’s notation: (f(g(x))’= f ’(g(x))·g’(x) dy/dx u
Proof: By definition dy = d(f(u))= f ’(u)·du
du/dx
But du =u’(x)dx
x
Replacing du …          dy = f ’(g(x)) u’(x)dx
Dividing by dx we get dy = f ’(g(x)) g’(x)
.                           dx
Ex 1: Find dy /dx if y = √3x2 –5                          y= √u
Let u = 3x2 – 5 , so y = √u                                        dy/du = 1/(2√u)
dy/dx
Then dy/du = 1/(2√u). & du/dx = 6x                      u = 3x2– 5
Using the chain Rule dy/dx =[ 1/(2√u )]6x                          du/dx = 6x
But u = 3x2– 5 ,we get dy          .    3x     .            x
=
.                           dx     √(3x2 – 5)
Remark: (f(g(x))’= f ’(g(x)) g’(x) says that to get the derivative of
“nested functions” you multiply the derivative of each one starting from
left to right and so on). (beware that sin 2(3x-1) means [sin(3x-1)]2
Ex 1. Find y ’ for y = (5x2 –17x+3)5
= 5(5x2 –17x+3 )4 • (10x-17)
Ex 2: Find y’(1) for y = (3x2-2)3( 5x3-x-3)4
y ’= 3(3x2 -2)2 (3x2-2)’( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 ( 5x3-x-3)’
y ’= 3(3x2 -2)2 (6x) ( 5x3-x-3)4 + (3x2-2)3 4( 5x3-x-3)3 (15x2-1)
y ’(1) = 3(3-2)2 (6)    (5-1-3)4 + (3-2)3 4 (5-1-3)3 (15-1) = 74
. 2x-1           dy .
Ex 3: For y =              , find      when x=1
√5x2+4           dx
.
y ’ = (2x–1)’ √5x 2 +4 – (2x-1) ((5x2 +4)1/2 )’

.                   √(5x2 +4)2
y ’ = (2)√5x2+4 – (2x-1) ((½(5x2+4)-1/2(10x)
.                    (5x2 +4)
13 .
y ’(1) = 2√9 -(2-1) [½(9) ](10) =
-1/2

.           9                   27
APPROXIMATING f by df
error
Since when x0 both error = f - df  0 and             0,
. x
we can say that for small values of x, df is a good approximation of f.
We’ll call error = f - df the relative error of approximating f - df
. f(x)  . f(x)
Ex 1: Use the differential to approximate √(27) and estimate the
relative error percent

Let f(x) = √x and x=25 (since 25 is the perfect square of 5 and is
near to 27), then x=2 .
Substituting x=25 and x=2 in the approximating formula
f(x+x)- f(x)  f ’(x) x we get f(x+x)  f(x)+ f’(x) x
√(25+2)  √25 + 1/(2 √25) (2) = 5 + 1/5 = 5.2 . So √(27)  5.2
Since (5.2 -5)/5.2= 0.038 then % error = 3.8 %
Ex 2: Given f(x) = x3 – 5x2 +8x +9, use the differential to estimate
value for f(3.74)

Let’s take x=4 and x = - 0.26.
Since f’(x) = 3x2 – 10x+8 , f’(4) = -16.
So f(3.74) = f(4) + f ’(4)x = 25 +16(-0.26) = 20.84

Since f(3.74) = 21.23 we have % error  (0.39 / 21.23) = 1.84%

REMARK: As you see when working with the differential df we are
replacing f(x) by the equation of the tangent line at x and f (x+x) =
f(x) + f ’(x) x is the (linear) approximation for f.
Besides        lim error = lim f- df = df – df = 0
.              x0 x     x0 x      dx dx
which means, that for small values of x , df is a good estimation of f .
Ex 3: Use linear approximation to estimate the change of the volume
of a cylinder when the radio increase from 5 in to 5.2 in and the height
decreases from10 in to 9.7 in. ( Volume of a cylinder is given by
V(r,h)= r2h) Find the relative error.
V  dV =(2rdrh+r2dh)
Now we replace r=5, h=10, dr=0.2 & dh = -0.3
dV= [ 2(5)(0.2)(10) + (5)2(-.3) = (20-3)]=17 =53.4
Relative error = 17 = 6.8 %
.                 250
Ex 4: A hemispherical bowl of radius 10in. is filled with water to a
depth of x in. Assume that you measure the depth of water in the
bowl to be 4 inch. with a maximum error of 1/20 in. Estimate the
maximum error in the calculated volume of water in the bowl.
(Volume of a hemispherical is V=(3rx2 – x3)/3 )
V (x+x)-V(x)  V’(x)x = (60x – 3x2)/3 ·x = (20x – x2)·(1/20)
= (80 – 16)/20 =   64/20=  10.05in3
So, V(4)=(301.59 10.05)in3
MAXIMA AND MINIMA OF FUNCTIONS ON CLOSED INTERVAL
If c is in a closed interval [a,b], then f( c) is called the minimum value of
f(x) on [a,b] if f( c) ≤ f(x) for all x in [a,b]. Similarly, if d in [a,b], then f(d) is
called the maximum value of f(x) on [a,b] if f(d) ≥ f(x) for all x in [a,b].
It can be proved that a continuous function on a closed interval [a,b]
attains a minimum value f( c) and a maximum value f(d).
Y                                              Y

f(d)
f(b)
f(c)                    f(a)
X                                         X
a   d          c          b                a                 b

We say that the value of f( c) is a local minimum value of f(x)
if f(c) ≤ f(x) for all x sufficiently near c ; more precisely if there exists
an open interval containing c where the inequality holds. A local
maximum value of f(x) is defined similarly.
APLICATIONS OF FIRST AND SECOND DERIVATIVES

Theorem: Let f be a differentiable function on (a,b) the
1) if f ’(x) is positive then f is increasing on (a,b)
2) if f ’(x) is negative then f is decreasing on (a,b)
Proof: 1) Let x1 & x2 be any two numbers in (a,b) and x2 > x1 , using
the Mean Value Theorem there exist a real number c in (x1 , x2 )
such that: f ’(c ) = [f(x ) – f(x )] / [x - x ]
2       1       2   1
+                          +
But by hypothesis f ’(c ) & x2 - x1 are positive so f(x2)–f(x1) is positive.
This means that f(x2) -f(x1)> 0 for any x1 & x2 in (a,b).
In other words f is increasing on (a,b)
Similarly 2) is true.
Ex 1: Prove that f(x) = 2x3+6x -13 is and increasing for every real x
Using the Theorem we only have to prove that f ’(x) is always positive
Since f ’(x)=6x2+6 =6(x2+1) and x2+1 is always positive we get that
f ’(x) > 0 on (- ∞ , + ∞).
Ex 2: Find the increasing and the decreasing open intervals for the function
.     f(x) = 6x5-240x3-750x+1
f ’(x)=30x4-720x2-750 = 30(x4-24x2-25) and we need study the sign of f ’
Factoring … f ’(x) = 30(x2-25)(x2+1) = 30(x+5)(x-5)(x2+1)
Note that x2 +1 is always positive so 5 and -5 are the only critical numbers

+                          +
-6             0            6
-5                5
Take -6 as a testing value for the sign of f ’ on (- ∞ ,-5): f ’(-6) is +
Take 0 as a testing value for the sign of f ’ on (-5 ,-5): f ’(0) is -
Take 6 as a testing value for the sign of f ’ on (6, + ∞): f ’(6) is +
So f is increasing on (- ∞ ,-5) & (6, + ∞) and decreasing on (-5,5)
Theorem: Suppose that f is differentiable at c in an open interval
containing c. If f(c) is either a local maximum value or a local
minimum value of f, then f ’(c)=0
Proof: Assume f is differentiable and has a local maximum at x=c.
If f ’(c) > 0 then f is increasing and cannot have a local max. at x=c.
Similarly if f ’(c) < 0 , then f is decreasing and cannot have a local min.
So, f ’(c) has to be equal to zero
REMARK: If a function f continuous on (a,b) has a local maximum
or a local minimum at x=c for then:
1) f ’(c)=0 or
2) f ’(c) is undefined at x=c (f has a corner).
The real numbers satisfying the above two conditions are called critical
numbers.
Theorem: Suppose that a continuous function f is defined on a closed
interval [a,b], then the absolute maximum and the absolute minimum of
f occur at a critical value or at the extreme of the interval.
Ex 1: Find the maximum and minimum values for the function
f(x)= x3-3x2-9x+20 on the interval [-2,5].

First find f ’(x) … f ’(x) =3x2-6x-9
Since f ’(x) exists for every x in [-2,5], the critical values are given by
f ’(x) =3x2-6x-9 =3(x+1)(x-3)=0  x= -1 or x=3
The maximum and minimum values are located at the critical
numbers -1 or 3 or at the ends -2 or 5
Select the maximum and the minimum values using the table below.

x      f(x)
Max {18, 25, -7, 25} = 25
-2    f(-2)=18
-1    f(-1)=25      Min {18, 25, -7, 25} = -7
3    f(3) = -7
5    f(5) = 25

So the maximum value is 25 and is obtained at x= -1 and x = 5.
And the minimum value is -7 and is obtained at x=3.
Ex 2: We are given a fence of 200 ft long and we want to build a
rectangular enclosure along a straight wall. If the side along the wall
needs no fence, find the largest area of such rectangle.

See figure on the right                    WALL
Let’s call x & y the
y                                         y
sides of the rectangle
as shown in the figure                      x
So, length of the fence = x+2y =200  x=200-2y
and area of the rectangle = xy = A  A = (200-2y)y=200y-2y2
We need to maximize A, so we need dA/dy … dA/dy = 200-4y
From 200-4y =0 we get y=50 (the only critical number)
From 200 = x+2y we get that 0 ≤ y ≤ 100 , so [0,100] is the interval
So, the maximum area is the maximum value of A for y= 0,100 & 50
y    A=200y-2y2
The maximum area is A=5000 ft2
. 0      . 0
100      . 0
. 50     5000
Ex 2: A can (cylinder) with volume of 240 cm3 has to be constructed.
Find the dimension of the cheapest (using the least material) can.
(Assume that the radius r has to satisfy the inequality 1cm ≤ r ≤ 5 cm).
See figure below.
We have to minimize the total surface of the cylinder.
Isolating h from V=240=r2h we get h= 240 / r2
r
Replacing h in S formula we get S= 2(240 / r )+ 2r2
Find the critical numbers … dS/dr = - 480/r2 + 4r = 0
So S=480/r + 2r2 and dS/dr = - 480/r2+4r                            h

Solving … - 480 +4r3= 0  r =(120/ )1/3  3.4cm
Since r has to be in [1,5], making the table:
V=
r      S(r)                                             r2h + 2r2
S=2rh
1.0    486.3
213.8        The radius of the cheapest can is r  3.4cm
3.4
5.0    253.1
LOCAL MAXIMUM AND LOCAL MINIMUM
Let f(x) be a continuous function. We say that x=c is a critical value of
f(x) if f ’(x) is 0 or if f ’(c) is not defined. In other word, if the slope of the
tangent line to y=f(x), at x=c, is 0 or is not defined.
We say that f(x) has a local maximum at x=c if there exists an open
interval (c-h , c+h) where f(c) is the maximum value of f(x). In a similar
way we define a local minimum.
Ex 1: The graph of y=f(x) is shown below. Find :
1) critical values,                     1) f ’(x)=0 at x=h with b  h  c , d, e & g
2) local max and min. Y                     f ’(x) DNE at x= a, b, c & f
So, the critical values are x=a, b, c, d, e, g & h

X
a     b h c        d                    e       f     g

2) Local maximum at x =d & f
Local minimum at x= a & e
Remark : The above example shows that local maximum and
minimum occur at the values of x that are critical values ( the
reverse is not true).
Test for Local Extrema: Let f(x) be continuous on an interval I and
differentiable there except possibly at the interior point c of I:
1) f ’(x) 0, for x< c & f ’(x) 0for x>c, then f(c ) is local minimum of f(x).
2) f ’(x)0,for x<c & f ’(x)0 for x>c, then f(c) is local maximum of f(x).
3) f ’(x) has the same sign both to the left and to the right of c, then f(c)
is neither a maximum nor a minimum value of f(x).
Ex 1: Find local Extrema for y= x2(x-3)2/3
y’ = 2x(x-3)2/3 + x2(2/3)(x-3)-1/3 = 2x(x-3)+x (2/3)
2

.   (x-3)1/3
1) y’ = 0  6x2-18x+2x2 =0  x= 0 or x=9/4 =2.25            Critical numbers
0, 2.25 &3
2) y’ DNE  (x-3)1/3 =0  x=3
Study of sign for y’ …
+                 +
Using -1,1,4 & 10 as                 -1       1       4         10
checking points….                         0       3       8.6
So using the Test for Extrema we get:             Y
Local minimum at x=0 & x=9/4
Local Maximum at x=3
Note that at x=3 the graph has a
cusp, see its shape on the right.                                      X
3

Ex 2: Find local Extrema for y= (x-3)3(x2-4x-4)
f ’(x) = 3(x-3)2(x2-4x-4)+(x-3)3(2x-4) = (x-3)2(3x2-12x-12+2x2-4x-6x+12 )
= (x-3)2(5x2-22x) =x (x-3)2(5x-22)
From f ’(x) = 0 we get the critical values 0,3 & 4.4
Find the sign of y’ …           +                       +
Using the Test …                  0        3          4.4
Local maximum at x=0
Neither a Local maximum nor a minimum at x=3
Local minimum at x=4.4
See the graph in the next slide
Y               Neither a local max
Note that f ’(3) exists e.g. the                          nor a local min
graph has an horizontal tangent
line at x=3 but it has neither a local
maximum nor minimum at there.                                   4.4        X
See the shape of the graph on the                     3
right

Second Derivative Test for Extrema: Suppose that f is twice
differentiable on an open interval I containing a critical value c at which
f ’(c)=0. Then
1) If f ”  0 on I, then f(c) is the local minimum number of f(x) on I.
2) If f ”  0 on I, then f(c) is the local maximum number of f(x) on I.
Proof: We will prove only part 1).
Since f ’(c)=0 and f ”(x)0 on I  f ’(x) is increasing on I, we may
conclude that f ’(x)  0 for x  0 and f ’(x) 0 for x  0.
Using the First Derivative Test, we conclude that f(c) is a local
minimum value
Ex 1: Find the minimum of f(x)=x2+128x-1 for 0 x+∞
1) Find critical values. f ’(x) = 2x-128x-2 = 0  2x3 -128 =0  x=4
2) Find second derivative f ”(x) = 2 +256x-3
3) Applying Second Test … Since f ”(4) = 6  0 on (0, +∞) , we
know that f(4) = 16+32=48 is the maximum value of f(x) on (0, +∞)

CONCAVITY POINTS AND INFLECTION POINTS

Theorem: Suppose that a function f is twice differentiable on an open
interval I and T is tangent line to the graph y=f(x) at (a,f(a)) then
1) If f ”(a)  0 then tangent line T at a lies above T .
We say that the function f is concave upward )
2) If f ”(a)  0 then tangent line T at a lies below T.
( We say that the function f is concave downward)
Let f be a continuous function. We say that (c,f(c)) is an inflection point
of f when for x c the concavity of f is the opposite of the one for x c.
In other words, f has a change of concavity at x = c.
A useful Theorem for absolute max or min over an open interval
Theorem: Let c in (a,b) and f(x) a continuous function on (a,b).
If c is the only critical number for f(x) then the local maximum
(minimum) is the absolute maximum (minimum).
Proof: 1) Assume that at x=c f has a local minimum
y
On (a,c)& (cb) f’(x) is or increscent everywhere
or increscent except at their inflexion points.
In similar way can be proved the rest.
x
Ex: Find the absolute maximum of                        a        c   b
. x2 .
f(x)=         on (0,+∞)
.1+x4
Look for critical numbers …
2x(1+x4)-x2(4x3)   2x(1-x4) 
f ’(x)=                  =              f ’(x) = 0 if x=0 or x=1
. (1+x4)2          .(1+x4)2
So on (0,+∞) the only critical number is x=1
It’s easy to see that at x=1 f(x) has a local min.
So at x=1 f(x) has the absolute maximum on (0,+∞)
Proof: See figure on the right                        Y
y=f(x)
Since y=T(x)= f(a)+ f ’(a)(x-a) , we define
g(x) = f(x)-T(x) then g ’(x) = f ’(x) – f ’(a)                       g(x)
For x=a we get g’(a) = 0 i.e. x=a is a
critical number for g(x).                                            y=T(x)
But g”(a) = f ”(a)  0, so at x=a, g(x) has                   f(a)
a minimum value 0 (since g(a)=0)                            a      x        X
So the graph of y=f(x) lies above of T
Ex 2: For f(x) = x6-3x5 -20x4+90x3-135x2+2x+1, find the intervals that
f is concave up , concave down, and identify the inflection points.
f ’(x)= 6x5-15x4-80x3+270x2-270x+2
f ”(x) = 30x4-60x3-240x2+540x-270= 30(x4-2x3-8x2+18x-9)
Possible rational zeroes for f ”(x) are 1, 3, 9 , working we get:
f ”(x)=30(x-1)2(x+3)(x-3)                      +    _          _        +
Sign diagram for f ” …..                         -3       1         3
Concave up on (-∞ , -3)(3 ,+ ∞ ) and Concave down on (-3,3).
So, f(x) has inflection points at x=-3 , x=3
Ex 3: Find the dimensions of the rectangle with maximum area that
can be inscribed in the ellipse 16x2+9y2=144 .
y
See figure on the right.                                 4 16x2+9y2=144
Let P=(x,y) be the vertex of the rectangle in the            P(x,y)
first quadrant. So x is in the interval (0,3).
Area of rectangle is A=2x(2y)=4xy
Isolating y from the ellipse … y= (4/3) (9 -x2 )1/2           3
x

So A= (2x) 2(4/3) (9 - x2 )1/2 = (16/3)x (9 - x2 )1/2
A’= (16/3)(9 - x2 )1/2 + (16/3)x (½ )(9 -x2 ) -1/2 (-2x)
A’= (16/3)(9 - x2 )-1/2 [9 - 2x2 ]
A’=0  9-2x2 =0  x= 1.5√2  2.12 (is the only critical value in (0,3) )
Using the 1st derivative test …
For x= 1.5√2 there is a local maximum.           +               -
2      2.12 2.2
But being this the only critical point in
(0,3) it has to be a absolute maximum.
The dimension of the rectangle are b=3√2 & h=4√2.
Ex 1: If the consumer demand for a commodity is D(p)= 10,000e-0.02p
units per week, where p is the selling price , find the price that will
maximize the consumer expenditure. ( Hint: E(p)= pD(p) )
E(p)=10000p e-0.02p
E’(p)= 10000e-0.02p +10000pe-0.02p(-0.02) = 200e-0.02p (50-p)
Since 200e-0.02p cannot be zero E’(p)=0  50-p=0  p= 50
E’’(p) = 200e-0.02p(-0.02)(50-p)+200e-0.02p (-1)
E’’(50)= -200e-1 < 0. So expenditure is maximized when p= \$50
Ex 2: If a certain drug is injected intramuscularly, the concentration of
the drug in the bloodstream after t hours will be C(t)= .75(e-0.2t –e-0.6t ).
Find the time of maximum concentration.
C’(t)= .75(-0.2 e-0.2t +0.6 e-0.6t ) =.15(-e-0.2t +3e-0.6t )
So C’(t) =0  -e-0.2t +3e-0.6t =0  3e-0.6t = e-             e-0.4t = 1/3
0.2t
Solving for t … -0.4t = ln(1/3) = -ln3  t = 2.5ln3 = ln 35/2  2.75
Let find C’’ to apply the 2nd derivative Criteria for locals max and min
C’’(t)=1.5 (0.2 e-0.2t – 1.8e-0.6t ) = 0.3e-0.2t (1- 9e-0.4t )
Since C’’(ln(35/2))  0, the maximum concentration occurs at t 2.75 hr.
Ex 4: A long gutter is to be made from a 12 inch-wide strip of metal
by folding up the edges. How much of each edges should be folded
up in order to maximize the capacity of the gutter. (Hint: maximizing the
capacity means maximizing the cross-sectional area). See figure below
Let x be the base and y the height of edge.
So, we want to maximize the area A= xy            y         area          y
Isolating x from the restraint x+2y =12 …
A=(12-2y)y = 12y -2y2
x
A’=12 -4y  Critical value y=3
A’’ = -4  0  for y=3 the area has a local maximum.
Since there is only one critical value, the local
maximum is the absolute maximum.
So the height of each edge y that maximizes the capacity is 3in.
Ex 5: When you cough you are using a high-speed stream of air to
When you cough the velocity of the air stream is given by
V(r)=c(3-r)r2, where r is the radius of the contracted trachea and c is
a positive constant (depending on the elasticity of the trachea).
Find the radius r that maximizes V.

V(r)=3cr2-cr3  V’(r)= 6cr-3cr2  critical values are 0 & 2
We discard r = 0 (has no meaning) and work on the interval (0,3)
where r = 2 is the only one critical value.

V’’(r)=6c-6cr  V’’(2) = -6c  0  V(r) has a local maximum at r =2
Since r=2 is the only critical value on (0,3) and is a local maximum,
we conclude that V(r) has absolute maximum at r=2.
So, the velocity of the airs tream is maximized when the radius
of the contracted trachea is 2cm.
Ex 6: It costs a Car Company \$ 18,000 to produce a car. The
company’s weekly selling price function is p(x)=30,000 – 60x
when exactly x items are sold. The fixed cost depends on the
amount of production is \$25,000. How many cars should be
produced each week to maximize the company’s profit?
Let x be the number of cars produced each week.
The Cost is C(x)= 18000 x +25000
The Revenue is R(x)=(30000 – 60x )x
So Profit P(x) = R(x)-C(x) = 30000x-60x2 – (18000x+25000)
P(x) = -60x2 +12000x - 25000
P’(x)= -120x +12000  Critical value is x=100
Since P’’(x) = -120  0 , P(x) has a local maximum at x=100
But since this is the only critical value for P(x), it has to be a maximum
The profit is maximized when 100 cars are produced weekly.
TRIGONOMETRIC FUNCTION DERIVATIVES
Using a geometric approach we will prove that lim sinx = 1
x0 x
The figure below shows a unitary circle with an acute central
angle x (in rad) and its corresponding arc BC = x .
D         From figure: AC  arc BC  BD
so          sin x  x       tan x
C
Dividing by sin x .…

tan x
1       x
1  x        1 .
sin x
x                                        . sin x   cos x

O           A           B          If x 0 …   1  lim sinx  1
x0 x
So lim sinx = 1
sin2x              x0 x
1-cosx = (1-cos x)(1+cosx) = . sin x
REMARK: Since . sin x   . sin x (1+cos x)  1+cos x
When x0 ….   lim 1-cos x = lim . sin x = 0
x0     x        x0 1+cos x
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

1) (sin x)’ = cos x
Since sin (x+x) – sin x = sin x cos x +cos x sin x – sin x
.       x               .              x
= sin x (cos x – 1) +cos x sin x
So when x0 …                    .   0            x          1
So (sin x)’ = (sin x) lim cosx – 1 . + (cos x) lim sin x. = cos x
x0      x                  x0 x
2) (cos x)’ = - sin x
(cos x)’ = [sin(/2 – x)]’ = cos (/2 – x)] (-1) = - sin x
3) (tan x)’ = 1/cos2 x = sec2x

1-sin x
Ex 1: Find y’ and y’’ for y=
1+sinx
- 2 cosx
y’ = (-cosx)(1+sinx)-(1-sinx)( cos x) . =
(1+sin x)2                           (1+sinx)2
2(2-sin x)
(2sinx)(1+sinx)2 - (-2cosx)[2(1+sinx)cos x] =
y’’ =                                               (1+sin x)2
.                 (1+sin x) 4 3
Using Quotient Rule we get:
4) (sec x)’= sec x tan x
5) (cot x)’ = - csc2 x
6) (csc x)’ = - csc x cot x
1-sin x
Ex 2: Let y=           for 0≤ x ≤ 3/2
1+sinx
Find local extrema, concavity up and down intervals & inflection points.
From Ex 1 above we get           y’ = - 2 cosx         y’’ =
2(2-sin x)
(1+sinx)2             (1+sin x)2
Since (1+sinx)2 ≥ 0 the sign of y’ depends only on -2cosx, so:
+                 y is decreasing on (0, /2))
0            /2                 3/2     y is increasing on (/2,3/2)
So, y has a local min at x = /2
Since (1+sinx)2 ≥ 0 the sign y’’ deepens only on 2(2-sinx), so:
+
0
y is concave up on (0, 3/2)
Since y’’ > 0 on [0, 3/2] there is no inflection point.
Ex 3: The position function of a particle is given by s=(1+cos 3t)2
(1+sin3t)3, find the velocity of the particle when t =/6.
s’ = [(1+cos 3t)2 ]’ (1+sin3t)3 + (1+cos 3t)2 [(1+sin3t)3 ]’
2(1+cos 3t)(-sin 3t)(3)                          3(1+sin3t)2 (cos 3t)(3)
s’ = [2(1+cos 3t)(-sin 3t)(3) ] (1+sin3t)3 +(1+cos 3t) 2 [3(1+sin3t)2 (cos 3t)(3)]
Since for x= /6: sin(3t)= sin /2 = 1 & cos(3t)= cos /2 =0
v(/6 )= s’(/6) = - 6 (1) (1+0) (1+1)3+ 9(1+0)2 (1+1)2 (0) = -48 +0 = - 48

Ex 3: The position function of a particle is given by s=(1+sin5t)2 (1-sin5t)2
Find the values of t in [0,/2] that the velocity is zero.
First simplify the expression s= (1-sin2 5t)2 = cos4(5t)
v=s’ = 4cos3(5t) (-sin 5t) (5) = -20 sin 5t cos3(5t)
v=0  sin 5t =0 or cos 5t =0
sin 5t = 0  5t = 0  2k = 0 or 5t =   2k  t= 0, 0.2, 0.4
cos5t =0  5t = /2  2k = 0 or 5t = 3/2  2k  t = 0.1, 0.3 , 0.5
So v= 0 when t=0, 0.1, 0.2, 0.3 , 0.4, 0.5
RELATED RATES
A related rates problem involves two or more quantities that vary
with time and an equation that expresses some relationship
between these quantities. In general, the chain rule is the key to
find the related rates.

Ex 1: A 15 ft ladder is leaning against a house when its base
starts to slide away. By the time the base is 12 ft away from the
house, the base is moving away at the rate of 6 ft/sec.
Determine each of the following:
(a) How fast is the top of the ladder sliding down the
wall then?
(b) At what rate is the area of the triangle formed by
the ladder, wall, and ground changing then?
(c) At what rate is the angle between the ladder and
the ground changing then?
Sol: a) Using Pythagoras Theorem we get:
dx     dy     dy    x dx
x 2  y 2  152  2 x  2 y    0    
dt     dt     dt    y dt
dy                      12
x 12
       6   8
dt    y 9               9
dx / dt 6

Thus the top of the ladder is sliding down the wall at the rate of 6
ft/sec when the base of the ladder is 9 ft away from the house.
1          dA 1  dx       dy 
b) Area of triangle = A   xy         y x 
2          dt 2  dt       dt 
dA                      1
       9  6  12  8    21
2                   
x 12
dt y 9
dx / dt 6
dy / dt 8

The area of the triangle is decreasing at the
rate of 21 ft2/s when x = 12 ft.
(c) We need an equation that involve  allow us
to get d/dt
y        d 1 dy
sin    cos    
15         dt        15 dt

Isolating d/dt …          d    1    dy

dt 15cos  dt
When x=12 the value of cos  is 12/15 and from b) dy/dt = - 8 ft/s

Evaluating d/dt in that moment …
d                       1  15           2
            8    0.67
dt   cos 12 /15       15  12           3
dy / dt 8

The angle  is decreasing at the rate of 0.67 rad/s when
x = 12 feet away from the house.
Ex 2: The radius of a spherical balloon is increasing at the rate
of 2 cm/min. How fast is the volume changing when the radius is
5 cm?
Sol: See figure on the right .
The volume is increasing at constant                       r
rate dV/dt = 2 cm/min.
Differentiating both sides of V= 4/3 r3, with
respect to t.                                       V= 4/3 r3
4         dV 4  2 dr 
V   r3          3r    
3         dt 3      dt 
Evaluating for the moment when r=5
dV               4
r 5
       3  25  2  200  628.32
dt   dr          3
2
dt

Thus the volume is increasing at the rate of 628.32 cm3/min
when the radius is 5 cm.
Ex 5: A plane flying parallel to level ground at a constant rate of 500
mi/hr approaches a radar station. If the altitude of the plane is 2 mi,
how fast is the distance between the plane and the radar station
decreasing when the horizontal distance between them is 1.5 miles?
Sol: See figure on the right.
Using Pythagoras Theorem: s  x  2
2   2  2

Differentiating on both sides of the equation
with respect to t:
ds     dx ds x dx
2s       2x  
dt     dt dt s dt
When x=1.5 mi we have that s 2  x2  22 = 1.52 +22 = 6.25  s=2.5
ds               1.5
x
     (500)  300
dt dx 1.5 500 2.5
/ dt
s  2.5

The rate of change of the distance between the plane and the radar
station, in the instant when x = 1.5 mi, is decreasing at 300 mi/hr.
DERIVATIVES OF EXPONENTIAL & LOGARITHMIC FUNCTIONS
1
1) (Ln x)’ =
x
(Ln x)’ = lim Ln(x+x) – Ln x = lim Ln(1+x/x)
x0       x             x0   x
= lim Ln (1+ x/x)1/x = Ln ( lim (1+ x/x)(x/x)(1/x) .
. x0                     x0
= Ln (e1/x) = 1/x
. 1 . 1
2) (log x)’ = [ (Ln x )/(Ln 10) ]’ =
. Ln 10 x
3) (ex)’ = ex
Differentiating ln(ex) = x , we get     (1/ex) (ex)’ =1. So (ex)’ = ex
4) (ax)’ = ax Ln(a)
Using ax = e xLn(a)  (ax )’ = (e xLn(a) )’ = e xLn(a) Ln(a) = ax Ln(a)
IMPLICIT DIFFERENTIATION
There are two ways to define functions, implicitly and explicitly.
The functions we have seen so far are described by expressing
one variable explicitly in terms of another variable. For example,
y  3x  5; or y  f ( x), where f ( x)  3x  5
Some functions are implied by an equation and it can not be
readily to express y in terms of x. For example, x 2  2 ln y  xy 5
If this is the case, it may still possible to find the derivative implicitly
by considering y as a function of x and apply the chain rule to it.
Ex 1: Find dy/dx by implicit differentiation: x  6 xy  y  5
5         3       4

Sol: Treat the dependent variable y as a function of the independent
variable x and apply the chain rule, and then solve for dy/dx.
d 5
dx
 x  6 xy 3  y 4    5  5 x 4  6(1 y 3  x  3 y 2 y)  4 y 3 y  0
d
dx
Grouping y’ terms on the left side … (4 y  18 xy ) y  6 y  5 x
3           2             3   4

6 y3  5x4
Isolating y’ …         y  3
4 y  18 xy 2
Remark: The calculation of derivatives of complicated
functions involving powers, products, or quotients can often
be simplified by taking logarithms.
Ex 3: Find the derivative of the function y  (sin(7 x))
2x

Sol: Taking ln on both sides in order to take down the exponent:
ln y  ln(sin(7 x)) 2 x  2 x  ln sin(7 x)
Now applying the logarithmic differentiation, the product
rule and then the chain rule, we have:
d            d                           d
ln y   2 x  ln sin(7 x)  2  x  ln sin(7 x)
dx          dx                          dx
y                              1       d          
 2 1 ln sin(7 x)  x                sin(7 x) 
y                           sin(7 x) dx            
y                           1                d      
 2 ln sin(7 x)  x            cos(7 x)  (7 x) 
y                         sin(7 x)            dx     
y
 2  ln sin(7 x)  x  cot(7 x)  7       y=2 y lnsin(7 x)  7 x cot(7 x)
y
Thus we have:                y=2(sin(7 x)2 x ln sin(7 x)  7 x cot(7 x)
sin 5 x  cos 7 x
Ex4: Find the derivative of the function: y 
(2 x 2  7 x  1)3
sin 5 x  cos 7 x
Sol: Taking ln on both sides of the equation:                             ln y  ln
(2 x 2  7 x  1)3
Simplify by using the properties of the logarithmic functions:
ln y  ln  sin5 x   ln  cos7 x   ln(2 x2  7 x 1)3
ln y  5ln sin x   7ln  cos x   3ln(2x2  7 x 1)

Using logarithmic differentiation, we have:
y   5 d             7 d                 3        d
        sin x           cos x   2              (2 x 2  7 x  1)
y sin x  dx        cos x  dx        2 x  7 x  1 dx
y    5             7                     3
      cos x          sin x   2           (4 x  7)
y sin x          cos x              2x  7 x 1
y                        3(4 x  7)                                        3(4 x  7) 
 5cot x  7 tan x  2                  y  y 5cot x  7 tan x  2
y                      2x  7 x 1                                       2 x  7 x  1

sin x  cos x 
5      7
3(4 x  7) 
Thus we have: y  2                           5cot x  7 tan x  2
(2 x  7 x  1)3 
                   2 x  7 x 1

Ex : A sphere of ice with r = 6cm is sitting on an open box with
dimensions 12cm by 12cm by 7/4cm. If the ice is melting at rate 20
cm3/sec. (We assume that during the melting process, the shape is
always a sphere.) Find the rate of the radius at the moment when the
water is beginning to run over the edge of the box.
Replacing dV/dt by - 20 …
V=(4/3)r3  dV/dt = 4r2 dr/dt
Differentiating V respect to t :
-20 = 4r2 dr/dt
Finding the corresponding sphere radius at the moment …
The volume lost = Box Volume =(12)2(7)/4 = 252
Let’s find the new volume of the sphere after it has melting 252 cm3
New Volume = (4/3)(6)3 - 252 = 36