TEACH YOURSELF VISUALLY - CALCULUS

4Derivatives by Rule chapter

DERIVATIVE OF QUOTIENT OF RADICAL FUNCTIONS 1
Find f '(x) for f ^xh = 2x - 5.
=^2x -5h 2
3 3x + 1
1
1 Rewrite each radical expression as a power.

f ^xh

^3x +1h3

2 Find f '(x) using the d d f n = f l$ g -gl$ f
Quotient Rule. dx g 2

Note: You found each _gi
derivative on the top by
using the General Power < d ^2x -5h 1 F^3x +1h 1 -< d ^3x +1h 1 F^2x -5h 1
Rule. dx 2 3 dx 3 2
2
f l ^ xh = 1

<^3x +1h3 F

1 -5h- 1 1 1 +1h- 2 1
$ $ $ $=2^2x 2 2 ^3x +1h 3 - 3 ^3x 3 3
^2x -5h 2

2

^3x +1h 3

^2x -5h- 1 ^3x +1h 1 -^3x +1h- 2 ^2x -5h 1
2 3 3 2
2
=

^3x +1h 3

87

The Quotient ^2x -5h- 1 ^3x +1h- 2 8^3x +1h -^2x -5hB
Rule (continued) 2 3

3 Next take out the common factor and then =2
simplify the resulting expression.
^3x +1h 3
4 Last, rewrite the denominator in
radical form. =^2x -5h- 1 ^3x +1h- 2 ^ x +6h
2 3
2

^3x +1h 3

= x +6 4

1

^2x -5h 2^3x +1h 3

f l^xh = x +6
2x -5 3 ^3x +1h4

f l^xh = x +6

2x -5^3x +1h 3 3x +1

88

Rolle’s Theorem and the 4Derivatives by Rule chapter
Mean Value Theorem

This section covers two important theorems that relate continuity and differentiability:
Rolle’s Theorem and the Mean Value Theorem.

Rolle’s Theorem

Let f be a function satisfying the following 3 conditions:
1 f is continuous on the closed interval [a,b]
2 f is differentiable on the open interval (a,b)
3 f(a) = f(b)

Then, there exists at least one number c in (a,b) for which f'(c) = 0. f(x)
f(x)

ab c x
x a b

c f(x)
f(x)

c2 x x
b a cb
a c1
c is any number in (a, b)

89

Rolle’s Theorem and the Mean
Value Theorem (continued)

For the function f(x) = x2 – 8x + 19, find the value of c in the open interval (2, 6) that is mentioned in
Rolle’s Theorem.

1 To make use of Rolle’s Theorem, Condition #1: Since f(x) = x2 – 8x + 19 is a
you must first show that f(x) polynomial, it is everywhere continuous —
satisfies all 3 conditions mentioned so it is continuous on [2, 6].
in the theorem:
Condition #2: Since f(x) = x2 – 8x + 19 is a
polynomial, it is everywhere differentiable —
so it is differentiable on (2, 6).

Condition #3: After finding the values for f(2)
and f(6), you can see that f(2) = 7 = f(6).

2 Since f(x) = x2 – 8x + 19 satisfies the 3 conditions listed in the f(x) = x2 –8x + 19
theorem, you can apply the conclusion: f'(x) = 2x – 8

There is at least one number c in (2, 6) for which f'(c) = 0.

Find f'(x).

3 Substitute c for x, set f '(c) = 0, and finish by f'(c) = 2c – 8
solving for c. 0 = 2c – 8
4=c

4 is in the open interval (2, 6) and
f'(4) = 0.

4 At the right is the graph of f(x) = x2 – 8x + 19, showing the f(x)
f(x) = x2 − 8x + 19
horizontal tangent at x = 4 (where f'(x) = 0 ) in the interval (2, 6).

x
24 6
90

4Derivatives by Rule chapter

The Mean Value Theorem

Let f be a function which satisfies the following two conditions:
1 f is continuous on the closed interval [a,b]
2 f is differentiable on the open interval (a,b)

f(x)

Then there exists at least one number c in (a,b) for which

f l ^ch = f ^bh -f ^ah .
b -a
(b, f(b))
(a, f(a))

(c, f(c))

ac x

b

Note that the term f'(c) is just the slope of the tangent to the graph of f(x) at the point with coordinates
(c, f(c)).

The other term on the right f ^bh -f ^ah is the slope of the secant line containing the points (a, f(a) ) and
b -a

(b, f(b) ).

For some number c in (a,b), the tangent line and the secant line have equal slope.

For f(x) = x3 – x2 – 2x, find the value of c in the interval (–1,1), which is mentioned in the Mean Value
Theorem.

1 To make use of the Mean Value Theorem, you must first show that f(x) satisfies both of the conditions
mentioned in the theorem

Condition #1: Since f(x) = x3 – x2 – 2x is a polynomial, it is everywhere continuous, so it is
continuous on (–1,1).

Condition #2: Since f(x) = x3 – x2 – 2x is a polynomial, it is everywhere differentiable, so it is
differentiable on (–1,1).

91

Rolle’s Theorem and the Mean f(x) = x3 – x2 – 2x
Value Theorem (continued) f '(x) = 3x2 – 2x – 2
f '(c) = 3c2 – 2c – 2
2 Next, find f'(x) and then replace the x with the c to
get f'(c). f(a) = f(–1)
= (–1)3 – (–1)2 – 2(–1)
3 The interval in the problem is (–1,1) so that a = –1 and b = 1 =0
Find the values of f(a) and f(b).
Using f(x) = x3 – x2 – 2x, you find that: and
f(b) = f(1)

= (1)3 – (1)2 – 2(1)
= –2

4 Now put all the pieces together and solve for c. f l ^ch = f ^bh -f ^ah
The c = 1 is not in the open interval (–1,1). b -a

3c2 -2x -2 =1--2^--10h

3c2 -2x -2 =-1

3x2 -2x -1 =0

^3c +1h^c -1h =0

Therefore c = -1 or c =1
3

92

Limits: Indeterminate Forms 4chapter
and L’Ho^pital’s Rule

In Chapter 2 you encountered the indeterminate forms 0 and 3 when trying to calculate limits.
0 3

These forms were dealt with by using tedious algebraic methods. L’Hôpital’s Rule gives you a

quicker alternative.

L’Ho^pital’s Rule

If lim f ^xh is one of the indeterminate forms 0 and 33, then lim f ^xh =lim f l^xh .
g^xh 0 g^xh x"c gl ^ x h
x"c x"c

The indeterminate form 3 may be one of the forms:
3

3 or -33 or 3 -3
3 -3 -3

TIP

Do not confuse this rule with the Quotient Rule.
Here you are merely finding the derivative of the
top and then the derivative of the bottom function
and then finding the limit of their ratio.

L’HO^ PITAL’S RULE: EXAMPLE 1

Determine lim x12 --11.
x11
x "1

1 Direct substitution leads to the indeterminate form 00. lim x12 -1 =lim 12x11
x11 -1 x "1 11x10
Apply L’Hôpital’s Rule x "1

2 Now use direct substitution, 1 for x. $$=1112 111
In Chapter 2, you would have divided all terms by x12. 110

=1121

93

Limits: Indeterminate Forms
and L’Ho^pital’s Rule (continued)

L’HO^ PITAL’S RULE: EXAMPLE 2

Determine lim 1 + 13
x +3
x " -3 x

1 The indeterminate form 0 results from direct substitution.
0

First rewrite the term 1 as a power in preparation for finding the 1 + 13
x x +3
x
derivative of the top and the bottom. lim

x " -3

x -1 + 1
3
=lim x +3
x " -3

2 Apply L’Hôpital’s Rule, taking the derivative of the top and the bottom. $=lim -1 x -2
x " -3 1

=lim -1
x " -3 x2

3 Substitute x = –3. = -1
In Chapter 2, you would have found a common denominator. 9

L’HO^ PITAL’S RULE: EXAMPLE 3

Determine lim 2- x .
4 -x
x"4

1 After encountering the indeterminate form 00, rewrite x as a power. lim 2- x
4 -x
x"4

1
2 -x 2
=lim 4 -x
x"4

94

4Derivatives by Rule chapter

2 Apply L’Hôpital’s Rule – derivative of top and then derivative of bottom. -1 x - 1
2 2

lim -1

x"4

=lim 1
2x " 4 x

3 Substitute x = 4 and simplify. =1
24
In Chapter 2, you would have multiplied the numerator and denominator
by the conjugate of the numerator. = 1
4

L’HO^ PITAL’S RULE: EXAMPLE 4

Calculate lim x2 -x -2
x -2
x"2

1 Direct substitution leads to the indeterminate form 00. lim x2 -x -2
Use L’Hôpital’s Rule. x -2
x"2

=lim 2x -1
x"2 1

=lim ^2x -1h
x"2

2 Put 2 in for x and then simplify. =2•2–1
=3
In Chapter 2, you would have factored and reduced.

FAQ

How do I know when to use L’Hôpital’s Rule versus
the techniques shown in Chapter 2?
If direct substitution leads to one of the indeterminates and you
can find the derivative of both numerator and denominator, use
L’Hôpital’s Rule to find the limit.

95

5chapter

Derivatives of
Trigonometric

Functions

In this chapter, you will greatly expand your Derivatives of Sine, Cosine,
ability to find derivatives—specifically and Tangent . . . . . . . . . . . . . . . . . . . . . 97
derivatives of trigonometric and inverse
trigonometric functions. L’Hôpital’s Rule returns, Derivatives of Secant, Cosecant,
and you are introduced to the Chain Rule (finding and Cotangent . . . . . . . . . . . . . . . . . . 100
the derivative of a composite function).
L’Hôpital’s Rule and Trigonometric
Functions. . . . . . . . . . . . . . . . . . . . . . . 102

The Chain Rule . . . . . . . . . . . . . . . . . . 104

Trigonometric Derivatives
and the Chain Rule . . . . . . . . . . . . . . 109

Derivatives of the Inverse
Trigonometric Functions . . . . . . . . . . 110

Derivatives of Sine, 5chapter
Cosine, and Tangent

This section covers how to find the derivatives of three of the trigonometric functions: sine, cosine, and
tangent. To the right is a “unit circle,” which gives the cosine and sine of radian measures in the interval
[0,2π].

y

(– 1 , √3 ) (0, 1) ( 1 , √3 )
2 2 90° 2 2
π
(–√22 , √2 ) 2π 2 π (√22 √2
2 3 120° 60° 3 2
, )
3π
√3 1 π
2 2 4
(– , ) 5π 135° 4 π ( √3 , 1 )
45° 2 2

65
150° 30°

(–1, 0) 180° 0°
π x

2π (1, 0)

210° 330°
11π
7π
315° 6
√3 6 225° (√23 1
(– 2 , – 1 ) 5π 7π , – 2 )
2 4 4
240° 300°
(–√22 √2 4π 5π (√22 , – √2 )
2 2
, – ) 3 3π 3
270° 2
1 √3
(– 1 , – √3 ) ( 2 , – 2 )
2 2
(0, –1)

Sine and Cosine d ^sin xh = cos x and d ^cos xh = -sin x
dx dx
Listed at right are the formulas for finding the
derivatives of the sine and cosine functions.
Following this text are some examples of how
these two derivatives can be used.

97

Derivatives of Sine, Cosine, f(x) = 3sinx + 2cosx
and Tangent (continued)

DERIVATIVE OF A SUM
Find f'(x) for f(x) = 3sinx + 2cosx.

1 Start with the original function.

2 Find the derivative using the Constant Multiple f l ^ xh = 3 $ d ^sin xh + 2 $ d ^cos xh
and Sum Rules (see Chapter 4): dx dx

= 3 cos x + 2 ^-sin xh

f l^xh = 3 cos x - 2 sin x

DERIVATIVE OF A PRODUCT

Find d ^cos x sin x h.
dx

1 Begin with the original expression. d ^cos x sin xh
dx

2 Apply the Product Rule. = ; d cos x E sin x + ; d sin xE cos x
dx dx

3 Find the derivatives of cosine and sine and then = –sinx • sinx + cosx • cosx
simplify. = –sin2x + cos2x

98

5Derivatives of Trigonometric Functions chapter

DERIVATIVE OF PRODUCT OF ALGEBRAIC AND TRIGONOMETRIC FUNCTION
Find f '(x) for f(x) = x2sinx.

1 Start with the original function. f(x) = x2sinx

2 Apply the Product Rule (see Chapter 4). f '(x) = (2x)sinx + (cosx) • x2

3 Simplify. = 2xsinx +x2cosx

DERIVATIVE OF A QUOTIENT

Find f ' (x) for f ^xh = 1 + cos x .
sin x

1 Beginning with the original function, apply f ^ xh = 1 + cos x
the Quotient Rule (see Chapter 4). sin x

f l ^ x h = ^- sin x h sin x -^cos x h^1 + cos x h
^sin xh2

2 Simplify. = - sin2 x - cos x - cos2 x
sin2 x
3 Group the –sin2x and –cos2x together and then
use the Pythagorean Identity: sin2x + cos2x = 1. - sin 2 x - cos2 x - cos x
sin2 x
=

-_sin2 x + cos2 xi - cos x
= sin2 x

= -1 - cos x
sin 2x

99

Derivatives of Secant,
Cosecant, and Cotangent

This section covers the derivatives of the other four trigonometric functions: tangent, secant,
cosecant, and cotangent.

Formulas for Derivatives of Secant, Cosecant, and Cotangent

d tan x = s ec2 x
dx

d sec x = sec x tan x
dx

d csc x = -csc x cot x
dx

d cot x = -csc2 x
dx

DERIVATIVE OF SECANT AT SPECIFIC VALUE

For f ^xh = sec x, find f l b π l.
3

1 Start with the original function and find its derivative. f(x) = secx
f'(x) = secx tanx

2 Find f l b π l . f l b π l = sec b π l tan b π l
3 3 3 3

1 $ sin b π l
cos b 3
=
π π
3 l cos b 3 l

3

= 1 $ 2
1 1

22

f l b π l = 2 3
3

100

5Derivatives of Trigonometric Functions chapter

DERIVATIVE OF ANOTHER PRODUCT f ^xh = sec x tan x
Find f'(x) for f(x) = secx tanx.

1 Find the derivative using the Product Rule.

f l ^ xh = ; d sec x E tan x + ; d tan x E sec x
dx dx

f ^xh = sec x tan x +sec2x sec x

2 Simplify and factor. = secx tan2x + sec3x
Note: Although it is not required that you factor your = secx(tan2x + sec2x)
derivatives, you will find in later chapters that it can be
very helpful. y = 2 + tan x

DERIVATIVE OF A RADICAL TRIGONOMETRIC FUNCTION 1
Find y' for y = 2 + tan x.
1 Rewrite the original function as a power. = ^2 + tan xh2

2 Find y' using the General Power Rule. yl= 1 ^2 + tan xh- 1 sec2 x
3 Simplify the result. 2 2

= sec2 x

1

2^2 + tan xh2

= sec2 x
2 2 + tan x

101

L’Hôpital’s Rule and
Trigonometric Functions

This section covers L’Hôpital’s Rule and its use in determining the limits of some of the
trigonometric functions.

EXAMPLE 1 sin x
x
Determine lim .

x"0

Using direct substitution leads to the indeterminate form 0⁄0.

1 Apply L’Hôpital’s Rule elim fl o . lim sin x
gl x
x"0

= lim cos x
1
x"0

2 Use direct substitution, 0 for x. = cos ^0h
1

=1

102

5Derivatives of Trigonometric Functions chapter

EXAMPLE 2

Determine lim 1 - sin x .
cos x
x" π
2

Using direct substitution results in an indeterminate form.

1 Apply L’Hôpital’s Rule. lim 1 - sin x
cos x
x " π
2

= lim - cos x
π - sin x
x " 2

= lim cos x
π sin x
x " 2

2 Use direct substitution, π for x. cos b π l
2 2 l

= π
2
sin b

= 0
1

=0

103

The Chain
Rule

Frequently in your calculus studies, you will need to find the derivative of a composite function

f (g(x)). This section discusses the Chain Rule—the tool to do this.

The Chain Rule: First Form

If f and g are both differentiable and H is the composite function defined by H ^xh = f _g^xhi, then H is
differentiable and H' (x) is given by:

Hl^xh = 9 f l_g^xhiC $ gl^xh

Stated another way, you can write the Chain Rule as:

der. of outer evaluated at der. of outer evaluated at der. of inner
inner funct. Hfunction inner funct. times function
C function
ddx D[ f _Hg x i] = [ f l H_g x i A$ Hgl x)^ h
^h

In shorthand notation, it can be written as:
If y = f(g), then y' = f '(g) • g'

CHAIN RULE (FIRST FORM): EXAMPLE 1

Find d ^sin 3xh.
dx

1 Begin with the original expression. d ^sin 3xh
dx

104

5Derivatives of Trigonometric Functions chapter

2 Identify the outer and the inner functions. d 8sin ^3x hB
dx

3 Apply the Chain Rule and then simplify. der. of der. of

$C C A Csin at 3x times 3x

= cos ^3xh 3

= 3 cos 3x

CHAIN RULE (FIRST FORM): EXAMPLE 2

Find d _cos x 3i.
dx

1 Beginning with the original expression, identify the outer and d _ cos x 3i
inner functions. dx

= d _cos x 3i
dx

2 Apply the Chain Rule and simplify. $der. of Cat x3 Atimes der. of
_ x 3i
Ecos C3x2
= - sin 3x 2

= -3x2 sin x3

CHAIN RULE (FIRST FORM): EXAMPLE 3

Find d sin x.
dx

1 Once again, identify the outer and inner functions. d sin x
dx

= d sin x
dx

= d ^sin xh 1
dx 2

105

The Chain Rule 6d4e4r. 7 44of ] g1/28
Hat sin x der. of
(continued) 1 xh- 1 Atimes
2 2 Dsin x
2 Apply the Chain Rule and then simplify. cos x
Note: You could also have just used the General Power Rule.

$=
^sin

= cos x
2 sin x

CHAIN RULE (FIRST FORM): EXAMPLE 4

Calculate d sin ^tan xh.
dx

1 Rewrite the original expression, noting the outer and inner functions. d sin ^tan xh
dx

= d sin ^tan xh
dx

2 Apply the Chain Rule and simplify. = cos(tanx) • sec2x
= cos(tanx)sec2x

The Chain Rule: Second Form

If y = f(u) is a differentiable function of u and if u = g(x) is a differentiable function of x, then the

composite function y = f _g^xhi is a differentiable of x, and dy = dy $ ddux .
dx du

Stated another way:

The derivative of y with respect to x equals the product of the derivative of y with respect to u and the
derivative of u with respect to x.

106

5Derivatives of Trigonometric Functions chapter

CHAIN RULE (SECOND FORM): EXAMPLE 1

Find dy for y = sin3x.
dx

1 Start with the original function and write it as a composite of two functions. y = sin3x
y = sin(3x)

2 Let y = the outer function and let u = the inner function. y = sin(u) u = 3x

3 Find dy and ddux . dy = cos ^uh du = 3
du du dx

4 Apply the Chain Rule: Second Form. dy = dy $ du
dx du dx

= cos^uh $ 3

= 3 cos^uh

5 Substitute 3x for u. = 3cos(3x)

Therefore, dy = 3 cos 3x.
dx

107

The Chain Rule

(continued)

CHAIN RULE (SECOND FORM): EXAMPLE 2

For y = sin3x, find dy .
dx

1 Rewrite the original function as a composite of two functions. y = sin3x
y = (sinx)3
2 Let y = the outer function and let u = the inner function.
y = u3 u = sinx

3 Find dy and ddux . dy = 3u2 du = cos x
du du dx

4 Apply the Chain Rule: Second Form. dy = dy $ du
dx du dx

= 3u2$ cos x
= 3u2$ cos x

5 Substitute sin x for u. = 3(sinx)2cosx

= 3cosxsin2x

Therefore, dy = 3 cos x sin2 x .
dx

108

Trigonometric Derivatives 5chapter
and the Chain Rule

After the use of the Chain Rule, the trigonometric derivatives, with u as a function of x, can now be
written as follows.

& d ^sin uh = cos u du d ^sin 5xh = cos ^5x h $ 5 = 5 cos x 5x
dx dx

d $d x3 3x2= -3x2 sin x3
dx
dx
& ^cos uh = -sin u du cos x3= -sin

d sec2 u $d 1 = sec2 x
dx 2x 2x
dx
& ^tan uh = du tan x = sec2 x

& d ^csc uh = -csc u cot u du d csc 3x = -csc 3x cot 3x $ 3 = -3 csc 3x cot 3x
dx dx

& d ^sec uh = sec u tan u du $d sec x 2 = sec x 2 tan x 2 2x = 2x sec x2 tan x2
dx
dx

d $d
dx
dx
& ^cot uh = -csc2 u du cot _ x4+ 7i = -csc _x4+ 7i 4x3 csc _ x4+ 7i

109

Derivates of the Inverse
Trigonometric Functions

Another set of derivatives you will need are those of the six inverse trigonometric functions.

INVERSE TRIGONOMETRIC FUNCTIONS

Read the last equation of “if sin y = u, then y = arcsinu,” as “arc sine of u” or “inverse sine of u”

(sometimes written as y = sin–1u). For example, since cos b π l = 21, you can write π = arccos b 1 l, or
3 3 2
π 1
3 = cos- 1b 2 l.

INVERSE TRIGONOMETRIC DERIVATIVES
If u is a function of x, then the derivative forms are as follows.

d ^arcsin uh = du d _arcsin x 2 i = 2x
dx 1 - u2 dx 1 - x4

u = x 2, so du = 2x

& d ^arccos uh = -du d ^arccos 3xh = -3
dx 1 - u2 dx 1 - 9x2

u = 3x, so du = 3

& d ^arctan uh = du d `arctan xj = 1 = 1 with u = x then, du = 1
dx 1 + u2 dx 2x 2 x^1 + xh 2x

1 + ` x2j

110

5Derivatives of Trigonometric Functions chapter

& d ^arc cot uh = - du d _arc cot x3i = - 3x2 with u= x3, du= 3x 2
dx 1 + u2 dx 1 + x6

& d ^arc sec uh = u du d ^arc sec 3 xh = 3 = x 1 1 for u = 3x, du= 3
dx u2- 1 dx ^3x h2 - 9x 2 -
3x 1

& d ^arc csc uh = - du d _arc csc x 2i = x2 - 2x 1 = x2 - 2x 1 Let u = x2 so that du = 2x
dx u u2- 1 dx x4- x4-

111

6chapter

Derivatives of
Logarithmic and
Exponential Functions

The first theme of this chapter is that of Derivatives of Natural Logarithmic
differentiating logarithmic functions—natural Functions. . . . . . . . . . . . . . . . . . . . . . . 113
logarithmic as well as other base logarithmic
functions. L’Hôpital’s Rule is visited again, and Derivatives of Other Base
then the chapter concludes with the derivative of Logarithmic Functions . . . . . . . . . . . 119
exponential functions.
Logarithms, Limits, and
L’Hôpital’s Rule . . . . . . . . . . . . . . . . . 123

Derivatives of Exponential
Functions. . . . . . . . . . . . . . . . . . . . . . . 125

Derivatives of Natural 6chapter
Logarithmic Functions

The natural logarithmic function, written as lnx, has as its base the number e. The number e is
defined many ways, and its approximate value is 2.71828.

At right are two of the more common ways of 1 n xh1/x
defining the number e. n
e = lim b1 + l or e = lim ^1 +

n"3 x"0

In each case, you end up with an expression: f(x) y = lnx
(1+ really small number)really big power. l
x
Instead of writing logex, you just write lnx. l e

Properties of Natural Logarithms

Listed below are some properties of natural logarithms. These can be used to alter the form of a
logarithmic expression or equation.

1 If lnx = n, then en = x. This shows the relationship between If ln x = 3, then x = e3.
natural log equation and an exponential equation. If x = e–2, then ln x = –2.

2 ln(xy) = ln x + ln y ⇒ the log of a product property. ln(2x) = ln2 + lnx
ln(x2) + ln(y3) = ln(x2y3)

113

Derivatives of Natural Logarithmic
Functions (continued)

3 ln b x l = ln x - ln y ⇒ the log of a quotient property.= ln b 2 l = ln 2 - ln 3
y 3

the log of a quotient property. x
5
ln x - ln 5 = ln b l

4 ln xn = n • ln x ⇒ the log of a power property. ln x3= 3 ln x

ln x = ln x 1/2 = 1 ln x
2

2 ln x = ln x =2/3 ln 3 x2
3

5 ln x = logb x ⇒the change of base property. ln 5 = log10 5
logb e log10 e

log712 = ln 12
log7 e

Derivative of the Natural Logarithm Function

Listed below are the formulas used to find the derivative of lnx, or lnu where u is some function of x.
Following these formulas are some examples showing their uses in a variety of applications.

1 d ^ln xh = 1
dx x

2 If u is a differentiable function of x, then d ^ln uh = du .
dx u

DERIVATIVE OF A NATURAL LOG OF A POWER f(x) = ln(x2) where u = x2
Find f '(x) for f(x) = ln(x2). and then du = 2x

1 This can be done one of two ways. Let’s use derivative form
number 2 (listed above) first.

Identify the u function.

114

6Derivatives of Logarithmic and Exponential Functions chapter

2 Apply derivative form number 2 (see p. 114, “Derivative f l^xh= 2x % this is the du
of the Natural Logarithm Function). x2 u

= 2
x

A second way to approach the same problem is to take advantage of your log properties—specifically

the log of a power property. f(x) = ln(x2)

1 Rewrite the original function using the log of a power property. = 2 • lnx

2 Use the natural log derivative form number 2 from above. f l ^ xh = 2 $ 1
x

= 2
x

DERIVATIVE OF A LOG OF A RADICAL

Find d `ln x + 1j.
dx

1 Using the log of a power property, rewrite the d `ln x + 1j
original function. dx

= d ln ^ x + 1h1/2
dx

$= d 8 1 ln^ x + 1hB
dx 2

Letting u = x + 1, you have du = 1

2 Find the derivative using the du form. = 1 $ x 1 1
u 2 +

= 2 ^ 1 1h
x+

115

Derivatives of Natural Logarithmic
Functions (continued)

DERIVATIVE OF A QUOTIENT CONTAINING A NATURAL LOG

Find f'(x) for f ^xh= ln x .
x

1 Write the original function. f ^xh = ln x
x

2 Use the Quotient Rule to find the derivative. f l^xh = 1 $ x-1$ ln x
3 Simplify the result. x
^ x h2

f l^xh= 1 - ln x
x2

DERIVATIVE OF A POWER OF A NATURAL LOG f(x) = (ln x)3
Find f '(x) for f(x) = (ln x)3.

1 Start with a given function.

2 Use the General Power Rule (or Chain Rule) to differentiate. $f l^xh = 3^ln xh2 1
x

f l^xh = 3^ln xh2
x

116

6Derivatives of Logarithmic and Exponential Functions chapter

DERIVATIVE OF A COMPLICATED NATURAL LOG EXPRESSION
R V
S x_x2+ 2 W
Find f '(x) for f ^xh = ln S W.
S 1i W

2x 3 - 1 R 2 V
S x_x2+ W
TX 1i

Here’s where the real power of the natural log properties comes into use. You can write ln S 2x3- 1 WWas
a sum and/or a difference and/or a multiple of natural log expressions. S X
T

1 Begin with the function. R 2 V
S W
x _ x 2 + 1i

f ^xh = ln S 2x3- 1 W
S W

2 Use the log of a quotient property. = ln : x _ x 2 + 2 D - ln 2x3- 1

1i

3 Use the log of a product property. = ln x + ln _ x 2 + 2 - ln 2x3- 1

1i

4 Rewrite the last term as a power. = lnx + ln(x2 + 1)2 – ln(2x – 1)1/2

5 Use the log of a power property. = ln x + 2 ln _ x 2 + 1i - 1 ln _2x 3 - 1i
2

6 Finally, find the derivative of each ln expression. f l^xh = 1 + 2 c 2x 1m - 1 d 6x 2 1n
x x2- 2 2x 3 -
Remember the du for each ln derivative.
u 3x 2
1 4x 2x 3 -
f l^xh = x + x 2- 1 - 1

117

Derivatives of Natural Logarithmic
Functions (continued)

DERIVATIVE OF A TRIGONOMETRIC FUNCTION EVALUATED AT A NATURAL LOG
Find f'(x) for f(x) = cos(ln x).

1 Start with the original function. f ^xh = cos^ln xh let u = ln x, then du = 1
x

2 Find the derivative using d ^cos uh = -sin u $ du. f l^ xh = -sin ^ln xh $ 1
dx x

= -sin ^ln xh
x

118

Derivatives of Other Base 6chapter
Logarithmic Functions
y y = log2x
This section covers derivatives of logarithmic functions with bases other 1
than e, the base of the natural logarithmic function. You write these x
logarithms as logax and read them as “the logarithm of x in base a” or as 1 2
“the logarithm in base a of x.”
y
TIP y = log10x

Logn1 = 0 for all positive bases n 1
Log71 = 0 since 7° = 1
Log31 = 0 since 3° = 1 x
1 10

Properties of Logarithms log101,000 = 3 because 103= 1,000

If x and y are positive numbers and a > 0, then the 1 4 116 , 1
following properties of logarithms can be used. 2 16
Notice the similarity of these properties to those of Since b l = then log1/2 b l = 4
the natural logarithmic properties.
If log 3 b 1 l = -2, then 3- 2 = 1
Property #1: If logax = n, then an = x. 9 9

Property #2: loga(xy) = log10(99) = log10(9 • 11) = log109 + log1011 = log9 + log11
logax + logay
Note: Logarithms in base 10 are called “common logs.”
Instead of writing log10x, you just write logx.
log23 + log25 = log2(3 • 5) = log215

119

Derivatives of Other Base
Logarithmic Functions (continued)

Property #3: loga b x l = loga x - loga y log3b170 l = log310 - log3 7
y log817 - log813 = log8 b1173l

Property #4: logaxn = n • logax $log2 x5= 5 log2 x

log3 x + 1 = log3 ^ x + 1h1/2 = 1 log 3 ^ x + 1h
2
J N
KK x 3 y 2 OO=
L z P
log5 log _ x 3 y 2 i - log z

5 5

= log5 x3+ log5 y2- log5 z1/2

= 3 log5 x + 2 log5 y - 1 log5 z
2

Property #5: loga x = logb x log311 = log1011 = log 11
logb a log10 3 log 3

log215 = log1315
log213

Derivatives of Logarithmic Functions

Listed below are the formulas for finding the derivative of just logax, or logau where u is some function
of x. Following these derivative formulas are some examples of how those formulas can be put to use.

$1d 1 1
dx _loga xi = ln a x

2 If u is a differentiable function of x, then d _loga ui = 1 du
x ln a u

120

6Derivatives of Logarithmic and Exponential Functions chapter

DERIVATIVE OF A LOG OF A POLYNOMIAL f(x) = log (x2 + 3)
Find f '(x) for f(x) = log5(x2 + 3).

1 Start with the original function.

2 Identify the u and du. Let u = x2 + 3 and du = 2x

3 Find the derivative. f l ^ xh = 1 c 2x m
ln 5 2+
x 3

= _ x 2 2x ln 5
+ 3i

DERIVATIVE OF LOG OF A QUOTIENT

Find f '(x) for f ^xh = log 3 d x3 1n.
2x -

1 Rewrite the original function using the log of a quotient f ^xh = log3 d x3 1n
property. 2x -

= log3 x3- log3^2x - 1h

2 Use the log of a power property on the first term. = 3log3x – log3(2x – 1)

121

Derivatives of Other Base
Logarithmic Functions (continued)

3 Find the derivative of each term using 1 $ du . f l^xh = 3 $ 1 $ 1 - 1 $ 2
ln a u ln 3 x ln 3 -
2x 1

= x 3 3 - ^2x 2 ln 3
ln - 1h

= 1 b 3 - 2x 2 1l
ln 3 x -

DERIVATIVE OF A LOG OF A RADICAL FUNCTION f ^xh = log2 3 x3- 5
Find f '(x) for f ^xh = log2 3 x3- 5.
1 Rewrite the radical as a power. = log2 _ x 3- 1/3

2 Use the log of a power property. 5i

3 Find the derivative. = 1 log2 _ x 3- 5i
3

$f l ^ xh = 1 1 d 3x 2 5 n
3 ln 2 x3-

= _ x 3 x2 ln 2
- 5i

122

Logarithms, Limits, 6chapter
and L’Ho^pital’s Rule

L’Hôpital’s Rule returns here. You apply it to limits of natural log and a variety of other functions.
It is restated below for your use:

lim f ^xh = lim f l^xh if the first limit is one of the indeterminate forms 0 or 33.
g ^xh gl ^ x h 0
x"c x"c

LIMITS AND LOGS: EXAMPLE 1

Determine lim ln ^3xh
x"3 x2 .

1 Direct substitution leads the form 0 , so apply L’Hôpital’s Rule. lim ln ^3xh
0 x2
x"3

= lim 3/ ^3x h
2x
x"3

2 Simplify and then use direct substitution. = lim 1
2x
x"3 2

= 1
3

=0

123

Logarithms, Limits, and
L’Ho^pital’s Rule (continued)

LIMITS AND LOGS: EXAMPLE 2

Determine lim sin x .
l
x"π ln b x
π

1 A 0 form results when replacing x with π. L’Hôpital’s Rule is lim sin x
0
x"π x
applicable. ln b π l

= lim cos x

x"π b 1 l ÷b πxl
π

2 Simplify and substitute π for x. = lim x cos x
x"π

= π cos π
= π • –1
= –π

124

Derivatives of Exponential 6chapter
Functions
y y = 2x y = 2–x y
This section introduces you to additional techniques
of differentiating functions, such as f(x) = 2 x, f(x) 2 2
= ex2, and f(x) = 3sin x. These are called exponential 1 1
functions, with the base being a constant and an
exponent that contains a variable. x x
1
–1

y y = ex y = e–x y

e e
1 1

x x
1
–1

Rules for Differentiating Exponential Functions

There are two rules for differentiating exponential functions—one involves bases other than e, and the
other involves e as the base.

Case I: The base is other than e.

If u is a differentiable function of x, then d _aui = au$ du $ ln a.
dx
d
Using this formula, find dx _35xi. $ $Bder; of@au@au?du Aln a
35x 35x 5 ln 3
d =
dx

125

Derivatives of Exponential
Functions (continued)

Case II: The base is e.

If u is a differentiable function of x, then d _eui = eu du.
dx
d
Using this formula, find dx ` e x3j. $d `e jx3 = e x3 3x 2

dx

Finding the Derivative of Some Exponential Functions

Provided below are some additional examples that illustrate the use of the derivatives of exponential
functions.

y = 2sin x f(x) = 3x2 + 5x
y' = 2sin x • cos x • ln 2 f '(x) = ex2 + 5x • (2x + 5) • ln3

f(x) = 72x + 3 y = 3 ln x
f '(x) = 72x + 3 • 2 • ln7 = 3(ln x)1/2

or = 3(1/2) ln x
f '(x) = 72x + 3 • ln72
$ $yl= 3(1/2) ln x < 1 b 1 lF ln 3
= 72x + 3 ln49 2 x

g(x) = x3e2x $= ln x
f '(x) = [3x2] • e2x + [e2x • 2] • x3 ln 3 3
2x
= 3x2e2x + 2x3e2x
= x2e2x(3 + 2x) y = 2x
ex
126
$ $ $ $yl= _2x
1 ln 2i ex-_exi 2x

_e x 2

i

2 x e x ln 2 - e x 2 x

= _e x 2

i

2 x ln 2 - 2 x
ex
=

6Derivatives of Logarithmic and Exponential Functions chapter

Using L’Ho^pital’s Rule Revisited

We return to L’Hôpital’s Rule and apply it to limits which involve exponential functions. L’Hôpital’s
Rule is restated below for your use:

lim f ^xh = lim gf ll^^xxhh, if the first limit is one of the indeterminate forms 0 or 33.
g^xh 0
x"c x"c

MORE LIMITS AND L’HO^ PITAL’S RULE: EXAMPLE 1

Find lim 2x- 7x.
x
x"0

1 Substituting 0 for x results in the indeterminate form 00. Apply 2 x - 7x
L’Hôpital’s Rule. x
lim

x"0

2 x ln 2 - 7 x ln 7
1
= lim

x"0

2 Simplify. = lim _2x ln 2 - 7x lim 7i
3 Use direct substitution and simplify again. x"0

= 20 ln 2 - 70 ln 7

= ln 2 - ln 7

= ln b 2 l
7

127

Derivatives of Exponential
Functions (continued)

MORE LIMITS AND L’HO^ PITAL’S RULE: EXAMPLE 2

Determine lim e3x .
x3
x"3

1 The original limit results in a 3 indeterminate form, so apply lim e3x
3 x3
x"3
L’Hôpital’s Rule.
$= e3x 3
lim 3x 2

x"3

2 Simplifying leads to another 3 form, so apply L’Hôpital’s Rule again. = lim e3x
3 x2
x"3

$= e3x 3
lim 2x

x"3

= lim 3e3x
2x
x"3

3 Simplifying leads to another 3 form, so apply L’Hôpital’s Rule one $3_e3x 3i
3
= lim
more time. x"3 2

= lim 9e3x
2
x"3

=3

Therefore, no
limit exists.

128

7chapter

Logarithmic and
Implicit Differentiation

This chapter introduces a technique called Logarithmic Differentiation . . . . . . . . 130
logarithmic differentiation—finding the
derivative of a function having both a variable base Techniques of Implicit
and a variable exponent. Chapter 7 also covers Differentiation . . . . . . . . . . . . . . . . . . 134
implicit differentiation—finding the derivative of
an equation having 2 or more variables for which it Applications of Implicit
may be difficult or impossible to express one of the Differentiation . . . . . . . . . . . . . . . . . . 139
variables in terms of the other variables.

Logarithmic
Differentiation

Up to this point, you have been able to determine d _ x 3i and d _3xi in which the variable is in
dx dx

either the base or the exponent, but not in both places. A function of the form y = [f(x)]g(x), such as

f(x) = xsin x or y = (cos x)3x, can be differentiated using logarithmic differentiation—taking the

natural log on both sides and then differentiating both sides.

EXAMPLE 1 y = (3x)x2
Find y' for y = (3x)x2.

1 Start with the original equation.

2 Take the natural log of both sides. lny = ln(3x)x2

3 Use the log of a power property on the right. lny = x2ln(3x)

4 Differentiate both sides—ln on the left, product rule on yl = 6 2x @ $ ln ^3xh + ; 3 E $ x 2
the right. y 3x

5 Simplify. yl = 2x $ ln ^3xh+ x
y

130

6 Multiply both sides by y. 7Logarithmic and Implicit Differentiation chapter
7 Substitute (3x)x2 for y.
y' = y(2xln(3x) + x)
EXAMPLE 2
y' = (3x)x2(2xln(3x) + x)
Find d ^sin x hcos x
dx or
.
y' = (3x)x2(ln(3x)2x + x)
1 Let y = (sin x)cos x.
y = (sin x)cos x

2 Take ln of both sides. ln y = ln(sin x)cos x

3 Use the log of a power property on the right. ln y = cos x • ln(sin x)

4 Take the derivative of both sides—ln on the yl = 6-sin x @ $ ln ^sin xh + ; cos x E $ cos x
left, Product Rule on the right. y sin x

131

Logarithmic Differentiation

(continued)

5 Simplify. yl = - sin x $ ln ^sin xh+ cot x $ cos x
y

6 Multiply both sides by y. y' = y[–sin x • ln(sin x) + cot x • cos x]

7 Substitute (sin x) cosx for y. y' = (sin x)cos x[–sin x • ln(sin x) + cot x • cos x]
y' = (sin x)cos x[cot x • cos x – sin x • ln(sin x)]
y' = y' = (sin x)cos x[cot x • cos x – ln(sin x)sin x

EXAMPLE 3 f(x) = (ln x)x
Find f '(x) for f(x) = (ln x)x. ln f(x) = ln(ln x)x

1 Starting with the original function, take the natural log of
both sides.

2 Use the log of a product property and then ln f ^xh = x $ ln^ln xh SSRS ln1xx V
differentiate the result.
WW$
f l^xh = 61@ $ ln ^ln xh + x
f ^xh W

TX

132

7Logarithmic and Implicit Differentiation chapter

3 Simplify and then multiply both sides by f(x). f l^xh = ln ^ln xh + 1
f ^xh ln x

f l ^ xh = f ^ xh;ln ^ln xh + 1 E
ln x

4 Substitute (ln x)x for f(x). f l ^ x h = ^ln x h x ; ln ^ln x h + 1 E
ln x

FAQ

How do you know when to use “logarithmic differentiation”?

Use logarithmic differentiation when you are finding the derivative of
a function such as [f (x)]9(x) and discover that both the base and
exponent contain variables.

133

Techniques of Implicit
Differentiation

Up to this point, the functions you encountered were expressed in an explicit form—that is, writing

one variable in terms of another: y = x2 + 3x, s(t) = t2 – t2 + 15t – 7, or V ^r h = 4 πr 3. Unfortunately,
3

many relationships are not written explicitly and are only implied by a given equation: x2 + y2 = 25,

xy = 7, or x + xy + 2y3 = 13. These equations are written in implicit form. It may not be possible to

change an implicit form into an explicit form. For those cases, you use implicit differentiation.

Y AS A FUNCTION OF X ( dy AS THE DERIVATIVE)
dx

Find dy for x3 + xy – y2 = 12.
dx

1 Start with the original equation. x3 + xy – y2 = 12

2 Find the derivative of each term, treating y as d _ x 3i + d _ xyi - d _y2i = d ^12h
a function of x. dx dx dx dx

6 4p4rod7uct r4ule4 8
$ $power dy dy
dx - dx =
Brule
3x 2
+1 y + x 2y 0

3 Isolate all dy terms of the left. x dy - 2y dy = - 3x2- y
dx dx dx

4 Factor out dy . _x - 2yi dy = - 3x2- y
dx dx

134

7Logarithmic and Implicit Differentiation chapter

5 Solve for ddxy. dy = -3x2- y
dx x - 2y
Note: We took to the opposite of both the top and the bottom,
thus using fewer symbols in the final answer. or

dy = 3x2+ y
dx 2y - x

Y AS A FUNCTION OF X (Y’ AS THE DERIVATIVE) (ln x) • y3 = ey • x2
Find y' for (ln x) • y3 = ey • x2.

1 Start with the original equation.

2 Differentiate implicitly, treating y as b 1 l $ y 3+ _3y 2 $ yli $ ln x = _e $ yli $ x2+^2xh $ ey
a function of x. Use the Product Rule x
on the left and on the right.
$ $y3 yl+ 2xey
3 Simplify.
x
4 Isolate all y' terms on the left. + 3y2 ln x yl= ey x2

5 Factor out y' on the left. $ $3y2 ln x 2xe y - y3
yl- ey x2 yl= x

_3y2 ln x - ey x2i yl= 2xe y - y3
x

135

Techniques of Implicit _3y 2 ln x - ey x 2i yl= 2xe y - y3
Differentiation (continued) x

6 Solve for y'.

yl= xey- y3
x
3y2 ln x - ey 2
x

X AND Y AS FUNCTIONS OF AN UNKNOWN VARIABLE (DX AND DY AS THE DERIVATIVES)

Find dy for exy + sin x = ln y.
dx

1 Write the given equation. exy + sin x = ln y

2 Differentiate implicitly, treating x and y $ $ $@eu 6 4 447du 4 448 dx = dy
as functions of an unknown variable. y
exy _dx y + dy xi + cos x
3 Simplify.
exy ydx + exy xdy + cos xdx = dy
4 Put all dx terms on the left and all dy y
terms on the right.
e xy ydx + e xy xdy + cos xdx = dy
y

e xy ydx + cos xdx = dy - e xy xdy
y

136