Part (i) y = 3 cos(3x), y = ’9 sin(3x)

Part (ii) y = ’8x sin(4x2 ), y = ’8[sin(4x2 ) + x · (8x) · cos(4x2 )]

Part(iii) y = 3 sec2 (3x), y = 3[2 sec(3x) · sec(3x) tan(3x) · 3]

y = 18 sec2 (3x) tan(3x)

Part(iv) y = ’5 csc2 (5x), y = ’10 csc(5x)[(’ csc 5x cot 5x) · 5]

y = 50 csc2 (5x) cot(5x)

Part(v) y = 2 sec(2x) tan(2x)

y = 2[(2 sec(2x) tan(2x)) · tan(2x) + sec(2x) · (2 sec2 (2x))]

y = 4 sec(2x) tan2 (2x) + 4 sec3 (2x)

Part(vi) y = ’2x csc(x2 ) cot(x2 )

y = ’2[1 · csc(x2 ) cot(x2 ) + x(’2x csc(x2 ) cot(x2 )) · cot(x2 )

3.5. HIGHER ORDER DERIVATIVES 139

+ x csc(x2 ) · (’2x csc2 (x2 ))]

= ’2 csc(x2 ) cot(x2 ) + 4x2 csc(x2 ) cot2 (x2 ) + 4x2 csc3 (x2 )

Example 3.5.2 Compute the second order derivative of each of the follow-

ing functions:

(ii) y = cosh(x2 )

(i) y = sinh(3x) (iii) y = tanh(2x)

(iv) y = coth(4x) (v) y = sech(5x) (vi) y = csch(10x)

Part (i) y = 3 cosh(3x), y = 9 sinh(3x)

Part (ii) y = 2x sinh(x2 ), y = 2 sinh(x2 ) + 2x(2x cosh x2 ) or

y = 2 sinh(x2 ) + 4x2 cosh(x2 )

Part (iii) y = 2 sech2 (2x), y = 2 · (2 sech(2x) · (’sech(2x) tanh(2x) · 2)),

y = ’8 sech2 (2x) tanh(2x)

Part (iv) y = ’4 csch2 (4x), y = ’4(2(csch(4x)) · (’csch(4x) coth(4x) · 4))

y = 32 csch2 (4x) coth(4x)

Part (v) y = ’5 sech (5x) tanh(5x)

y = ’5[’5 sech(5x) tanh(5x) · tanh(5x) + sech(5x) · sech2 (5x) · 5]

y = 25 sech(5x) tanh2 (5x) ’ 25 sech3 (5x).

Part (vi) y = ’10 csch(10x) coth(10x)

y = ’10[’10 csch(10x) coth(10x) · coth(10x)

140 CHAPTER 3. DIFFERENTIATION

+ csch(10x)(’10 csch2 (10x))]

y = 100 csch(10x) coth2 (10x) + 100 csch3 (10x)

Example 3.5.3 Compute the second order derivatives for the following func-

tions:

2

(i) y = ln(x2 ) (ii) y = ex (iii) log10 (x2 + 1)

2

(iv) y = 10x (v) y = arcsin x (vi) y = arctan x

2x 2

= = 2x’1

Part (i) y =

x2 x

’2

y = ’2x’2 = .

x2

2 2 2 2

Part (ii) y = 2xex , y = 2ex + 4x2 ex = (2 + 4x2 )ex .

(x2 + 1) · 1 ’ x · 2x

1 2x 2

·2

Part (iii) y = ,y = ,

(x2 + 1)2

ln 10 x + 1 ln 10

1 ’ x2

2

·

y=

ln 10 (x2 + 1)2

2

Part (iv) y = 10x · (ln 10) · 2x

2 2

y = 2 ln 10[10x + x · 10x ln 10 · 2x]

2

y = 10x [2 ln 10 + (2 ln 10)2 x2 ]

1

= (1 ’ x2 )’1/2

Part (v) y = √

1 ’ x2

3.5. HIGHER ORDER DERIVATIVES 141

’1

(1 ’ x2 )’3/2 (’2x)

y=

2

x

y= .

(1 ’ x2 )3/2

1

= (1 + x2 )’1

Part (vi) y = 2

1+x

’2x

y = ’1(1 + x2 )’2 · 2x =

(1 + x2 )2

Example 3.5.4 Compute the second derivatives of the following functions:

(i) y = arcsinh x (ii) y = arccosh x (iii) y = arctanh x

From Section 1.4, we recall that

√

arcsinh x = ln(x + 1 + x2 )

√

arccosh x = ln(x + x2 ’ 1) , x ≥ 1

1 1+x 1

= [ln(1 + x) ’ ln(1 ’ x)], |x| < 1.

arctanh x = ln

1’x

2 2

Then

Part (i)

1

y =√

1 + x2

d2 d

(1 + x2 )’1/2

(arcsinh x) =

2

dx dx

’1

(2x)(1 + x2 )’3/2

=

2

x

=’ .

(1 + x2 )3/2

142 CHAPTER 3. DIFFERENTIATION

Part (ii)

1

y =√ , x>1

x2 ’ 1

d2 d

(x2 ’ 1)’1/2

(arccosh x) =

2

dx dx

’1

(2x)(x2 ’ 1)’3/2

=

2

x

=’ 2 , x>1

(x ’ 1)3/2

Part (iii)

1

, |x| < 1.

y=

1 ’ x2

d2 d

(1 ’ x2 )’1

(arctanh x) =

dx dx

= (’1)(1 ’ x2 )’2 (’2x)

x

, |x| < 1.

=

(1 ’ x2 )2

Example 3.5.5 Find y for the equation x2 + y 2 = 4.

First, we ¬nd y by implicit di¬erentiation.

x

2x + 2yy = 0 ’ y , .

y

Now, we di¬erentiate again with respect to x.

y · 1 ’ xy

y=

y2

y ’ x(’x/y)

(replace y by ’x/y)

=’

y2

y 2 + x2

=’ (Why?)

y3

4

(since x2 + y 2 = 4)

=’ 3

y

3.5. HIGHER ORDER DERIVATIVES 143

Example 3.5.6 Compute y for x3 + y 3 = 4xy.

From Example 25 in the last section we found that

4y ’ 3x2

if 3y 2 ’ 4x = 0.

y= 2

3y ’ 4x

To ¬nd y , we di¬erentiate y with respect to x to get

(3y 2 ’ 4x)(4y ’ 3x2 ) ’ (4y ’ 3x2 )(6yy ’ 4)

, 3y 2 ’ 4x = 0.

y=

3y 2 ’ 4x

In order to simplify any further, we must ¬rst replace y by its computed

value. We leave this as an exercise.

Example 3.5.7 Compute f (n) (c) for the given f and c and all natural num-

bers n:

(i) f (x) = sin x, c = 0 (ii) f (x) = cos x, x = 0 (iii) f (x) = ln(x), c = 1

(iv) f (x) = ex , c = 0 (v) f (x) = sinh x, x = 0 (vi) f (x) = cosh x, x = 0

To compute the general nth derivative formula we must discover a pattern

and then generalize the pattern.

Part (i) f (x) = sin x, f (x) = cos x, f (x) = ’ sin x, f (x) = cos x, f 4 (x) =

sin x. Then the next four derivatives are repeated and so on. We get

f (4n) (n) = sin x, f (4n+1) (x) = cos x, f (4n+2) (x) = ’ sin x, f (4n+3) (x) = ’ cos x.

By evaluating these at c = 0, we get

f (4n) (0) = 0, f (4n+2) (0); f (4n+1) (0) = 1 and f (4n+3) (0) = ’1,

for n = 0, 1, 2, · · ·

Part (ii) This part is similar to Part (i) and is left as an exercise.

Part (iii) f (x) = ln x, f (x) = x’1 , f (x) = (’1)x’2 , f (3) (x) = (’1)(’2)x’3 , . . . .,

f (n) (x) = (’1)(’2) . . . (’(n ’ 1))x’n = (’1)n’1 (n ’ 1)!x’n , f (n) (1) =

(’1)n’1 (n ’ 1)!, n = 1, 2, . . .

144 CHAPTER 3. DIFFERENTIATION

Part (iv) f (x) = ex , f (x) = ex , f (x) = ex , . . . , f (n) (x) = ex , f (n) (0) =

1, n = 0, 1, 2, . . .

Part (v) f (x) = sinh x, f (x) = cosh x, f (x) = sinh x, . . . f (2n) (x) = sinh x,

f (2n+1) (x) = cosh x, f (2n) (0) = 0, f (2n+1) (0) = 1, n = 0, 1, 2, . . .

Part (vi) f (x) = cosh x, f (x) = sinh x, f (x) = cosh x, . . . , f (2n) (x) =

cosh x, f (2n+1) (x) = sinh x, f (2n) (0) = 1, f (2n+1) (0) = 0, n = 0, 1, 2, . . .

Exercises 3.5 Find the ¬rst two derivatives of each of the following func-

tions f .

1. f (t) = 4t3 ’ 3t2 + 10 2. f (x) = 4 sin(3x) + 3 cos(4x)

3. f (x) = (x2 + 1)3 4. f (x) = x2 sin(3x)

5. f (x) = e3x sin 4x 6. f (x) = e2x cos 4x

x2

8. f (x) = (x2 + 1)10

7. f (x) =

2x + 1

9. f (x) = ln(x2 + 1) 10. f (x) = log10 (x4 + 1)

11. f (x) = 3 sinh(4x) + 5 cosh(4x) 12. f (x) = tanh(3x)

14. f (x) = x2 ex

13. f (x) = x tan x

15. f (x) = arctan(3x) 16. f (x) = arcsinh (2x)

18. f (x) = (x2 + 1)100

17. f (x) = cos(nx)

Show that the given y(x) satis¬es the given equation:

19. y = A sin(4x) + B cos(4x) satis¬es y + 16y = 0

20. y = A sinh(4x) + B cosh(4x) satis¬es y ’ 16y = 0

3.5. HIGHER ORDER DERIVATIVES 145

21. y = e’x (a sin(2x) + b cos(2x)) satis¬es y ’ 2y + 2y = 0

22. y = ex (a sin(3x) + b cos(3x)) satis¬es y ’ 2y + 10y = 0

Compute the general nth derivative for each of the following:

23. f (x) = e2x 24. f (x) = sin 3x

25. f (x) = cos 4x 26. f (x) = ln(x + 1)

27. f (x) = sinh(2x) 28. f (x) = cosh(3x)

1+x

29. f (x) = (x + 1)100 30. f (x) = ln 1’x

Find y and y for the following equations:

31. x4 + y 4 = 20 32. x2 + xy + y 2 = 16

Chapter 4

Applications of Di¬erentiation

One of the important problems in the real world is optimization. This is the

problem of maximizing or minimizing a given function. Di¬erentiation plays

a key role in solving such real world problems.

4.1 Mathematical Applications

De¬nition 4.1.1 A function f with domain D is said to have an absolute

maximum at c if f (x) ¤ f (c) for all x ∈ D. The number f (c) is called the

absolute maximum of f on D. The function f is said to have a local maximum

(or relative maximum) at c if there is some open interval (a, b) containing c

and f (c) is the absolute maximum of f on (a, b).

De¬nition 4.1.2 A function f with domain D is said to have an absolute

minimum at c if f (c) ¤ f (x) for all x in D. The number f (c) is called the

absolute minimum of f on D. The number f (c) is called a local minimum

(or relative minimum) of f if there is some open interval (a, b) containing c

and f (c) is the absolute minimum of f on (a, b).

De¬nition 4.1.3 An absolute maximum or absolute minimum of f is called

an absolute extremum of f . A local maximum or minimum of f is called a

local extremum of f .

146

4.1. MATHEMATICAL APPLICATIONS 147

Theorem 4.1.1 (Extreme Value Theorem) If a function f is continuous

on a closed and bounded interval [a, b], then there exist two points, c1 and c2 ,

in [a, b] such that f (c1 ) is the absolute minimum of f on [a, b] and f (c2 ) is

the absolute maximum of f on [a, b].

Proof. Since [a, b] is a closed and bounded set and f is continuous on [a, b],

Theorem 4.1.1 follows from Theorem 2.3.14.

De¬nition 4.1.4 A function f is said to be increasing on an open interval

(a, b) if f (x1 ) < f (x2 ) for all x1 and x2 in (a, b) such that x1 < x2 . The

function f is said to be decreasing on (a, b) if f (x1 ) > f (x2 ) for all x1 and

x2 in (a, b) such that x1 < x2 . The function f is said to be non-decreasing

on (a, b) if f (x1 ) ¤ f (x2 ) for all x1 and x2 in (a, b) such that x1 < x2 . The

function f is said to be non-increasing on (a, b) if f (x1 ) ≥ f (x2 ) for all x1

and x2 in (a, b) such that x1 < x2 .

Theorem 4.1.2 Suppose that a function f is de¬ned on some open interval

(a, b) containing a number c such that f (c) exists and f (c) = 0. Then f (c)

is not a local extremum of f .

1

Proof. Suppose that f (c) = 0. Let = |f (c)|. Then > 0.

2

Since > 0 and

f (x) ’ f (c)

f (c) = lim ,

x’c

x’c

there exists some δ > 0 such that if 0 < |x ’ c| < δ, then

f (x) ’ f (c) 1

’ f (c) < |f (c)|

x’c 2

f (x) ’ f (c)

1 1

’ |f (c)| < ’ f (c) < |f (c)|

x’c

2 2

f (x) ’ f (c)

1 1

f (c) ’ |f (c)| < < f (c) + |f (c)|.

x’c

2 2

The following three numbers have the same sign, namely,

1

1

|f (c)| and f (c) + |f (c)|.

f (c), f (c) ’

2 2

148 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Since f (c) > 0 or f (c) < 0, we conclude that

f (x) ’ f (c) f (x) ’ f (c)

0< or <0

x’c x’c

for all x such that 0 < |x ’ c| < δ. Thus, if c ’ δ < x1 < c < x2 < c + δ, then

either f (x1 ) < f (c) < f (x2 ) or f (x1 ) > f (c) > f (x2 ). It follows that f (c) is

not a local extremum.

Theorem 4.1.3 If f is de¬ned on an open interval (a, b) containing c, f (c)

is a local extremum of f and f (c) exists, then f (c) = 0.

Proof. This theorem follows immediately from Theorem 4.1.2.

Theorem 4.1.4 (Rolle™s Theorem) Suppose that a function f is continuous

on a closed and bounded interval [a, b], di¬erentiable on the open interval

(a, b) and f (a) = f (b). Then there exists some c such that a < c < b and

f (c) = 0.

Proof. Since f is continuous on [a, b], there exist two numbers c1 and c2

on [a, b] such that f (c1 ) ¤ f (x) ¤ f (c2 ) for all x in [a, b]. (Extreme Value

Theorem.) If f (c1 ) = f (c2 ), then the function f has a constant value on [a, b]

1

and f (c) = 0 for c = 2 (a + b). If f (c1 ) = f (c2 ), then either f (c1 ) = f (a)

or f (c2 ) = f (a). But f (c1 ) = 0 and f (c2 ) = 0. It follows that f (c1 ) = 0 or

f (c2 ) = 0 and either c1 or c2 is between a and b. This completes the proof

of Rolle™s Theorem.

Theorem 4.1.5 (The Mean Value Theorem) Suppose that a function f is

continuous on a closed and bounded interval [a, b] and f is di¬erentiable on

the open interval (a, b). Then there exists some number c such that a < c < b

and

f (b) ’ f (a)

= f (c).

b’a

Proof. We de¬ne a function g(x) that is obtained by subtracting the line

joining (a, f, (a)) and (b, f (b)) from the function f :

f (b) ’ f (a)

g(x) = f (x) ’ (x ’ a) + f (a) .

b’a

4.1. MATHEMATICAL APPLICATIONS 149

The g is continuous on [a, b] and di¬erentiable on (a, b). Furthermore, g(a) =

g(b) = 0. By Rolle™s Theorem, there exists some number c such that a < c < b

and

0 = g (c)

f (b) ’ f (a)

= f (c) ’ .

b’a

Hence,

f (b) ’ f (a)

= f (c)

b’a

as required.

Theorem 4.1.6 (Cauchy-Mean Value Theorem) Suppose that two functions

f and g are continuous on a closed and bounded interval [a, b], di¬erentiable

on the open interval (a, b) and g (x) = 0 for all x in (a, b). Then there exists

some number c in (a, b) such that

f (b) ’ f (a) f (c)

= .

g(b) ’ g(a) g (c)

Proof. We de¬ne a new function h on [a, b] as follows:

f (b) ’ f (a)

h(x) = f (x) ’ f (a) ’ (g(x) ’ g(a)).

g(b) ’ g(a)

Then h is continuous on [a, b] and di¬erentiable on (a, b). Furthermore,

h(a) = 0 and h(b) = 0.

By Rolle™s Theorem, there exist some c in (a, b) such that h (c) = 0. Then

f (b) ’ f (a)

0 = h (c) = f (c) ’ g (c)

g(b) ’ g(a)

and, hence,

f (b) ’ f (a) f (c)

=

g(b) ’ g(c) g (c)

as required. This completes the proof of Theorem 4.1.6.

150 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Theorem 4.1.7 (L™Hospital™s Rule, 0 Form) Suppose f and g are di¬eren-

0

tiable and g (x) = 0 on an open interval (a, b) containing c (except possibly

at c). Suppose that

f (x)

= L,

lim f (x) = 0 , lim g(x) = 0 and lim

g (x)

x’c x’c x’c

where L is a real number, ∞, or ’∞. Then

f (x) f (x)

lim = lim = L.

g(x) x’c g (x)

x’c

Proof. We de¬ne f (c) = 0 and g(c) = 0. Let x ∈ (c, b). Then f and g are

continuous on [c, x], di¬erentiable on (c, x) and g (y) = 0 on (c, x). By the

Cauchy Mean Value Theorem, there exists some point y ∈ (c, x) such that

f (x) ’ f (c)

f (x) f (y)

= = .

g(x) ’ g(c)

g(x) g (y)

Then

f (x) f (y)

lim = lim = L.

g(x) y’c+ g (y)

+

x’c

Similarly, we can prove that

f (x)

lim = L.

g(x)

’

x’c

Therefore,

f (x) f (x)

lim = lim = L.

g(x) x’c g (x)

x’c

Remark 12 Theorem 4.1.7 is valid for one-sided limits as well as the two-

sided limit. This theorem is also true if c = ∞ or c = ’∞.

Theorem 4.1.8 Theorem 4.1.7 is valid for the case when

lim f (x) = ∞ or ’ ∞ and lim g(x) = ∞ or ’ ∞.

x’c x’c

Proof of Theorem 4.1.8 is omitted.

4.1. MATHEMATICAL APPLICATIONS 151

Example 4.1.1 Find each of the following limits using L™Hospital™s Rule.

sin 3x tan 2x sin x

(i) lim (ii) lim (iii) lim

sin 5x tan 3x x

x’0 x’0 x’0

1 ’ cos x

x

(iv) lim (v) lim (vi) lim x ln x

sin x x

x’0 x’0 x’0

We compute these limits as follows:

sin 3x 3 cos 3x 3

(i) lim = lim =

sin 5x x’0 5 cos 5x 5

x’0

2 sec2 x

tan 2x 2

(ii) lim = lim =

x’0 3 sec2 3x

x’0 tan 3x 3

sin x cos x

(iii) lim = lim =1

x 1

x’0 x’0

x 1

(iv) lim = lim =1

sin x x’0 cos x

x’0

1 ’ cos x sin x

(v) lim = lim =0

x 1

x’0 x’0

1

ln x x

(vi) lim x ln x = lim = lim = lim (’x) = 0.

’1

1

x’0 x’0 x’0 x’0

x2

x

Theorem 4.1.9 Suppose that two functions f and g are continuous on a

closed and bounded interval [a, b] and are di¬erentiable on the open interval

(a, b). Then the following statements are true:

(i) If f (x) > 0 for each x in (a, b), then f is increasing on (a, b).

(ii) If f (x) < 0 for each x in (a, b), then f is decreasing on (a, b).

(iii) If f (x) ≥ 0 for each x in (a, b), then f is non-decreasing on (a, b).

(iv) If f (x) ¤ 0 for each x in (a, b), then f is non-increasing on (a, b).

(v) If f (x) = 0 for each x in (a, b), then f is constant on (a, b).

152 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

(vi) If f (x) = g (x) on (a, b), then f (x) = g(x)+C, for constant C, on (a, b).

Proof.

Part (i) Suppose a < x1 < x2 < b. Then f is continuous on [x1 , x2 ] and

di¬erentiable on (x1 , x2 ). By the Mean Value Theorem, there exists some c

such that a < x1 < c < x2 < b and

f (x2 ) ’ f (x1 )

= f (c) > 0.

x2 ’ x1

Since x2 ’ x1 > 0, it follows that f (x2 ) ’ f (x1 ) > 0 and f (x2 ) > f (x1 ). By

de¬nition, f is increasing on (a, b). The proof of Parts (ii)“(v) are similar

and are left as an exercise.

Part (vi) Let F (x) = f (x) ’ g(x) for all x in [a, b]. Then F is continuous on

[a, b] and di¬erentiable on (a, b). Furthermore, F (x) = 0 on (a, b). Hence,

by Part (v), there exists some constant C such that for each x in (a, b),

F (x) = C, f (x) ’ g(x) = c, f (x) = g(x) + C.

This completes the proof of the theorem.

Theorem 4.1.10 (First Derivative Test for Extremum) Let f be continuous

on an open interval (a, b) and a < c < b.

(i) If f (x) > 0 on (a, c) and f (x) < 0 on (c, b), then f (c) is a local maxi-

mum of f on (a, b).

(ii) If f (x) < 0 on (a, c) and f (x) > 0 on (c, b), then f (c) is a local minimum

of f on (a, b).

Proof. This theorem follows immediately from Theorem 4.1.9 and its proof

is left as an exercise.

Theorem 4.1.11 (Second Derivative Test for Extremum) Suppose that f, f

and f exist on an open interval (a, b) and a < c < b. Then the following

statements are true:

(i) If f (c) = 0 and f (c) > 0, then f (c) is a local minimum of f .

(ii) If f (c) = 0 and f (c) < 0, then f (c) is a local maximum of f .

4.1. MATHEMATICAL APPLICATIONS 153

(iii) If f (c) = 0 and f (c) = 0, then f (c) may or may not be a local extremum.

Proof.

Part (i) If f (c) > 0, then by Theorem 4.1.2, there exists some δ > 0 such

that for all x in (c ’ δ, c + δ),

f (x) ’ f (c)

f (c)

= > 0.

x’c x’c

Hence, f (x) > 0 on (c, c + δ) and f (x) < 0 on (c ’ δ, c). By the ¬rst

derivative test, f (c) is a local minimum of f .

Part (ii) The proof of Part (ii) is similar to Part (i) and is left as an exercise.

Part (iii) Let f (x) = x3 and g(x) = x4 . Then

f (0) = g (0) = f (0) = g (0).

However, f has no local extremum at 0 but g has a local maximum at 0.

This completes the proof of this theorem.

De¬nition 4.1.5 (Concavity) Suppose that f is de¬ned in some open inter-

val (a, b) containing c and f (c) exists. Let

y = g(x) = f (c)(x ’ c) + f (c)

be the equation of the line tangent to the graph of f at c.

(i) If there exists δ > 0 such that f (x) > g(x) for all x in (c’δ, c+δ), x = c,

then the graph of f is said to be concave upward at c. If the graph of f is

concave upward at every c in (a, b), then it is said to be concave upward

on (a, b).

(ii) If there exists δ > 0 such that f (x) < g(x) for all x in (c’δ, c+δ), x = c,

then the graph of f is said to be concave downward at c. If the graph of

f is concave downward at every c in (a, b), then it is said to be concave

downward on (a, b).

(iii) The point (c, f (c)) is said to be a point of in¬‚ection if there exists some

δ > 0 such that either

154 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

(i) the graph of f is concave upward on (c ’ δ, c) and concave downward

on (c, c + δ), or

(ii) the graph of f is concave downward on (c ’ δ, c) and concave upward

on (c, c + δ).

Remark 13 The ¬rst derivative test, second derivative test and concavity

test are very useful in graphing functions.

Example 4.1.2 Let f (x) = x4 ’ 4x2 , ’3 ¤ x ¤ 3

(a) Locate the local extrema, and point extrema and points of in¬‚ections.

(b) Locate the intervals where the graph of f is increasing, decreasing, con-

cave up and concave down.

(c) Sketch the graph of f . Determine the absolute maximum and the abso-

lute minimum of the graph of f on [’3, 3].

Part (a)

(i) f (x) = x4 ’ 4x2 = x2 (x2 ’ 4) = 0 ’ x = 0, x = ’2, x = 2 are zeros of f .

√

√

(ii) f (x) = 4x3 ’ 8x = 0 = 4x(x2 ’ 2) = 0 ’ x = 0, x = ’ 2 and x = 2

are the critical points of f .

1 1

1

(iii) f (x) = 12x2 ’ 4 = 12 x2 ’ = 0 ’ x = ’ √ and x = √ are

3 3 3

the x-coordinates of the points of in¬‚ections of the graph of f , since f

changes sign at these points.

(iv) f (0) = 0, f (0) = ’4 ’ f (0) = 0 is a local minimum of f .

√ √ √

f (’ 2) = 0, f (’ 2) > 0 ’ f (’ 2) = ’8 is a local minimum of f .

√ √ √

f ( 2) = 0, f ( 2) > 0 ’ f ( 2) = ’8 is a local minimum of f .

1 ’11

1

(v) f (x) changes sign at x = ± √ and hence ± √ , are the points

39

3

of in¬‚ection of the graph of f .

4.1. MATHEMATICAL APPLICATIONS 155

√ √

Part (b) The function f is decreasing on (’∞, ’ 2) ∪ (0, 2) and is increasing

√ √ ’1

on (’ 2, 0) ∪ ( 2, ∞). The graph of f is concave up on ’∞, √ ∪

3

’1 1

1

√ , ∞ and is concave down on √ , √ .

3 33

(c) f (’3) = f (3) = 45 is the absolute maximum of f and is obtained at the

end points of the interval.

√ √

Also, f (’ 2) = f ( 2) = ’8 is the absolute minimum of f on [’3, 3].

We note that f (0) = 0 is a local maximum of f . The graph is sketched

with the above information.

graph

Example 4.1.3 Consider g(x) = x2 ’x2/3 , ’2 ¤ x ¤ 3. Sketch the graph of

g, locating extrema, zeros, points of in¬‚ection, intervals where f is increasing

or decreasing, and intervals where the graph of f is concave up or concave

down.

Let us compute the zeros and critical points of g.

(i) g(x) = x2/3 (x4/3 ’ 1) = 0 ’ x = 0, ’1, 1.

3/4

1 1

2

g (x) = 2x ’ x’1/3 = 2x’1/3 x4/3 ’ =0’x=± .

3 3 3

3/4

1

We note that g (0) is unde¬ned. The critical points are, 0, ± .

3

2

(ii) g (x) = 2 + x’4/3 > 0 for all x, except x = 0, where g (x) does not

9

exist.

3/4 3/4

1 1

’∞, ’

The function g is decreasing on and 0, .

3 3

3/4 3/4

1 1

’ ∪ ,∞ .

The function g is increasing on ,0

3 3

156 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

(iii) The point (0, 0) is not an in¬‚ection point, since the graph is concave up

everywhere on (’∞, 0) ∪ (0, ∞).

Exercises 4.1 Verify that each of the following Exercises 1“2 satis¬es the

hypotheses and the conclusion of the Mean Value Theorem. Determine the

value of the admissible c.

1. f (x) = x2 ’ 4x, ’2 ¤ x ¤ 2

2. g(x) = x3 ’ x2 on [’2, 2]

3. Does the Mean Value Theorem apply to y = x2/3 on [’8, 8]? If not, why

not?

4. Show that f (x) = x2 ’ x3 cannot have more than two zeros by using

Rolle™s Theorem.

5. Show that f (x) = ln x is an increasing function. (Use Mean Value The-

orem.)

6. Show that f (x) = e’x is a decreasing function.

7. How many real roots does f (x) = 12x4 ’ 14x2 + 2 have?

8. Show that if a polynomial has four zeros, then there exists some c such

that f (c) = 0.

A function f is said to satisfy a Lipschitz condition with constant M if

|f (x) ’ f (y)| ¤ M |x ’ y|

for all x and y. The number M is called a Lipschitz constant for f .

9. Show that f (x) = sin x satis¬es a Lipschitz condition. Find a Lipschitz

constant.

10. Show that g(x) = cos x satis¬es a Lipschitz condition. Find a Lipschitz

constant for g.

In each of the following exercises, sketch the graph of the given function over

the given interval. Locate local extrema, absolute extrema, intervals where

the function is increasing, decreasing, concave up or concave down. Locate

the points of in¬‚ection and determine whether the points of in¬‚ection are

oblique or not.

4.2. ANTIDIFFERENTIATION 157

x2

12. f (x) = x2 (1 ’ x)2 , [’2, 2]

11. f (x) = 2 , [’1, 1]

2x + 1

1

14. f (x) = 2x2 +

13. f (x) = |x ’ 1| + 2|x + 2|, [’4, 4] , [’1, 1]

x2

15. f (x) = sin x ’ cos x, [0, 2π] 16. f (x) = x ’ cos x, [0, 2π]

2x

18. f (x) = 2x3/5 ’ x6/5 , [’2, 2]

17. f (x) = , [’4, 4]

x2 ’ 9

2

19. f (x) = (x2 ’ 1)e’x , [’2, 2] 20. f (x) = 3 sin 2x + 4 cos 2x, [0, 2π]

Evaluate each of the following limits by using the L™Hospital™s Rule.

x + sin πx

sin 3x

22. lim

21. lim

x ’ sin πx

tan 5x x’0

x’0

ex ’ 1

x ln x

23. lim 24. lim

x’1 1 ’ x x’0 ln(x + 1)

ex ’ 1 10x ’ 1

25. lim 26. lim

x x

x’0 x’0

sin 3x 1

’ csc x

27. lim 28. lim

sinh(5x) x

x’0 x’0

(1 ’ x2 )

x + tan x

29. lim 30. lim

x’1 (1 ’ x3 )

x’0 x + sin x

4.2 Antidi¬erentiation

The process of ¬nding a function g(x) such that g(x) = f (x), for a given

f (x), is called antidi¬erentiation.

De¬nition 4.2.1 Let f and g be two continuous functions de¬ned on an

open interval (a, b). If g (x) = f (x) for each x in (a, b), then g is called an

antiderivative of f on (a, b).

158 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Theorem 4.2.1 If g1 (x) and g2 (x) are any two antiderivatives of f (x) on

(a, b), then there exists some constant C such that

g1 (x) = g2 (x) + C.

Proof. If h(x) = g1 (x) ’ g2 (x), then

h (x) = g1 (x) ’ g2 (x)

= f (x) ’ f (x)

=0

for all x in (a, b). By Theorem 4.1.9, Part (iv), there exists some constant c

such that for all x in (a, b),

C = h(x) = g1 (x) ’ g2 (x)

g2 (x) = g1 (x) + C.

De¬nition 4.2.2 If g(x) is an antiderivative of f on (a, b), then the set

{g(x)+C : C is a constant} is called a one-parameter family of antiderivatives

of f . We called this one-parameter family of antiderivatives the inde¬nite

integral of f (x) on (a, b) and write

f (x)dx = g(x) + C.

The expression “ f (x)dx” is read as “the inde¬nite integral of f (x) with

respect to x.” The function “f (x)” is called the integrand, “ ” is called the

integral sign and “x” is called the variable of integration. When dealing with

inde¬nite integrals, we often use the terms antidi¬erentiation and integration

interchangeably. By de¬nition, we observe that

d

f (x)dx = g (x) = f (x).

dx

Example 4.2.1 The following statements are true:

xn+1

1

3

x dx = x4 + c n

+ c, n = ’1

1. 2. x dx =

4 n+1

4.2. ANTIDIFFERENTIATION 159

1

dx = ln |x| + c sin x dx = ’ cos x + c

3. 4.

x

’1

5. sin(ax)dx = cos(ax) + c 6. cos x dx = sin x + c

a

1

tan x dx = ln | sec x| + c

7. cos(ax)dx = sin(ax) + c 8.

a

1

ln | sec(ax)| + c cot x dx = ln | sin x| + c

9. tan(ax)dx = 10.

a

1

ex dx = ex + c

ln | sin(ax)| + c

11. cot(ax)dx = 12.

a

1 ax

e’x dx = ’e’x + c eax dx =

13. 14. e +c

a

15. sinh xdx = cosh x + c 16. cosh x dx = sinh x + c

tanh x dx = ln | cosh x| + c

17.

coth x dx = ln | sinh x| + c

18.

1

19. sinh(ax) = cosh(ax) + c

a

1

20. cosh(ax)dx = sinh(ax) + c

a

1

ln | cosh ax| + c

21. tanh(ax)dx =

a

1

ln | sinh(ax)| + c

22. coth (ax)dx =

a

160 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

sec x dx = ln | sec x + tan x| + c

23.

csc x dx = ’ ln | csc x + cot x| + c

24.

1

ln | sec(ax) + tan(ax)| + c

25. sec(ax)dx =

a

’1

ln | csc(ax) + cot(ax)| + c

26. csc(ax)dx =

a

sec2 xdx = tan x + c

27.

1

sec2 (ax)dx =

28. tan(ax) + c

a

csc2 x dx = ’ cot x + c

29.

’1

csc2 (ax)dx =

30. cot(ax) + c

a

tan2 x dx = tan x ’ x + c

31.

cot2 x dx = ’ cot x ’ x + c

32.

1 1 sin 2x

sin2 x dx = (x ’ sin x cos x) + c = x’

33. +c

2 2 2

1 1 sin 2x

cos3 xdx =

34. (x + sin x cos x) + c = x+ +c

2 2 2

35. sec x tan x dx = sec x + c

4.2. ANTIDIFFERENTIATION 161

csc x cot x dx = ’ csc x + c

36.

Each of these inde¬nite integral formulas can be proved by di¬erentiating

the right sides of the equation. We show some details in selected cases.

Part 3. Recall that

|x|

x

d

(|x|) = = , x = 0.

|x|

dx x

Hence,

|x|

1 1

d

(ln |x| + c) = · +0 = .

|x|

dx x x

The absolute values are necessary because ln(x) is de¬ned for positive num-

bers only.

1

d

· (sec x tan x + sec2 x)

(ln | sec x + tan x|) =

Part 23.

dx sec x + tan x

sec x(tan x + sec x)

=

(sec x + tan x)

= sec x.

d

(tan x ’ x + c) = sec2 x ’ 1 = tan2 x.

Part 31.

dx

d 1

(x ’ sin x cos x) + c

Part 33.

dx 2

d x sin 2x

’

= (Trigonometric Identity)

dx 2 4

1 2 cos 2x

’

=

2 4

1

(1 ’ cos x)

=

2

= sin2 x (Trigonometric Identity)

162 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

d 1

Part 34. (x + sin x cos x) + c

dx 2

d x sin 2x

= +

dx 2 4

11

= + cos 2x

22

1

= (1 + cos 2x)

2

= cos2 x (Trigonometric Identity)

Example 4.2.2 The following statements are true:

√

1 x

√ √ dx = ’ 1 ’ x2 + c

1. dx = arcsin x + c 2.

1 ’ x2 2

1’x

√

1 1

√ √ dx = 1 + x2 + c

3. dx = arcsinh x + c 4.

1 + x2 1 + x2

√

= ln(x + 1 + x2 ) + c

√

x

1

√ √ dx = x2 ’ 1 + c

dx = arccosh x + c 6.

5.

2’1 x2 ’ 1

x √

= ln |x + x2 ’ 1| + c

1 1

7. dx = arctan x + c 8. dx = arctanh x + c

1 + x2 1 ’ x2

1+x

1

+c

= ln

1’x

2

bx

1 x

√

9. dx = arcsec x + c 10. b dx = + c, b > 0, b = 1

ln b

2’1

|x| x

All of these integration formulas can be veri¬ed by di¬erentiating the right

sides of the equations.

4.2. ANTIDIFFERENTIATION 163

Remark 14 In the following exercises, use the substitution to reduce the

integral to a familiar form and then use the integral tables if necessary.

Exercises 4.2 In each of the following, evaluate the inde¬nite integral by

using the given substitution. Use the formula:

f (g(t))g (t)dt = f (u)du, where u = g(t), du = g (t)dt.

1 1

√ √

1. dx, x = 2 sin t 2. dx, x = 2 cosh t

4 ’ x2 4 + x2

1 1

√ √ dx, x = 3 sec t

3. dx, x = 3 tan t 4.

x x2 ’ 9

9 + x2

2

xe’x dx, u = ’x2

5. 6. sin(7x + 1)dx; u = 7x + 1

sec2 (3x + 1)dx, u = 3x + 1 cos2 (2x + 1)dx, u = 2x + 1

7. 8.

x sin2 (x2 )dx, u = x2 tan2 (5x + 7)dx, u = 5x + 7

9. 10.

sec(2x ’ 3) tan(2x ’ 3)dx, u = 2x ’ 3

11. 12. cot(5x + 2)dx, u = 5x + 2

x

x(x2 + 1)10 dx, u = x2 + 1 dx, u = x2 + 1

13. 14. 2 + 1)1/3

(x

e2x ’ e’2x

1

dx, u = e2x + e’2x

dx, u = ex

15. 16.

x + e’x 2x + e’2x

e e

sin3 (2x) cos 2x dx, u = sin 2x esin 3x cos 3x dx, u = sin 3x

17. 18.

sec2 x tan x dx, u = sec x tan10 x sec2 x dx, u = tan x

19. 20.

164 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

x ln(x2 + 1) x

dx, u = ln(x2 + 1) dx, u = 4 + x2

√

21. 22.

2+1

x 4 + x2

x dx x

, u = 4 ’ x2 dx, u = 9 + x2

√

23. 24.

9 + x2

4 ’ x2

1 1

√ √

25. dx, u = 2 cosh x

dx, u = 2 sinh x 26.

x2 ’ 4

4 + x2

4.3 Linear First Order Di¬erential Equations

De¬nition 4.3.1 If p(x) and q(x) are de¬ned on some open interval, then

an equation of the form

dy

+ p(x)y = q(x)

dx

is called a linear ¬rst order di¬erential equation in the variable y.

Example 4.3.1 (Exponential Growth). A model for exponential growth is

the ¬rst order di¬erential equation

dy

= ky, k > 0, y(0) = y0 .

dx

To solve this equation we divide by y, integrate both sides with respect to x,

dy

replacing dx by dy as follows:

dx

1 dy

dx = k dx

y dx

1

dy = kx + c

y

ln |y| = kx + c

|y| = ekx+c = ec ekx

y = ±ec ekx .

4.3. LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS 165

Next, we impose the condition y(0) = y0 to get

y(0) = ±ec = y0

y = y0 ekx .

The number y0 is the value of y at x = 0. If the variable x is replaced by the

time variable t, we get

y(t) = y(0)ekt .

If k > 0, this is an exponential growth model. If k < 0, this is an example of

an exponential decay model.

Theorem 4.3.1 (Linear First Order Di¬erential Equations) If p(x) and q(x)

are continuous, then the di¬erential equation

dy

+ p(x)y = q(x) (1)

dx

has the one-parameter family of solutions

y(x) = e’ p(x)dx

p(x)dx

q(x)e dx + c .

p(x)dx

Proof. We multiply the given di¬erential equation (1) by e , which is

called the integrating factor.

p(x)dx dy p(x)dx p(x)dx

e + p(x)e y = q(x)e . (2)

dx

Since the integrating factor is never zero, the equation (2) has exactly the

same solutions as equation (1). Next, we observe that the left side of the

equation is the derivative of the product the integrating factor and y:

d p(x)dx p(x)dx

e y = q(x)e . (3)

dx

By the de¬nition of the inde¬nite integral, we express equation (3) as follows:

p(x)dx p(x)dx

e y= q(x)e dx + c. (4)

166 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Next, we multiply both sides of equation (4) by e’ p(x)dx

:

y = e’ p(x)dx

p(x)dx

q(x)e dx + c . (5)

Equation (5) gives a one-parameter family of solutions to the equation. To

pick a particular member of the family, we specify either a point on the curve,

or the slope at a point of the curve. That is,

or y (0) = y0 .

y(0) = y0

Then c is uniquely determined. This completes the proof.

Example 4.3.2 Solve the di¬erential equation

y + 4y = 10 , y(0) = 200.

Step 1. We multiply both sides by the integrating factor

4dx

= e4x

e

dy

e4x + 4e4x y = 10e4x . (6)

dx

Step 2. We observe that the left side is the derivative of the integrating factor

and y.

d

(e4x y) = 10e4x . (7)