1

Lesson 5

Derivatives of Logarithmic Functions and Exponential Functions 　

5A •  Derivative of logarithmic functions

Course II

2

Review of the Logarithmic Function

( )1,0 ≠> aaxay =Exponential function

Logarithmic function

xay =

We replace the notation

yax = xy alog=

Fig.1

Fig.2

Fig.3 xO

y xy alog=

Fig.1

x

y xay =

Fig.1

x

y

O

xay =

3

Derivative of the Logarithmic Function

From the definition

We put . When , .

?

( ) ( )

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛ +⋅=⎟⎠

⎞⎜⎝

⎛ +=

−+=ʹ

→→

→

xh

hx

xxh

h

hxhxx

ahah

aaha

1log1lim1log1lim

logloglimlog

00

0

kxh= 0→k0→h

( ) ( )⎭⎬⎫

⎩⎨⎧ +⋅=ʹ

→k

kxx aka 1log11limlog

0

( ) ( ) ⎥⎦⎤

⎢⎣⎡ +=+=

→→k

kakakk

xk

x

1

0

1

01limlog11loglim1

4

Napier’s Constant Trial

We expect that approaches one value as . Napier’s Constant

=⎟⎠

⎞⎜⎝

⎛ +=∞→

n

n ne 11lim

Important Mathematical Constants “π=3.1415…” was known 4000 years ago “e=2.7182…” was found in 17th century

5

Natural Logarithm

1

Then

If the base is , we have

Natural logarithm is the logarithm to the base .

e

e

Notation:

Summary x

xdxd

axx

dxd

a1)(ln,

ln1)(log ==

xxe lnlog →

6

Examples

[ Example 5.1] Find the derivative of the following functions.

Ans.

xxyxyxy 3ln)3(),23(log)2(,2ln)1( 2 =+==

2ln)23(3)23(

2ln)23(1)2(

+=ʹ+⋅

+=ʹ

xx

xy

13ln333ln)3(ln3ln)3( +=⋅+=ʹ+ʹ=ʹ xx

xxxxxxy

7

Examples

[ Example 5.2] Calculate the money which you can receive one year later using various compound systems. The principal is 10000 yen. (1) Annual interest is 100%. (2) Half a year interest is 50%, (3) Monthly interest is 100/12%, (4) Daily interest is 100/360%.

Ans. yen yen yen yen

(1) (2) (3) (4)

[ Note ] Napier’s Constant

7182.211lim ==⎟⎠

⎞⎜⎝

⎛ +∞→

en

n

n

148,27)365/11(10000130,26)12/11(10000500,22)2/11(10000

000,20)11(10000

365

12

2

1

=+×

=+×

=+×

=+×

8

Exercise

[Ex.5.1] Find the derivatives of the following functions.

(1) , (2) , (3)

Pause the video and solve the problem by yourself.

2)(ln xy = xxy ln= xy 10log=

9

Answer to the Exercise

[Ex.5.1] Find the derivatives of (1) and (2)

(1)

(2)

)1ln()(ln 32 +== xyxy

( )xxx

dxdxx

dxd ln2lnln2ln 2 =⋅=

( ) ( ) ( ) 1ln1lnlnlnln +=⋅+=ʹ⋅+⋅ʹ= xx

xxxxxxxxdxd

10lnlnlog 10xx =

( )xx

dxdx

dxdx

dxd

10ln1ln

10ln1

10lnlnlog10 ===

(3)

Ans.

10

Lesson 5

Derivatives of Logarithmic Functions and Exponential Functions 　

5B •  Derivative of exponential functions

Course II

11

Derivative of Inverse Functions

)(xfy =

Differentiate both sides of (1) by and from the chain rule, we have

Therefore

⎟⎟⎠

⎞⎜⎜⎝

⎛=

dydxdx

dy 1or

y

dyydg

dxxdf

dydx

dxxdf

dyxdf )()()()(1 ===

dyydg

dxxdf )()(1=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

dyydgdx

xdf)(

1)(

)(ygx =

Let and be inverse functions. Then gf

12

Derivative of the Exponential Function Exponential function of base e

xexfy == )( yygx ln)( ==

Therefore, from the previous slide we have

y

ydyydgdx

xdfdxdy

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛==

11

)(1)(

xay =Case

If , then . Therefore xea = ax ln=

( ) axxax eeay lnln ===

From the Chain Rule

( ) ( ) aaaxdtdee

dtda

dtd xaxaxx ln)ln(lnln ===

( ) aaadtd xx ln=

xx eedxd

=)(

13

Examples

[ Example 5.3] Find the derivative of the following functions.

Ans.

xx ayey 22 )2(,)1( −==

(1) Chain rule xx exey 22 2)2( =ʹ⋅=ʹ

(2) Chain rule

( ) aaxaay xx log2)2(log 22 −− −=ʹ−⋅=ʹ

14

Exercise

[Ex.5.2] Find the derivatives of the following functions.

(1) (2) (3)

xaxy = xy ln2=xx

xx

eeeey −

−

+−

=

Pause the video and solve the problem by yourself.

15

Answer to the Exercise

22

22

)(4

)()()(

xxxx

xxxx

eeeeeeeey −−

−−

+=

+−−+

=ʹ

[Ex.5.2] Find the derivatives of the following functions.

(1) (2) (3)

xaxy = xy ln2=xx

xx

eeeey −

−

+−

=

(1) Product rule

)log1(log)()( axaaaxaaxaxy xxxxx +=⋅+=ʹ+ʹ=ʹ

(2) Chain rule

xxy

xx 2ln2)1(2ln2

lnln =⋅=ʹ

(3) Quotient rule