Chapter Four
4
Applications of Derivatives
APPLICATIONS OF DERIVATIVES
1. Analysis of Functions: Increase, Decrease and Concavity:
a. Increase, Decrease and Relative Extrema:
DEFINITION: Let f be a function defined on an interval I and let x1 and x2
be any two points in I.
1. If f ( x1 )  f ( x2 ) whenever x1  x2 , then f is said to be increasing on I.
2. If f ( x1 )  f ( x2 ) whenever x1  x2 , then f is said to be decreasing on I.
3. If f ( x1 )  f ( x2 ) for all x1 and x2, then f is said to be constant on I.
y
y
y
decreasing
increasing
f(x1)
f(x1)
f(x2)
x1
x2
f(x1) < f(x2) if x1 < x2
Graph has +ve slope
(1)
constant
x
f(x1)
f(x2)
x1
x2
f(x1) > f(x2) if x1 < x2
Graph has -ve slope
(2)
x
f(x2)
x1
x2
f(x1) = f(x2) if for all x1 and x2
Graph has zero slope
(3)
Theorem 1: Suppose that f is continuous on [a,b] and differentiable on (a,b).
1. If f `(x) > 0 at each point x  (a, b) , then f is increasing on [a,b].
2. If f `(x) < 0 at each point x  (a, b) , then f is decreasing on [a,b].
3. If f `(x) = 0 at each point x  (a, b) , then f is constant on [a,b].
Note: A function f(x) has critical point like x=c if it is continues there and f`(c) =0
or f`(x) is not found.
Theorem 2: (First Derivative Test for Local Extrema)
Suppose that c is a critical point of a continuous function f, and that f is
differentiable at every point in some interval containing c except possibly
at c itself. Moving across c from left to right,
1. if f ` changes from negative to positive at c, then f has local minimum at c;
2. if f ` changes from positive to negative at c, then f has local maximum at c;
3. if f ` does not change sign at c (that is, f ` is positive on both sides of c or f `
is negative on both sides of c ), then f has no local extremum at c;
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Chapter Four
Sign of f`
Applications of Derivatives
a ------ c ++++ b
Sign of f`
a ++++ c -------- b
Local minimum
(1)
Sign of f`
a ------ c ------ b
Sign of f`
Local maximum
(2)
a ++++ c ++++
b
f has no local extremum
(3)
Theorem 3: (Second Derivative Test for Local Extrema)
Suppose f`` is continuous on an open interval that contains c;
1. If f `` (c) > 0, then f has local minimum at x = c.
2. If f `` (c) < 0, then f has local maximum at x = c.
3. If f `` (c) = 0, then the test fails. The function f may have a local
maximum, a local minimum, or neither.
Example 1: Find the intervals on which the following functions are increasing
and the intervals on which decreasing, and also locate the maximum
and the minimum.
(a) f ( x)  x 2  4 x  3
Sol.: f `( x)  2 x  4  2( x  2)
Put f `=0  2( x  2)  0
Sign of f`

x2
------ 2 ++++
decrease
increase
Local minimum
extrimum point (critical point)
Since f is continuous at x  2
 f is decreasing on (-∞,2].
And f is increasing on [2, ∞).
f (2)  2 2  4 * 2  3  1
 (2,-1) is minimum point.
Or by second derivative test:
f ``( x)  2  0 So f has minimum point at x  2
(b) f ( x)  x 3
Sol.: f `( x)  3x 2
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Applications of Derivatives
Put f `=0  3x 2  0

x  0 extremum point (critical point)
Since f is continuous at x  0
0
++++ ++++
Sign of f`
increase
increase
f has no local extremum
 f is increasing on (-∞,0].
And f is increasing on [0, ∞).
So f is increasing over entire interval (-∞,∞) thus f has no local extremum at
x0
Or by second derivative test:
f ``( x)  6 x

f ``( 0)  6(0)  0  no indication of critical point.
(c) f ( x)  3x 4  4 x3  12 x 2  2
Sol.: f `( x)  12 x3  12 x 2  24 x
Sign of x
------------------
Sign of (x+2) -∞ -------
12 x 3  12 x 2  24 x  0
12 x( x 2  x  2)  0
12 x( x  2)( x  1)  0
either 12x  0 
0
++++++ +∞
-2
+++++++++++++ +∞
+1
------------------------+++ +∞
Sign of (x-1) -∞
-2
+1
++++ 0------ ++++ +∞
-∞ --------Sign of f`
Put f `=0

-∞
decrease
x0
or
x2  0 
x  2
or
x 1  0 
x 1
increase
decrease
increase
min.
max.
min.
at x=-2
at x=-0
at x=1
since the function is continuous at x  2 , x  0 and x  1 then,
f
is decreasing on (-∞,-2] and [0,1].
and f is increasing on [-2,0] and (1, ∞).
at x  2 
 (-2,-30) is
 (0,2) is
relative min.

at x  0
f (2)  3(2) 4  4(2)3  12(2) 2  2  30
f (0)  3(0) 4  4(0)3  12(0) 2  2  2
relative max.
at x  1
 (1,-3) is

f (1)  3(1) 4  4(1)3  12(1) 2  2  3
relative min.
or by second derivative test
f ``( x)  36 x 2  24 x  24
f ``( 2)  36(2) 2  24(2)  24  72  0 so the function has min. point at x  2 .
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Applications of Derivatives
f ``(0)  36(0) 2  24(0)  24  24  0
so the function has max. point at x  0 .
f ``(1)  36(1) 2  24(1)  24  36  0
so the function has min. point at x  1.
(d) f ( x)  3x5 3  15x 2 3
5
2
Sol.: f `( x)  3 * x 2 3  15 * x 1 3
3
3
 5 x 2 3  10 x 1 3

Sign of 5(x-2) -∞
Sign of(x1/3)
Sign of f`
5( x  2)
x1 3
------------------
2
++++++ +∞
0
-∞ -------
+++++++++++++ +∞
0
2
+++++++ +∞
-∞ ++++++ ---------increase
decrease
max.
at x=-0
min.
at x=2
increase
5( x  2)
0
x1 3
Put f `=0 
either x  2
x0
or
since the function is continuous at x  0 and x  2 then,
 f is
increasing on (-∞,0] and [2,∞).
and f is decreasing on [0,2].

at x  0
 (0,0)
f (0)  3(0)5 3  15(0)2 3  0
is relative max.
at x  2

 (2,-14.287) is
f (2)  3(2) 5 3  15(2) 2 3  14.287
relative min.
or by second derivative test
f ``( x) 
10 1 3 10 4 3
10
10
10  x  1 
x  x
 13  43   43 
3
3
3x
3x
3x 
at x  0 f``(x) is not found, so this test is failed.
at x  2 f ``( 2) 
10  2  1 

  0.396  0 so the function has min. point at x  2 .
3  24 3 
b. Concavity and Inflection Points (I.P.):
DEFINITION: Concave up, Concave down
The graph of a differentiable function y=f(x) is:
 Concave up in an open interval I if f` is increasing in I.
 Concave down in an open interval I if f` is decreasing in I.
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Chapter Four
Applications of Derivatives
Theorem 1: The second derivative test for concavity
Let y=f(x) be twice-differentiable on an interval I.
1. If f `` > 0 on I, the graph of f over I is concave up.
2. If f `` < 0 on I, the graph of f over I is concave down.
DEFINITION: Point of Inflection
A point P on a curve y=f(x) is called an inflection point (I.P.) if f is
continuous there and the curve changes from concave upward to concave
downward or from concave downward to concave upward at P.
Theorem 2: Inflection points (I.P.)
Let y=f(x) be continuous and twice-differentiable at x=c then, if f ``(c)=0, the
function f has inflection point (I.P.) at x=c.
Examples: Find the intervals on which the following functions are concave up
and the intervals on which are concave down, then, if any, locate the
inflection points.
(a) f ( x)  x 2  4 x  3
Sol.: f `( x)  2 x  4
f ``( x)  2  0
Since f ``( x)  0 for all x, the function f is concave up on the interval (-∞,∞)
also f ``( x)  0 for all x, the function f dose not have I.P.
(b) f ( x)  x 3
Sol.: f `( x)  3x 2
f ``( x)  6 x
Put f ``( x)  0  6x  0  x  0  y  03  0
Sign of f``
I.P.
------ 0 ++++
Concave down
Concave up
So, f is concave down on (-∞,0),
and f is concave up on (0,∞).
(0,0) is the inflection point.
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I.P.
(c) f ( x)  sin x on [0,2]
Sign of f``
0 ------  ++++ 2
Concave down
Sol.: f `( x)  cos x
Concave up
f ``( x)   sin x
Put f ``( x)  0   sin x  0
x 

 y  sin   0
So, f is concave down on (0,),
and f is concave up on ( ,2).
(,0) is the inflection point.
(d) f ( x)  3x 4  4 x3  12 x 2  2
Sol.: f ' ( x)  12 x3  12 x 2  24 x
f " ( x)  36 x 2  24 x  24
Put f ``( x)  0

36 x 2  24 x  24  0

3x 2  2 x  2  0
Sign of f``
I.P.
I.P.
-1.22 ------0.55
++++
++++
Concave up
Concave down
Concave up
 B  B 2  4 AC
x
2A
Where A  3 , B  2 , C  2
 2  22  4 * 3 * (2)  1  7
x 

2*3
3
either x 
or
x
1  7
 1.22
3

y  16.36
1 7
 0.55
3

y  0.68
So, f is concave up on intervals (-∞,-1.22) and (0.55,∞)
and f is concave up on interval (-1.22,0.55).
It has I.P. at points (-1.22,-16.36) and (0.55,-0.68)
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Chapter Four
Applications of Derivatives
Example: If y 
x3 x 2
  6 x  8 find:
3 2
(a) Critical points.
(b) The intervals in which the function increase and decrease.
(c) Maximum and minimum values of (y).
(d) The intervals in which the function concave up and concave down.
(e) The inflection points.
Sol.: (a) Critical points:
y
x3 x 2
  6x  8
3 2
3x 2 2 x
y`

 6  x2  x  6
3
2
Put y ` 0 

( x  3)( x  2)  0
either x  3  0

x  3

y
43
2
x2  0

x2

y
2
3
x2  x  6  0
or
 The
critical points are (-3,
2
43
), (-3, ).
3
2
(b) The intervals in which the function increase and decrease.
 The function increases on
intervals: (-∞,-3]and[2,∞)
-3 +++++++++++++
+∞
2
-----------------++++++ +∞
Sign of (x-2) -∞
2 +++++++
-3
-∞ ++++++ ---------+∞
Sign of f`
Sign of(x+3) -∞ -------
 The function decrease on
decrease
increase
max.
at x=--3
interval: [-3,2]
min.
at x=2
increase
(c) Maximum and minimum values of (y).
 The function has max. value of (y) at x=-3  y=43/2
 The function has min. value of (y) at x=2 
y=2/3.
or by second derivative test.
y` x 2  x  6

y``( x)  2 x  1
y``( 3)  2(3)  1  5  0  there is max. value at x=-3.
y``( 2)  2(2)  1  5  0
 there is min. value at x=-2.
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Applications of Derivatives
(d) The intervals in which the function concave up and concave down.
y`` 2 x  1
Put y`` 0
 x
1
2

2x  1  0

y
I.P.

-∞ -----++++
Sign of f``
Concave down
133
12
∞
Concave up
1 133
).
2 12
The inflection points is ( ,
Example: If y  2  x 2 3 , find the critical points and recognize them.
2
3
Sol.: y` x 1 3 
2
3
3 x
 The derivative does not exit at x=0.
Note: ( x  0  D f ` and x  0  D f )
So there is critical point at x=0.
To recognizing the critical point, the first derivative
test must be used since the second derivative test is
-∞
Sign of f`
------ 0 ++++
decrease
increase
Local minimum
failed because that y" dose not exit at x=0 (where
2
 1 2
).
y``    x 4 3  3
 3 3
9 x4
f has min point at x = 0
Homework:
1. Find the max., min., inflection points and the intervals on which the
functions increase, decrease, concave up and concave down.
(a) y  4 x3  6 x 2  6 x  4
(b) y  x3  8
(d) y  ( x  2)2 3
(c) y  3x  ( x  2)2 5
2. Find the intervals on which f are:
(i) Increasing and decreasing.
(ii) Concave up and concave down.
and the x-coordinate of all critical and inflection points.
(a) y  x 2  5x  6
(b) y  ( x  1)3
(c) y  x 4  8x 2  16
x2
x2  2
(d) y  3x 4  4 x3
(e) y 
(g) y  3 x  2
(h) y  x1 3 ( x  2)
(f) y 
(i) y  x 4 3  x1 3
(j) y  cos x ;
[0,2]
(k) y  sin 2 2 x ;
(l) y  2 x  cot x ;
[0,]
(m) y  sin x cos x ; [0,]
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x
x 2
2
[0,]
……………………. Mathematics \1st class
∞
Chapter Four
Applications of Derivatives
(n) y  cos 2 x  2 sin x ; [0,2]
3. Show that
3
1
x  1  1  x if x > 0.
3
1
3
Hint: Show that the function f ( x)  1  x  3 x  1 is increasing on [0,∞).
4. Show that x  tan x if 0  x 

2
.
Hint: Show that the function f ( x)  tan x  x is increasing on [0,

].
2
c. Horizontal and Vertical Asymptotes:
DEFINITION:
 A line y=b is a horizontal asymptote of the graph of the function y=f(x) if:
f ( x)  b
either xlim

or
lim f ( x)  b
x
 A line x=a is a vertical asymptote of the graph of the function y=f(x) if:
either lim f ( x)   
x a 
or
lim f ( x)   
x a 
Examples: Find the asymptotes of the following curves:
1. y 
1
x 1
x=1
Sol.: (a) horizontal asymptotes:
lim f ( x)  lim
x
x
1
1x
1x
0
 lim
 lim

0
x

x

x 1
x x 1 x
11 x 1 0
1
1x
1x
0
 lim
 lim

0
x x  1
x x x  1 x
x 1  1 x
1 0
lim f ( x)  lim
x
y=0
 y  0 (x-axis) is horizontal asymptote.
(b) vertical asymptotes:
To find a, put the denominator equal zero,
x 1  0  x  1
a  1
lim f ( x)  lim
1
1
 
 
x 1 1 1
lim f ( x)  lim
1
1
 
 
x 1 1 1
x a
x a
x1
x1
x2
1
x3
 x2
1
 x  1 is vertical asymptote.
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2. y 
Applications of Derivatives
x3
1
1
x2
x2
Sol.: (a) horizontal asymptotes:
x3
x x3 x
1 3 x 1 0
 lim
 lim

1
x x  2
x x x  2 x
x 1  2 x
1 0
lim f ( x)  lim
x
lim f ( x)  lim
x
x
x3
x x3 x
1 3 x 1 0
 lim
 lim

1
x


x


x2
x x2 x
1 2 x 1 0
 y  1 is horizontal asymptote.
(b) vertical asymptotes:
To find a, put the denominator equal zero,
x  2  0  x  2
a  2
lim f ( x)  lim 
x3 23

 
x  2  2  2
lim f ( x)  lim 
x3
23

 
x  2  2  2
x a
x a 
x2
x2
 x  2 is vertical asymptote.
3. y 
8
x 4
2
Sol.: (a) horizontal asymptotes:
8
 8 x2
 8 x2
0
lim f ( x)  lim 2
 lim 2 2
 lim

0
2
2
x 
x x  4
x  x
x


1 0
x 4 x
1 4 x
8
 8 x2
 8 x2
0
lim f ( x)  lim 2
 lim 2 2
 lim

0
2
2
x
x x  4
x x
x


1 0
x 4 x
1 4 x
 y  0 is horizontal asymptote.
(b) vertical asymptotes:
To find a, put the denominator equal zero,
x 2  4  0  x  2
a  2
When a  2 lim f ( x)  lim
x2

x2

lim  f ( x)  lim 
x2
x2
8
8

 
 2
x  4 (2 )  4
2
8
8

 
 2
x  4 (2 )  4
2
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Chapter Four
Applications of Derivatives
When a  2 lim f ( x)  lim
8
8
  2
 
x  4 (2 )  4
lim f ( x)  lim
8
8
  2
 
x  4 (2 )  4
x 2

x 2
x 2 
x 2

2
2
 x  2 and x  2 are vertical asymptotes.
d. Oblique Asymptotes:
If the degree of the numerator of a rational function is one greater than
the degree of denominator, the graph has an oblique asymptote, that is, an
asymptote that is neither vertical nor horizontal.
Example: Find the asymptotes of the curve:
y
x2  3
2x  4
Sol.: To find asymptotes, oblique and otherwise, we
divide (2x-4) into (x2-3):
y 
x
1
2
2x  4 x2  3
x2  3 x
1
 1
2x  4 2
2x  4
linear
 x2  2x
2x  3
reminder
(a) Horizontal asymptotes:
 2x  4
1
x2  3
x2 x  3 x x  3 x   0
lim f ( x)  lim
 lim



x 
x 
2 x  4 x  2 x x  4 x 2  4 x  2  0
x2  3
x2 x  3 x x  3 x    0
lim f ( x)  lim
 lim


 
x  
x  
2 x  4 x 2 x x  4 x 2  4 x  2  0
So there is no horizontal asymptote.
(b) Vertical asymptotes:
To find a, put the denominator equal zero,
2x  4  0 
x2
a  2
lim f ( x)  lim
x2
x2
x2  3
22  3

 
2 x  4 2 * (2  )  4
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Chapter Four
Applications of Derivatives
lim f ( x)  lim
x2
x2
x2  3
22  3


2 x  4 2 * (2 )  4
 x  2 is vertical asymptote.
(c) Oblique asymptote:
As x→±∞, the reminder approaches zero and (the linear part is dominant)
x
2
thus f ( x)   1 .
x
2
So, the line y   1 is an oblique asymptote of the curve.
And as x→2, the reminder become large (the reminder part is dominant)
thus f ( x) 
1
2x  4
So, the line x  2 is a vertical asymptote of the curve.
Note: To find the oblique asymptote, do the following:
1. By long division, divide the equation of curve into two parts (linear part
and reminder part)
2. Put y equal the linear part, so the resulted equation represent equation of
inclined line and this line is the oblique asymptote of the curve.
Strategy for Graphing y=f(x):
1. Identify the domain of f.
2. Identify any symmetry the curve may have.
3. Find y` then find the critical points of f, and identify where the curve is
increasing and where it is decreasing.
4. Find y`` then find the points of inflection, if any occur, and determine the
concavity of the curve.
5. Identify any asymptotes.
6. Plot key points, such as intercepts and the points found in steps 3 and 4,
and sketch the curve.
x2  1
Example: Sketch the graph of y  f ( x) 
.
x
Sol.: 1. Domain and Range of f.
-Domain: x  0  Df = (-∞,∞)\{0}
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Applications of Derivatives
-Range: (put the function as x=f(y)).
y
x2  1
 xy  x 2  1
x
 x 2  xy  1  0
A  1 , B   y and C  1
x

 B  B 2  4 AC
2A
 ( y )  ( y ) 2  4(1)(1)
2(1)
y  y2  4

2
Sign of (y-2) -∞ -----------------------
 y  4  0  ( y  2)( y  2)  0
2
 R f  (,2]  [2, )
Sign of (y+2) -∞
Sign of (y-2) (y+2)
----------
+2
++++++ +∞
-2
+++++++++++++
+∞
+2
0
-2
-∞ +++++++ ------------ +++++++ +∞
2. Symmetry:
(  x) 2  1
x2  1
f (  x) 

 f ( x)
(  x)
x
 f ( x )  (
x2  1
)  f ( x)
x
So the curve has symmetry about the origin.
3. First derivative test:
x
x2  1
1
y
 x
x
x
x
 x2
1 x2  1
y` 1  2  2
x
x
Put y`=0

x2 1
1
x2 1
0
x2
( x  1)( x  1)
0
x2
 either x = 1  y= 2
or
x = -1  y= -2
but
x≠0

+1
++++++ +∞
-∞ ----------------------Sign of (x-1)
-1
-∞ ---------- +++++++++++++ +∞
Sign of (x+1)
+1
0
+++++++ -1------------ +++++++ +∞
∞
Sign of y``
rise
fall
max. at x=-1
fall
min. at x=1
So the curve rises on (-∞,-1] and [1, ∞)
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rise
Chapter Four
Applications of Derivatives
and it falls on [-1,0) and (0,1]
and has relative max. at point (-1,-2)
and has relative min. at point (1,2)
4. Second derivative test:
y``
2
x3
 y`` 0
Sign of y"
There is no inflection point because it is
-∞ - - - -  + + +
Concave down
∞
Concave up
not defined at x=0.
The curve is concave up on (0, ∞), and it is concave down on (-∞, 0)
5. Asymptotes:
 Horizontal asymptotes:
x2  1
x2 x  1 x
x 1 x    0
 lim
 lim

 
x 
x  
x 
x
x x
1
1
lim f ( x)  lim
x 
x2  1
x2 x  1 x
x 1 x   0
lim f ( x)  lim
 lim
 lim


x 
x 
x 
x 
x
x x
1
1
So there is no horizontal asymptote.
 Vertical asymptotes:
To find a put the denominator equal zero.
x=0  a=0
lim f ( x)  lim
x2  1 0  1
   
x
0
lim f ( x)  lim
x2  1 0  1
   
x
0
xa
xa
x0
x0
x  0
(y-axis)
y= 1/x
y= x+1/x
is
(1,2) min
vertical asymptote
 Oblique asymptotes: because of
that the function is a rational
(-1,-2) min
function with nominator is one
greater than the denominator.
 y  x is an oblique asymptote
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y= x
Chapter Four
Applications of Derivatives
Homework:
1. Find asymptotes of the following curves then graph them.
(a) y 
1
x2
(b) y 
x4
x5
(c) y 
x2  4
2x
(d) y 
x2  4
x 1
(e) y 
 x2  2x  4
x 1
(f) y 
x 1
x ( x  2)
(g) y 
8
2
x 4
(h) y 
4x
2
x 4
(i) y 
x2  x  6
x 1
(j) y 
x2
x2 1
(k) y 
x2  x  1
x 1
x2  2x  2
x2
2
2. Graph the following functions:
(a) y  9 x  x 2
(b) y  x3  3x 2  3
(c) y 
 x2
(d) y 
x 1
x2
(e) y  2
x 1
x2 1
(f) y 
x
(g) y  x1 3
(h) y 
8
4  x2
(i) y  4  x 2
(j) y  tan x  sin x
0  x  2
(k) y  sin x  2  x  2
(l) y  sin x  2  x  2
Note: When the function contains an absolute value, the function can be
graphed as two parts.
2. Related Rates of Changes:
Related Rate Problems Strategy:
1. Draw a picture and name the variables and constant. Use t for time.
Assume all variables are differentiable functions of t.
2. Write down the numerical information (in terms of symbols you have
chosen).
3. Write down what you are asked to find (usually a rate, expressed as a
derivative).
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Applications of Derivatives
4. Write an equation that relates the variables. You may have to combine two
or more equations to get single equation that relates the variable whose rate
you want to the variables whose rates you know.
5. Differentiate with respect to t. Then express the rate you want in terms of
the rate and variables whose values you know.
6. Evaluate. Use known values to find the unknown rate.
Example 1: How rapidly will the fluid level inside a vertical cylindrical tank if
we pump the fluid out at the rate of 3000 L/min.?
Sol.: Step 1: Draw a picture and name the variables and constant.
Assume the radius of the tank = r (constant),
height of the fluid = h (variable),
and the volume of the fluid = V (variable).
Step 2: Write down the numerical information.
dV
 1000 L/min.
dt
Step 3: Write down what you are asked to find.
dh
?
dt
Step 4: Write an equation that relates the variables.
V   .r 2 .h
Step 5: Differentiate with respect to t.
dV
dh
  .r 2 .
dt
dt
(where r is constant with respect to t)
Step 6: Evaluate.
dh dV dt
 3000
3



m/min. (where 1m3=1000 L).
2
2
dt
 .r
1000 .r
 .r 2
If we assume r=1m 
3
dh
3
3
3
 95 cm/min.
m/min=



2
2
100
dt  .r
 .(1)

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Chapter Four
Applications of Derivatives
Example 2: A hot air balloon rising straight up from a level field is tracked by a
range finder 500 ft from liftoff point. At the moment when the range finder's
elevation angle is /4, the angle is increasing at the rate of 0.14 rad/min. How fast
is the balloon rising at that moment?
Sol.: Step 1: Draw a picture and name the
variables and constant.
 = the angle in radians the range finder
makes with the ground (variable).
y = the height in feet of the balloon
(variable).
x=500 ft
x= the distance between the liftoff point and
the range finder (constant = 500 ft)
Step 2: Write down the numerical information.
d
 0.14 rad/min.
dt
when  

4
Step 3: Write down what you are asked to find.
dy
?
dt
when  

4
Step 4: Write an equation that relates the variables.
tan  
y
y

 y  500 tan 
x 500
(where x is constant with respect to t)
Step 5: Differentiate with respect to t.
dy
d
 500 sec 2  .
dt
dt
Step 6: Evaluate with  

4
and
d
dy
 0.14 to find
.
dt
dt
dy
 500( 2 ) 2 * 0.14  140
dt
( sec

4
 2)
At the moment in the question, the balloon is rising at a rate of 140 ft/min.
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Applications of Derivatives
Example 3: A police cruiser, approaching a right-angled intersection from the
north is chasing a speeding car that has turned corner and now is moving east.
When the cruiser is 0.6 mi north of intersection and the car 0.8 mi to the east, the
police determine with radar that the distance between them and the car is
increasing 20 mph. If the cruiser is moving at 60 mph at the instant of
measurement, what is the speed of the car?
Sol.: Step 1: Draw a picture and name
Cruiser
the variables and constant.
x = position of car at time t.
y = position of cruiser at time t.
s = distance between car and
cruiser at time t.
Car
Step 2: Write down the numerical
information.
x = 0.8 mi, y = 0.6 mi,
dy
ds
 60 mph and
 20 mph
dt
dt
Step 3: Write down what you are asked to find.
dx
?
dt
Step 4: Write an equation that relates the variables.
s2  x2  y2

x2  s2  y2
Step 5: Differentiate with respect to t.
2x
dx
ds
dy
 2s  2 y
dt
dt
dt

dx 1  ds
dy  1 
ds
dy 
  s  y    x2  y2
y 
dt x  dt
dt  x 
dt
dt 
Step 6: Evaluate with x = 0.8 mi, y = 0.6 mi,
find
dy
ds
 60 mph and
 20 mph to
dt
dt
dx
.
dt
dx 1  2
ds
dy 
1
  x  y2
y 
dt x 
dt
dt  0.8
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 (0.8)
2

 (0.6) 2 * 20  0.6(60)  70
……………………. Mathematics \1st class
Chapter Four
Applications of Derivatives
At the moment in the question, the car's speed is 70 mph.
Example 4: Water runs into a conical tank at the rate of 9 ft3/min. The tank
stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is
the water level rising when the water is 6 ft deep?
Sol.: Step 1: Draw a picture and name the variables and constant.
V = volume (ft3) of the water in the tank at
time t (min).
x = radius (ft) of the surface of the water at
time t.
y = depth (ft) of the surface of the water at
time t.
Step 2: Write down the numerical
information.
y = 6 ft and
dV
 9 ft3/min.
dt
Step 3: Write down what you are asked to find.
dy
?
dt
Step 4: Write an equation that relates the variables.
V 

3
r 2h 

3
x2 y
This equation involves x as well as V and y. Because no information is
given about x and dx/dt at the time in the question, we need to eliminate x.
the similar triangle give us away to express x in term of y.
x 5

y 10

x
y
2
Therefore,
V 

3
x2 y 
 y

( )2 y 
y3
3 2
12
Step 5: Differentiate with respect to t.
dV 
dy  2 dy
 *3y2
 y
dt 12
dt 4
dt
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Chapter Four
Applications of Derivatives
Step 6: Evaluate with y = 6 ft and
9

4
62 dy
dt

dV
dy
 9 ft3/min to find
.
dt
dt
dy 1
  0.32
dt 
At the moment in the question, the water level is rising at about 0.32 ft/min.
Homework:
1. When a circular plate of metal is heated in an oven its radius increases at
a rate of 0.01 cm/min. At what rate is the plate's area increasing when its
radius is 50 cm.
2. The length (l) of rectangle is decreasing at the rate of 2 cm/sec while the
width (w) is increasing at the rate of 2 cm/sec. When l=12 cm and w=5 cm,
find the rate of change of:
(a) the area
(b) the perimeter
(c) the length of the diagonal of the rectangle.
3. The commercial jets at 40 000 ft are flying at 520 mph along straight-line
courses that cross at right angles. How fast is the distance between the
planes closing when plane A is 5 mi from the intersection point and plane
B is 12 mi from the intersection point.
4. A 13-ft ladder is leaning against a house when its base is 12 ft from the
house, the base is moving at the rate of 5 ft/sec.
(a) How fast is the top of the ladder sliding
down the wall then?
(b) How fast is the area of the triangle
formed by the ladder, wall and the ground
changing then?
(c) At what rate is the angle  between the
ladder and the ground changing then?
5. Sand falls from a conveyer belt at the rate of 10 ft3/min onto a conical pile.
The radius of the base is always equal to half the pile's height. How fast is
the height growing when the pile is 5 ft high?
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Applications of Derivatives
6. Suppose that a drop of mist is a perfect sphere and that, through
condensation, the drop picks up moisture at a rate proportional to its
surface area. Show that under these circumstances the drop's radius
increases at a constant rate.
7. A balloon is rising vertically above a level,
straight road at a constant rate of 1 ft/sec. Just
when the balloon is 65 ft above the ground, a
bicycle passes under it, going 17 ft/sec. How fast
is the distance between the bicycle and balloon
increasing 3 sec later?
8. A spherical balloon is inflated with helium 0f 100 ft3/min. How fast is the
balloon's radius increasing at the instant the radius is 5 ft? How fast is the
surface area increasing?
9. A man 6 ft tall walks at the rate of 5 ft/sec towards a streetlight that is 16 ft
above the ground. At what rate is the tip of his shadow moving? At what
rate is the length of his shadow when he is 10 ft from the base of the light?
10. Two ships are steaming straight away
from a point O along routes that makes Ship B
120o angle. Ship A moves at 14 mph.
Ship B moves at 21 mph. How fast are
s

the ships moving apart when OA =5 mi
Ship A
and OB = 3mi?
11.A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and
9 ft deep at its deepest point. A cross-section is shown in the figure. If the
pool is being filled at a rate of 0.8
ft3/min, how fast is the water level
rising when the depth at the deepest
point is 5 ft?
12.Two carts, A and B, are connected
by a rope 39 ft long that passes over
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Applications of Derivatives
a pulley P (see the figure). The point Q is on the floor 12 ft directly
beneath P and between the carts. Cart A is being pulled away from Q at a
speed of 2 ft/s. How fast is cart B moving toward Q at the instant when cart
A is 5 ft from Q?
13.Coffee is poured at a uniform rate of 20 cm3/sec. into a
4 cm
shown below). If the upper and lower radii of the cup
are 4 cm and 2 cm and the height of the cup is 6cm, how
6 cm
cup whose inside shaped like a truncated cone (as
2 cm
fast will the coffee level be rising when the coffee
halfway up?
[Hint: Extend the cup downward to form a cone].
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Chapter Four
Applications of Derivatives
3. Optimization:
To optimize something means to make it as useful or effective as
possible. In the mathematical models in which we use differentiable functions to
describe things that interest us, this usually means finding where some function
has its greatest or smallest value. What is the size of the most profitable
production run? What is the best shape for an oil can? What is the stiffest beam
we can cut from a 12-in. log?
In this section we show where such functions come from and how to find
their extreme values.
Critical Points and Endpoints:
Our basic tool is the observation we made in previous section about local
maxima and minima. There we discovered that the extreme values of any
function f whatever can occur only at:
1. Interior points where f `=0,
named critical points.
2. Interior points where f ` does not exist,
3. Endpoints of the function's domain.
Strategy for Solving Max-Min Problems
1. Draw a picture. Label the parts are important in the problem.
2. Write an equation. Write an equation for the quantity whose maximum or
minimum value you want. If you can, express the quantities as a function
of single variable, say y=f(x). This may require some algebra and use of
information from the statement of the problem. Note the domain in which
the values of x are to be found.
3. Test the critical points and end points. The extreme value of f will be
found among the values f takes on at the endpoints of the domain and at
the points where f` is zero or fails to exist. List the values of f at these
points. If f has an absolute maximum or minimum on its domain, it will
appear on the list. You may have to examine the sign pattern of f ` or the
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Chapter Four
Applications of Derivatives
sign of f `` to decide whether a given value represents a maximum, a
minimum, or neither.
Example 1: Find the absolute maximum and minimum values of y  x 2 3 on the
interval -2 ≤ x ≤ 3.
Sol.: We evaluate the function at the critical points and endpoints and take the
largest and smallest of these values.
y  x2 3
y`
2 1 3
2
x  3
3
3 x
has no zeroes but is undefined at x=0. The values of the function at this one
critical point and the endpoints are:
y
3
absolute maximum;
also a local minimum
Critical points value: f(0) = 0
2/3
Endpoint values:
f(-2) = (-2)
local maximum
1/3
2
=4 ,
1
f(3) = (3)2/3 = 91/3
We conclude that the function's maximum
x
0
-3
-2
-1
value is 91/3, taken on at x=3.
0
1
2
3
absolute minimum;
also a local minimum
-1
The minimum value is 0, taken on at x=0
-2
Example 2: Find two positive numbers whose sum is 20 and whose product is
as large as possible.
Sol.: assume the first number is x
So, the second number is
20-x.
 f ( x)  x(20  x) ;
0  x  20 ( Positive )
 20 x  x
y
max.
100
2
y=x(20-x)
Critical points can be found from first derivative:
f `( x)  20  2 x
Put f `(x) = 0  20  2x  0  x 
50
20
 10
2
x
0
0
10
20
- Critical point value:
f (10)  10(20  10)  10 *10  100
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-
Applications of Derivatives
f (0)  0(20  0)  0 ;
Endpoints values:
f (20)  20(20  20)  0
We conclude that the max. value is f (10)  100
So the first number is 10
and the second number is 20-10=10
Example 3: A rectangle is to be inscribed in a semicircle of radius 2. What is the
largest area that the rectangle can have and what are the dimensions?
Sol.: Step 1: Draw a picture:

the length of rectangle = 2x;
and the height
= y.
Step 2: Write an equation:
Area  A( x)  2 x. y ; where y  r 2  x 2
 4  x2
0 xr
 A( x)  2 x 4  x 2 ;
positive
0 x2
Step 3: Test the critical points and end points:
dA
1
 2 x(4  x 2 ) 1 2  (2 x)  2 4  x 2
dx
2


 2x2
4 x
2
 2 4  x2
 2 x 2  2(4  x 2 )
4  x2

dA
0 
 Put
dx
 2x2  8  2x2
4  x2
8  4x2
4 x
2

8  4x2
4  x2
 0  8  4x2  0  x2 
8
2
4
x   2
We should neglect the negative root because it is out of the domain.
x  2
 And
dA
is not defined at
dx
4  x2  0  4  x 2  0
 x  2
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We also should neglect the negative root
 x  2 (it is also end point)
- Critical point value: A( 2 )  2 2 4  ( 2 )2  2 2 2  4
-
Endpoints values: A(0)  2 * 0 4  (0) 2  0 ;
A(2)  2 * 2 4  (2) 2  0

The max. area is 4 square units when the rectangle has
Length  2x  2 2 unit length;
and height  4  ( 2 )2  4  2  2 unit length.
Example 4: An open-top-box is to be made by cutting small congruent squares
from the corners of a 12-in-by 12-in sheet of tin and bending up the
sides. How large should the squares cut from the corners be to make
the box hold as much as possible?
Sol.: Step 1: Draw a picture:
Step 2: Write an equation:
V ( x)  x(12  2 x) 2
 144 x  48 x 2  4 x 3 ;
0 x
positive
12
2
0 x6
Step 3: Test the critical points and end points:
dV
 144  96 x  12 x 2
dx
 12(12  8x  x 2 )  12(2  x)(6  x)
Put
dV
 0  12(2  x)(6  x)  0
dx
either x  2 or x  6 (endpoint)
- Critical point value:
V (2)  2(12  2 * 2) 2  128 in 3
-
Endpoints values: V (0)  0(12  2 * 0)2  0 ;
V (6)  2(12  2 * 6) 2  0
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Applications of Derivatives
The max. volume is 128 in3. The cut out squares should be 2 in on a side.
Example 5: You have been asked to design a 1-liter can shaped like a right
circular cylinder. What dimensions will use the least material?
Solution: Volume of can: If r and h are measured in
centimeters, then the volume of the can in cubic centimeters is
r 2h = 1000.
1 liter = 1000 cm3
Surface area of can: A = 2r 2 + 2rh
Circular
ends
Circular
wall
How can we interpret the phrase “least material”? First, it is
customary to ignore the thickness of the material and the waste in
manufacturing. Then we ask for dimensions r and h that make the total surface
area as small as possible while satisfying the constraint r 2h = 1000.
To express the surface area as a function of one variable, we solve for one
of the variables in r 2h = 1000 and substitute that expression into the surface
area formula. Solving for h is easier:
h
1000
r 2
Thus
A = 2r 2 + 2rh
 1000 
 2r 2  2r  2 
 r 
 2r 2 
2000
r
Our goal is to find a value of r > 0
that minimizes the value of A. Figure below suggests that such a value exists .
Notice from the graph that for small r (a tall thin container, like a piece of pipe),
the term 2000/r dominates and A is large. For large r (a short wide container, like
a pizza pan), the term 2r2 dominates and A again is large.
Since A is differentiable on r > 0, an interval with no endpoints, it can have a
minimum value only where its first derivative is zero.
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dA
2000
 4r  2
dr
r
0  4r 
2000
r2
4r 3  2000
r3
500

(set dA/dr=0)
(Multibly by r2)
 5.24 cm
The second derivative:
d2A
4000
 4  3
2
dr
r
is positive throughout the domain of A. The graph is therefore everywhere
500
concave up and the value of A at r  3
an absolute minimum.

The corresponding value of h (after a little algebra) is
h
1000
500
 23
 2r.
2
r

The 1-L can that uses the least material has height equal to the diameter, here
with r  5.24 cm and h  10.48 cm.
Example 6: A drilling rig 12 mi offshore is to be connected by pipe to a refinery
onshore, 20 mi straight down the coast from the rig. If underwater pipe costs
$500,000 per mile and land based pipe costs $300,000 per mile, what combination
of the two will give the least expensive connection?
Solution: We try a few possibilities to get a feel for the problem:
(a) Smallest amount of underwater pipe
12
20
Underwater pipe is more expensive, so we use as little as we can. We run
straight to shore 12 mi and use land pipe for 20 mi to the refinery.
Dollar cost = 12(500,000) + 20(300,000)
= 12,000,000
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Applications of Derivatives
(b) All pipe underwater (most direct route)
We go straight to the refinery underwater.
Dollar cost = 500( 144  400 )
 11,661,900
This is less expensive than plan (a).
Now we introduce the length x of underwater pipe and the length y of landbased pipe as variables. The right angle opposite the rig is the key to
expressing the relationship between x and y, for the Pythagorean theorem
gives
x 2  12 2  (20  y ) 2
x  144  (20  y ) 2
Only the positive root has meaning in this model.
The dollar cost of the pipeline is
c = 500,000x + 300,000y
To express c as a function of a single variable, we can substitute for x,
c( y )  500,000 144  (20  y ) 2  300, y
Our goal now is to find the minimum value of c(y) on the interval 0  y  20 .
The first derivative of c(y) with respect to y according to the Chain Rule is
1 2(20  y )( 1)
c`( y )  500,000. .
 300,000
2 144  (20  y ) 2
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 500,000.
20  y
144  (20  y ) 2
 300,000
Setting c` equal to zero gives
500,000(20  y )  300,000 144  (20  y ) 2
5
(20  y )  144  (20  y ) 2
3
25
(20  y ) 2  144  (20  y ) 2
9
16
(20  y ) 2  144
9
3
(20  y )   .12  9
4
y  20  9
y  11 or
y  29
Only y  11 lies in the interval of interest. The values of c at this one critical
point and at the endpoints are
c(11) = 10,800,000
c(0) = 11,661,900
c(20) = 12,000,000
The least expensive connection costs $10,800,000, and we achieve it by running
the line underwater to the point on shore 11 mi from the refinery.
Example 7: Suppose a manufacturer can sell x-items a week for a revenue of r(x)
= 200 x-0.01x2 cents, and it costs c(x)=50x+20000 cents to make x-items.
Is there a most profitable number of items to make each week? If so,
what is it?
Sol.: profit = revenue – cost
p(x) = r(x) - c(x)
= (200 x-0.01x2) – (50x+20000)
=200 x-0.01x2-50x-20000
=150 x-0.01x2-20000
To find the critical point put

dp
 150  0.02 x  0
dx

dp
0
dx
x
150
 7500 items
0.02
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2
and d p2  0.02  0 for all values of x.
dx
 The
graph is concave down, so the critical point x=7500 is the location of
an absolute max.
To answer the question, then, there is a production level for max. profit,
and that level is x=7500 item per week.
Home work:
1. The sum of two non-negative numbers is 20. Find the numbers:
(a) If the sum of their squares is to be as large as possible.
(b) If one number plus to square root of the other is to be as large as possible.
2. What is the largest possible area of a right triangle whose hypotenuse is 5
cm long?
3. What is the smallest perimeter possible for a rectangle whose area is 16
in2?
4. You are planning to close off a corner of the first quadrant with a line
segment 20 units long running from (a,0) to (0,b). Show that the area of the
triangle enclosed by the segment is largest when a=b.
5. A rectangle plot of farmland will be bounded on one side by a river and
on the other three sides by a single-strand electric fence. With 800 m of
wire at your disposal, what is the largest area you can enclose?
6. A 216-m2 rectangular pea patch is to be enclosed by a fence
y and divided
into two equal parts by another fence parallel to one of the
sides. What
8
y=x^2
dimensions for the outer rectangle will require the smallest7 total length of
6
fence? How much fence will be needed?
5
4
7. Show that the value of x minimizes the square
3
of the distance, and hence the distance,
2
(x,x^2)
1
between point (x, x2) and (2,-1/2) in figure
d
x
0
below is a solution of the equation x 
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1
.
x 1
-3
-2
-1
0
1
2(2,-1/2)3
-1
2
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……………………. Mathematics \1st class
Chapter Four
Applications of Derivatives
8. You are planning to makes an open top rectangular box from an (8*15) in2
piece of cardboard by cutting squares from the corners and folding up the
sides. What are the dimensions of the box of largest
volume you can make this way?
9. Find the volume of the largest right circular cone that
can be inscribed in a sphere of radius 3?
10.What are the dimensions of the lightest open-top right
cylindrical can be holed a volume of 1000 cm3?
11.Show that the right circular cylinder of greatest volume
that can be inscribed in a right circular cone has volume
that is 4/9 the volume of the cone.
12. Find the dimensions of the largest cone can be inscribed
in the cone whose radius is 6in and height is 10in as
10"
shown in figure nearby.
6"
13. A trapezoid is inscribed in a semicircle of radius 2 so that one side is
along the diameter (as shown below). Find the maximum possible area
for trapezoid.

[Hint: Express the area of the trapezoid in term of ].

-2
0
2
14.A box shaped wire-frame consists of two identical
wire squares whose vertices are connected by four
straight wires of equal lengths. If the frame of length
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x
x
Chapter Four
Applications of Derivatives
(L=96 cm), what should the dimensions be to obtain a box of greatest
volume.
15.A steel pipe is being carried down a hallway 9 ft
wide. At the end of the hall there is a right-angled
turn into a narrower hallway 6 ft wide. What is the
length of the longest pipe that can be carried
horizontally around the corner?
16.The trough in figure nearby is to be made to the dimensions shown. Only
the angle can be varied. What value of will maximize the trough’s
volume?
17.A right triangle whose hypotenuse
3 long is revolved about one of its
legs to generate a right circular cone. Find the
radius, height, and volume of the cone of greatest
volume that can be made this way.
18.Compare the answers to the following two construction
Circumference = x
problems .
a . A rectangular sheet of perimeter 36 cm and
dimensions x cm by y cm to be rolled into a
x
y
y
cylinder as shown in part (a) of the figure.
(a)
What values of x and y give the largest volume?
b . The same sheet is to be revolved about one of
the sides of length y to sweep out the cylinder
as shown in part (b) of the figure. What values
of x and y give the largest volume?
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Chapter Four
Applications of Derivatives
4. Rolle’s Theorem
If y=f(x) is continuous on the closed interval [a,b] and differentiable on the
open interval (a,b) and also f(a) = f(b) and may be equal zero, then there is at least
one number c in (a,b) at which
f `(c)=0.
Example: Does Rolle's Theorem be applicable on the following functions. If so,
find the value or values of c.
1. y  2 x  x 2 ;
[0, 2]
Sol.: 1. The function is continuous on [0, 2].
2. y` 2  2 x is differentiable on (0,2).
3. f (0)  2 * 0  02  0 and f (2)  2 * 2  22  0 o.k.

Rolle's Theorem is applicable on this function on [0, 2].
To find the value of c: Put y`=0
 2  2x  0  2x  2  x 
2
1
2
c  1
2. y 
x3
 3x;
3
[-3, 3]
Sol.: 1. The function is continuous on [-3, 3].
2. y` x 2  3 is differentiable on (-3, 3).
3. f (3) 

(3) 3
 27
(3) 3
27
 3(3) 
 9  0 and f (3) 
 3(3) 
 9  0 o.k.
3
3
3
3
Rolle's Theorem is applicable on this function on [-3, 3].
To find the values of c: Put y`=0
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Applications of Derivatives
 x2  3  0  x2  3  x   3
 c1   3 and c2  3 .
Homework: Show if that the Rolle's Theorem is applicable on the following
functions.
[0,  ]
2. f ( x)  1  x ;
x2  1
;
x
[2, 3]
4. f ( x) 
5. f ( x)  x 3  1 ;
[1,1]
6. f ( x)  x 2  4 x  3 ;
1. f ( x)  sin x ;
3. f ( x) 
[3, 3]
x2 1
 ;
3 x
[2, 2]
[1, 3]
(b,f(b))
Finding Solution of Equations:
Corollary1: Suppose that:
1. f is continuous on [a,b], and differentiable on
a
(a,b).
b
2. f(a) and f(b) have opposite signs.
3. f `≠0 between a and b.
(a,f(a))
Then f has exactly one zero between a and b.
Example1: Show that the equation x 3  3x  1  0 has
exactly one real solution on the interval [-1,1].
Sol.: Let y  f ( x)  x 3  3x  1
Then the derivative f `( x)  3x 2  3 , so
1. f is continuous on [-1,1], and differentiable on
(-1,1).
2. f (a)  f (1)  (1)3  3(1)  1  3  0 (negative)
f (b)  f (1)  (1) 3  3(1)  1  5  0
(positive)
Thus f(-1) and f(1) have opposite signs.
3. f `( x)  3x 2  3 is never zero (because it is the sum of two positive
numbers).
So corollary1 is applicable on this function, so the above equation has
exactly one real solution on (-1, 1).
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Applications of Derivatives
Example 2: Show that the equation x 4  3x  1  0 has exactly one real root
between a=-2 and b=-1.
Sol.: Let y  f ( x)  x 4  3x  1
Then the derivative f `( x)  4 x 3  3 , so
3. f is continuous on [-2,-1], and differentiable on (-2,-1).
4. f (a)  f (2)  (2) 4  3(2)  1  11  0 (positive)
f (b)  f (1)  (1) 4  3(1)  1  1  0 (negative)
Thus f(-2) and f(-1) have opposite signs.
3. when f `( x)  4 x 3  3  0  x 3 
3
3
 x3
 0.91  (2,1)
4
4
Thus f`(x) ≠0 on (-1,-2)
So corollary1 is applicable on this function so the above equation has
exactly one real root on (-2,-1).
Homework: Show that the following equations have only one real root on the
corresponding intervals.
1. x 3  2 x  2  0 ;
2
x
3. x   0 ;
[-1, 1]
2.  x 3  3x  1  0 ;
[0, 1]
[1, 3]
4. 2x  cos x  0 ;
[-]
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Chapter Four
Applications of Derivatives
5. The Mean Value Theorem (M. V. T.):
Suppose y=f(x) is continuous on the closed
interval [a,b] and differentiable on the open interval
(a,b), Then there is at least one point c in (a,b) at which
f (b)  f (a)
 f `(c)
ba
Examples: Does the M. V. T. be applicable on the following functions. If so find
the value or values of c.
1. f ( x)  x  2 sin x ;
0  x  2
Sol.: 1. f ( x)  x  2 sin x
is continuous on [0, 2].
2. f `( x)  1  2 cos x is differentiable on (0, 2).

The M. V. T. is applicable on [0, 2].
To find c:
f `(c) 
f (b)  f (a )
ba
where f (b)  f (2 )  2  2 sin 2  2  0  2
f (a)  f (0)  0  2 sin 0  0  0  0
f `(c)  1  2 cos c , thus:
and
1  2 cos c 
c  
 c1 
n
;
2

2
2  0
2  0
 1  2 cos c  1
 2 cos c  0  cos c  0
n  1, 3, 5..
3
2
and c2 
2. f ( x)  x 2 3 ;
on the interval [0, 2].
[-8, 8]
Sol.: 1. f ( x)  x 2 3  3 x 2 is continuous on [-8, 8].
2
3
2. f `( x)  x 1 3 

2
3
3 x
is not differentiable x  0  (8,8) .
The M. V. T. is not applicable on [-8, 8].
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Applications of Derivatives
3. f ( x)  x 2 3 ;
[0, 8]
Sol.: 1. f ( x)  x 2 3  3 x 2 is continuous on [0, 8].
2
3
2. f `( x)  x 1 3 
2
is not differentiable x  0  (0,8)
3
3 x
So it is differentiable on (0,8)

The M. V. T. is applicable on [0, 8].
To find c:
f `(c) 
f (b)  f (a )
ba
where f (a)  f (0)  02 3  0
f (b)  f (8)  82 3  4
and
f `(c) 
2
3
3 c
, thus:
40
2
1
 3 

3
3 c 2
3 c 80
2
3

3
4
64
4
c
 2.3704
 c  
3
27
3
Note: If f`(x) is continuous on [a,b], the Max.-Min. Theorem for continuous
functions tells us that f` has absolute maximum value (max.f`) and absolute
minimum value (min.f`) on the interval, the equation:
f `(c) 
f (b)  f (a )
ba
gives us the inequality:
f `
f (b)  f (a)
 max . f `
ba
Example: Estimate f(1) if f `( x) 
1
and f(0)=2.
5  x2
min .
Sol.: a=0  f(a)=f(0)=2
b=1  f(b)=f(1)=?
min .
f `
f (b)  f (a)
 max . f `
ba
1
f (1)  2
1


2
50
1 0
5  12
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Applications of Derivatives
1
1
 f (1)  2 
5
4
0.2  2  f (1)  0.25  2
2.2  f (1)  2.25
Corollary2: If f` (x) = 0 for all x in an interval (a, b), then
f `(x)=0
f (x)=C
f(x)=C, for all x  (a, b), where C is a constant.
C
Corollary3: If f`(x) =g`(x) for all in an interval (a, b), then f-g
is constant on (a, b); that is f(x) = g(x)+C, where
C is a constant.
Homework:
1. Show the following equations have exactly one solution in the given interval:
x2 + 3x + 1 = 0
-2< x < -1
2. Find the value or values of c that satisfy the M.V.T. for the following
functions and intervals:
a. f ( x)  x 2  2 x  1
0  x 1
b. f ( x)  x  1
c. f ( x)  x 2 3
0  x 1
d. f ( x)  x 
1
x
1 x  3
1
x2
2
3. By applying Mean Value Theorem, Show that for any number a and b
sin b  sin a  b  a
4. By applying the inequality
min
f 
f (b)  f (a )
 max f
ba
Estimate f(0.1) when
a. f `( x) 
1
for 0  x  0.1 and that f(0)=1.
1  x cos x
4
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Chapter Four
b. f `( x) 
Applications of Derivatives
1
for 0  x  0.1 and that f(0)=2.
1  x4
5. Suppose that f(0)=3 and that f `(x) = 0 for all x. Use the Mean Value Theorem to
show that f(x) must be 3 for all x.
6. Suppose that f `(x) = 2 and that f `(0) = 5. Use the Mean Value Theorem to show
that f(x)=2x + 5 at every value of x.
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Chapter Four
Applications of Derivatives
6. L'Hopital's Rule:
Suppose that f(a) = g(a) = 0, and f'(a) and g'(a) exist, and that g'(a)  0, then:
lim
xa
f ( x)
f ' ( x) f ' (a)
 lim

g ( x) x  a g ' ( x) g ' (a )
Examples: Evaluate the following:
1. lim
x 0
3x  sin x 3(0)  sin 0 0


indeterminate quantity
x
0
0
= lim
x 0
2. lim
x 0
3  cos x
 3  cos 0  3  1  2
1
x  sin x 0

x3
0
= lim
x 0
0
1  cos x 0
 (still
, then apply the rule again)
2
0
3x
0
= lim
x 0
 ( sin x) 0
0
 (still
, then apply the rule again)
6x
0
0
= lim
x 0
cos x 1
 (A different result  then stop)
6
6
1 x 1 
3. lim
x 0
x2
x
2  0 (indeterminate quantity)
0
1
1
1
(1  x) 2  0 
2  0 (still 0 , then apply the rule again)
= lim 2
x 0
0
2x
0
3
3
1
1
 (1  x) 2  0  (1  0) 2  0
1
 4

= lim 4
x 0
2
2
8
4. lim
x 0
1  cos x 0
 (indeterminate quantity)
0
x  x2
= lim
x 0
sin x 0
  0 (If we continue to differentiate in an attempt to apply
1 2x 1
L'Hopital's rule once more, we get:)
= lim
x 0
1  cos x
sin x
cos x 1
 lim
 lim
 (which is wrong)
2
x 0 1  2 x
x 0
2
2
xx
Note: If we reach a point where one of the derivatives is zero and the other is
not, then the limit in question is either zero, or infinity as in the next example:
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Chapter Four
5. lim
x 0

Applications of Derivatives
sin x 0

x2
0
cos x 1
  
2x
0
= lim
x 0

Note: the form of

and  x 0  L'Hopital's rule applies to the indeterminate


0
as well as , if f(x) and g(x) both approaches infinity as x approaches a,
0

form
then:
lim
xa
f ( x) f ' (a )

g ( x) g ' (a )
6. lim
x


in the notation x  a (a may be either finite or infinite)
tan x


1  tan x 
2
= lim
x

sec 2 x
1
sec 2 x
2
Note: The form 0   can sometimes be handled by using algebra to get

0
or
0

instead.
 1
1
7. lim 
     
x 0 sin x
x

= lim
x 0
x  sin x 0

x sin x
0
= lim
1  cos x
sin x
sin x
0
 lim
 lim
 0
x

0
x

0
x cos x  sin x
 x sin x  cos x  cos x
2 cos x  x sin x 2
x 0
8. lim (1  tan x) sec 2 x  0  
x
4
= lim
x
4
(1  tan x) 0

cos 2 x
0
 
2
 sec 2 x
 2
2

 1
= lim
 2(1) 2
x   2 sin 2 x
4
csc x  cot x     
9. lim
x 0
1
cos x 
 1  cos x  0

  lim 

 sin x sin x  x 0  sin x  0

= lim

x 0
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Chapter Four
Applications of Derivatives

= lim

x 0
sin x  0
 0
 cos x  1

10. lim

x 0
1
1
 2 
2
 sin x x 
 x 2  sin 2 x 
0
= lim  2 2  
x 0
 x sin x  0
Hint: If we use L'Hopital's rule the problem will be complicated more
 x 2  sin 2 x 
 x 2  sin 2 x 
1
x2
x2




= lim 
2
  sin 2 x  x 2  lim
  lim
x 0
x 0 
x2
x4



 x 0 sin x
=1
 x 2  sin 2 x  0
   Use L'Hopital's rule
x4

 0
= lim 
x 0
2 x  2 cos x sin
4x3


= lim

x 0
x
 2 x  sin 2 x  0
  lim 
  (use L'Hopital's rule again)
4x3
 x 0 
 0
2  2 cos 2 x 
  2(2 sin 2 x) 
 4 sin 2 x 
  lim 
  lim 

2
x

0
x

0
24 x
 12 x



 24 x 

= lim

x 0
 8 cos 2 x  8 1
 lim 


x 0
 24  24 3
11. lim
x 
x  2 x2 

3x 2  5 x 
= lim
x 
12. lim
x 
1  4x
4 4 2
 lim


x


6x  5
6
6
3
3x 2  x  5
6x 1
6 3
 lim
 lim

2
5 x  6 x  3 x  10 x  6 x  10 5
Homework:
1. Find the limit of the following:
a. lim
x 1
x3  1
4 x3  x  3
b. lim
x 
6x  5
c. lim
x  3x  8
e. lim
x
2
sin t 2
d. lim
t 0
t
2x  
cos x
University of Kufa\Civil Eng. ………………
5 x 2  3x
7 x2  1
f. lim
x0
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sin 5 x
x
……………………. Mathematics \1st class
Chapter Four
Applications of Derivatives

g. lim   x  tan x
x
i. lim
x 0
2
2
h. lim
x0

x1  cos x 
x  sin x
1
10sin x  x 
x3

j. lim

x 0
1
1
 
 sin x x 
1 
k. lim   
x 0 x
x

2. Let f(x) =
g(x) =
Show that lim
x 0
x+2
when x  0
0
when x = 0
x+1
when x  0
0
when x = 0
f ' ( x)
1
g ' ( x)
but lim
x 0
f ( x)
2
g ( x)
Does not this contradict L'Hopital's rule? Explain? (Note 3)
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