Derivatives of Logarithmic and Exponential Functions

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

Derivatives of Logarithmic and Exponential Functions THEOREM: The function f (x) = loga x is differentiable and f ′ (x) =

1 x ln a

Proof: We have: d 1 x+h loga (x + h) − loga x (loga x) = lim = lim loga h→0 h→0 h dx h x

1 1 x h h = lim = lim loga 1 + · loga 1 + h→0 h h→0 x h x x " " x/h # 1/(h/x) # 1 h h 1 = lim loga 1 + = lim loga 1 + x h→0 x x h→0 x " 1/(h/x) # h i 1 h 1 1 ln e 1 = loga lim 1 + = lim (1 + u)1/u = e = loga e = = u→0 h→0 x x x x ln a x ln a COROLLARY: We have 1 d (ln x) = dx x REMARK: In general, (loga u)′ =

1 · u′ u ln a

and

(ln u)′ =

1 ′ ·u u

EXAMPLES: (a) If f (x) = log5 (x2 + 1), then f ′ (x) = [log5 (x2 + 1)]′ = (b) If f (x) = ln(ln x), then f ′ (x) = [ln(ln x)]′ = (c) Find f (x) if f (x) = log ′

x 1 + x2

.

1

(x2

2x 1 · (x2 + 1)′ = 2 . + 1) ln 5 (x + 1) ln 5

1 1 · (ln x)′ = . ln x x ln x



x (c) If f (x) = log , then 1 + x2 ′ x ′ = f (x) = log 1 + x2

1

x 1 + x2

′

· x · ln 10 1 + x2     ′   (1 + x2 ) · 1 x = · x  1 + x2   (1 + x2 ) ·  · ln 10 2 1+x ′ x 1 + x2 · = x ln 10 1 + x2 1 + x2 x′ (1 + x2 ) − x(1 + x2 )′ · = x ln 10 (1 + x2 )2 =

1 + x2 1 · (1 + x2 ) − x · 2x · x ln 10 (1 + x2 )2

=

1 + x2 1 + x2 − 2x2 · x ln 10 (1 + x2 )2

1 − x2 1 + x2 · = x ln 10 (1 + x2 )2 =

1 − x2 x(1 + x2 ) ln 10

or f (x) = log ′

x 1 + x2

′

′ = log x − log(1 + x2 ) =

1 1 − · (1 + x2 )′ 2 x ln 10 (1 + x ) ln 10

=

2x 1 − x ln 10 (1 + x2 ) ln 10

=

2x2 1 + x2 − x(1 + x2 ) ln 10 x(1 + x2 ) ln 10

=

1 + x2 − 2x2 x(1 + x2 ) ln 10

=

1 − x2 x(1 + x2 ) ln 10

2



THEOREM: We have d x (a ) = ax ln a dx In particular, if a = e, then (ex )′ = ex Proof 1: Let y = ax . We should prove that y ′ = ax ln a. We have y = ax

=⇒

ln y = ln ax = x ln a

therefore (ln y)′ = (x ln a)′ = ln a · (x)′ = ln a

=⇒

1 ′ · y = ln a y

so y ′ = y ln a = ax ln a Proof 2: Let y = ax , then loga y = x. We have (loga y)′ = x′

1 · y′ = 1 y ln a

=⇒

so y ′ = y ln a = ax ln a Proof 3: Let f (x) = loga x. Note that f −1 (x) = ax . Since (f −1 )′ (x) =

1 f ′ (f −1 (x))

we have (ax )′ =

and f ′ (x) = (loga x)′ =

1

=

f ′ (f −1 (a))

1 1 ax ln a

= ax ln a

REMARK: In general, (au )′ = au ln a · u′

and

EXAMPLES: 1. Find (2sin 3x )′ .

√

2. Find (xe

1−x ′

).

3

(eu )′ = eu · u′

1 x ln a



EXAMPLES: 1. (2sin 3x )′ = 2sin 3x ln 2 · (sin 3x)′ = 2sin 3x ln 2 · cos 3x · (3x)′ = 2sin 3x ln 2 · cos 3x · 3 √

2. (xe

1−x ′

√

) = x′ e

1−x

√

+ x(e

√

1−x ′

) =e

1−x

√

=e

1−x

√

+ xe

1−x

√

+ xe

1−x

√ · ( 1 − x)′

√

√ 1 xe 1−x · √ (1 − x)′ = e 1−x − √ 2 1−x 2 1−x

Logarithmic Differentiation Note that we can’t apply the rules (un )′ = nun−1 · u′ or (au )′ = au ln a · u′ to functions like f (x) = xx . EXAMPLES: 1. Let f (x) = xx . Find f ′ (x). Solution: We logarithm and then differentiate both sides of f (x) = xx . We have f (x) = xx therefore

=⇒

ln f (x) = ln xx = x ln x

=⇒

[ln f (x)]′ = [x ln x]′

1 1 · f ′ (x) = x′ ln x + x(ln x)′ = 1 · ln x + x · = ln x + 1 f (x) x

hence f ′ (x) = f (x)(ln x + 1) = xx (ln x + 1) 2. Let f (x) = (sin x)cos x . Find f ′ (x). Solution: We logarithm and then differentiate both sides of f (x) = (sin x)cos x . We have f (x) = (sin x)cos x

=⇒

ln f (x) = ln(sin x)cos x = cos x ln(sin x)

therefore [ln f (x)]′ = [cos x ln(sin x)]′ hence 1 · f ′ (x) = (cos x)′ ln(sin x) + cos x(ln(sin x))′ f (x) 1 = − sin x ln(sin x) + cos x (sin x)′ sin x 1 = − sin x ln(sin x) + cos x cos x sin x From this it follows that

= − sin x ln(sin x) + cos x cot x

f ′ (x) = f (x)(− sin x ln(sin x) + cos x cot x) = (sin x)cos x (− sin x ln(sin x) + cos x cot x) 3

3. Let f (x) = (sin 2x)x . Find f ′ (x). 4



3

3. Let f (x) = (sin 2x)x . Find f ′ (x). 3

Solution: We logarithm and then differentiate both sides of f (x) = (sin 2x)x . We have f (x) = (sin 2x)x therefore

3

=⇒

3

ln f (x) = ln(sin 2x)x = x3 ln(sin 2x)

′ [ln f (x)]′ = x3 ln(sin 2x)

hence

1 · f ′ (x) = (x3 )′ ln(sin 2x) + x3 (ln(sin 2x))′ f (x) = 3x2 ln(sin 2x) + x3

1 (sin 2x)′ sin 2x

= 3x2 ln(sin 2x) + x3

1 cos 2x(2x)′ sin 2x

= 3x2 ln(sin 2x) + x3

1 cos 2x · 2 sin 2x

= 3x2 ln(sin 2x) + 2x3 cot 2x From this it follows that 3 f ′ (x) = f (x) 3x2 ln(sin 2x) + 2x3 cot 2x = (sin 2x)x (3x2 ln(sin 2x) + 2x3 cot 2x) EXAMPLE: Find the derivative of

√ x x+1 y= √ 3 x + 2(x + 3)5

Solution: We have (see Appendix I, Example 5) √ x x+1 1 1 ln y = ln √ = ln x + ln(x + 1) − ln(x + 2) − 5 ln(x + 3) 3 5 2 3 x + 2(x + 3) therefore

′ 1 1 (ln y) = ln x + ln(x + 1) − ln(x + 2) − 5 ln(x + 3) 2 3 ′

so

1 1 1 1 1 1 1 5 1 ′ 1 1 ·y = + · − · −5 = + − − y x 2 x+1 3 x+2 x+3 x 2(x + 1) 3(x + 2) x + 3 It follows that √ 1 x+1 x 1 1 1 5 1 1 5 ′ y =y = √ + − − + − − 3 x 2(x + 1) 3(x + 2) x + 3 x + 2(x + 3)5 x 2(x + 1) 3(x + 2) x + 3 EXAMPLE: Find the derivative of

√ x2 3 7x − 14 y= (1 + x2 )4 5



EXAMPLE: Find the derivative of √ x2 3 7x − 14 y= (1 + x2 )4 Solution: We have (see Appendix I, Example 6) 2√ x 3 7x − 14 1 ln y = ln = 2 ln x + ln(7x − 14) − 4 ln(1 + x2 ) 2 4 (1 + x ) 3 therefore

′ 1 2 (ln y) = 2 ln x + ln(7x − 14) − 4 ln(1 + x ) 3 ′

so

1 ′ 2 7 2x 2 1 8x ·y = + −4 = + − 2 y x 3(7x − 14) 1+x x 3x − 6 1 + x2

It follows that ′

y =y

1 8x 2 + − x 3x − 6 1 + x2

√ 1 8x x2 3 7x − 14 2 = + − (1 + x2 )4 x 3x − 6 1 + x2

EXAMPLE: Find the derivative of √ 3

√ x2 − 8 x3 + 1 y=√ 1 − x(x + 2)−3 (x6 − 7x + 5)5 Solution: We have (see Appendix I, Example 7) ! √ √ 3 x2 − 8 x3 + 1 ln y = ln √ 1 − x(x + 2)−3 (x6 − 7x + 5)5 =

1 1 1 ln(x2 − 8) + ln(x3 + 1) − ln(1 − x) + 3 ln(x + 2) − 5 ln(x6 − 7x + 5) 3 2 2

therefore ′

(ln y) = so

′ 1 1 1 2 3 6 ln(x − 8) + ln(x + 1) − ln(1 − x) + 3 ln(x + 2) − 5 ln(x − 7x + 5) 3 2 2 1 ′ 2x 3x2 1 3 30x5 − 35 ·y = + + + − y 3(x2 − 8) 2(x3 + 1) 2(1 − x) x + 2 x6 − 7x + 5

It follows that 3x2 1 3 30x5 − 35 2x ′ + + + − y =y 3(x2 − 8) 2(x3 + 1) 2(1 − x) x + 2 x6 − 7x + 5 √ √ 3 3x2 1 3 30x5 − 35 2x x2 − 8 x3 + 1 =√ + + + − 1 − x(x + 2)−3 (x6 − 7x + 5)5 3(x2 − 8) 2(x3 + 1) 2(1 − x) x + 2 x6 − 7x + 5 6



Appendix I 1. ln (x(x + 1)) = ln x + ln(x + 1) x 2. ln = ln x − ln(x + 2) x+2 x(x + 1) = ln x(x + 1) − ln(x + 2) = ln x + ln(x + 1) − ln(x + 2) 3. ln x+2 x(x + 1) 4. ln = ln x(x + 1) − ln (x + 2)(x + 3) (x + 2)(x + 3) = ln x + ln(x + 1) − ln(x + 2) + ln(x + 3) = ln x + ln(x + 1) − ln(x + 2) − ln(x + 3) √ x(x + 1)1/2 x x+1 = ln 5. ln √ 3 (x + 2)1/3 (x + 3)5 x + 2(x + 3)5 = ln x(x + 1)1/2 − ln (x + 2)1/3 (x + 3)5 =

ln x + ln(x + 1)1/2 − ln(x + 2)1/3 + ln(x + 3)5

= ln x + ln(x + 1)1/2 − ln(x + 2)1/3 − ln(x + 3)5

1 1 ln(x + 1) − ln(x + 2) − 5 ln(x + 3) 2 3 2√ x 3 7x − 14 x2 (7x − 14)1/3 2 1/3 6. ln − ln(1 + x2 )4 = ln = ln x (7x − 14) (1 + x2 )4 (1 + x2 )4 = ln x +

= ln x2 + ln(7x − 14)1/3 − ln(1 + x2 )4

1 ln(7x − 14) − 4 ln(1 + x2 ) 3 ! √ √ 3 (x2 − 8)1/3 (x3 + 1)1/2 x2 − 8 x3 + 1 = ln 7. ln √ (1 − x)1/2 (x + 2)−3 (x6 − 7x + 5)5 1 − x(x + 2)−3 (x6 − 7x + 5)5 1/2 −3 6 5 2 1/3 3 1/2 − ln (1 − x) (x + 2) (x − 7x + 5) = ln (x − 8) (x + 1) = 2 ln x +

=

2

1/3

ln(x − 8)

3

1/2

+ ln(x + 1)

− ln(1 − x)

1/2

+ ln(x + 2)

−3

6

+ ln(x − 7x + 5)

= ln(x2 − 8)1/3 + ln(x3 + 1)1/2 − ln(1 − x)1/2 − ln(x + 2)−3 − ln(x6 − 7x + 5)5 =

1 1 1 ln(x2 − 8) + ln(x3 + 1) − ln(1 − x) + 3 ln(x + 2) − 5 ln(x6 − 7x + 5) 3 2 2

7

5



Appendix II The following problem was given in Fall 2013 (Calculus I, Midterm Exam II). None of the 86 test takers got the right answer. 1 , then f ′ (x) is Let f (x) = log2 1 + log3 x A

1 x(1 + log3 x) ln 2 ln 3

B

1 x(1 − log3 x) ln 2 ln 3

C

−

1 x(1 − log3 x) ln 2 ln 3

D

−

1 x ln 2 ln(3x)

E

None of the above

See the next page for the solution.

8



1 Let f (x) = log2 , then f ′ (x) is 1 + log3 x 1 A x(1 + log3 x) ln 2 ln 3 B

1 x(1 − log3 x) ln 2 ln 3

C

−

D

−

E

None of the above

1 x(1 − log3 x) ln 2 ln 3

1 ←− Correct x ln 2 ln(3x)

Solution: We first note that 1 = log2 (1 + log3 x)−1 = (−1) log2 (1 + log3 x) = − log2 (1 + log3 x) log2 1 + log3 x or 1 log2 = log2 1 − log2 (1 + log3 x) = 0 − log2 (1 + log3 x) = − log2 (1 + log3 x) 1 + log3 x Therefore ′ 1 log2 = [− log2 (1 + log3 x)]′ = −[log2 (1 + log3 x)]′ 1 + log3 x

9

=−

1 · (1 + log3 x)′ (1 + log3 x) ln 2

=−

1 1 · (1 + log3 x) ln 2 x ln 3

=−

1 (1 + log3 x)x ln 2 ln 3 1

= −

1+

= −

1 ln x · ln 3 x ln 2 1 · ln 3 + ln 3

ln x x ln 2 ln 3 ln 3

=−

1 (ln 3 + ln x) x ln 2

=−

1 ln(3x) x ln 2