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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka



 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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

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



Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka

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

Section 3.3 Derivatives of Logarithmic and Exponential Functions

2010 Kiryl Tsishchanka



 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