Chapter 3 Derivatives and Differentials

3.2 The rules for find derivative of a function

New Words

Addition 加 Sum 和

Subtraction 减 Difference 差

Multiplication 乘 Product 积

Division 除 Quotient 商

A composite function 复合函数

Differentiate 求导

This section develops methods for finding derivatives of

functions. As we known, functions can be built up from

simpler functions by addition, subtraction, multiplication,

and division. Thus, we first give the formula of

derivatives of sum, difference, product, and quotient of

two functions.

1.The Derivatives of the Sum, Difference, Product and Quotient

Theorem 1

If and are derivable, and is any constant, u x v x C

(1) ( ) ( ) ( ) ( )u x v x u x v x

(2) ( ) ( ) ( ) ( ) ( ) ( )u x v x u x v x u x v x

(3) ( ) ( )Cu x Cu x

2

( ) ( ) ( ) ( ) ( )(4) ( ( ) 0)

( ) ( )

u x u x v x u x v xv x

v x v x

then so is , , , and

. Its derivative is given by the formula

u x v x u x v x Cu x

u x

v x

Proof

To prove this theorem we must go back to the definition of the derivative.(1) Let ( ) ( ), we have to examiney u x v x

0 0

( ) ( ) ( ) ( )lim limx x

y u x x v x x u x v x

x x

0

( ) ( ) ( ) ( )limx

u x x u x v x x v x

x

0limx

u v

x

0lim( ) ( ) ( )x

u vu x v x

x x

Thus ( ) ( ) is derivable and

( ) ( ) ( ) ( )

u x v x

u x v x u x v x

A similar argument applies to ( ) ( ),

that is

( ) ( ) ( ) ( )

u x v x

u x v x u x v x

x 0

(2) Let ( ) ( ), then we express in terms

of and . Finally, we determine by

examining lim

y u x v x y

u v y x

y

x

0 0

( ) ( ) ( ) ( )lim limx x

y u x x v x x u x v xy

x x

0

[ ( ) ][ ( ) ] ( ) ( )limx

u x u v x v u x v x

x

0

( ) ( )limx

u x v v x u u v

x

0lim[ ( ) ( ) ]x

u v vv x u x ux x x

( ) ( ) ( ) ( )u x v x u x v x

Thus, ( ) ( ) is derivable and

( ) ( ) ( ) ( ) ( ) ( )

u x v x

u x v x u x v x u x v x

(3) is a special case of the formula (2)

( )(4) Let , the argument is similar to the

( )

proof concerning

u xy

v x

u x v x

0 0

( ) ( )( ) ( )

lim limx x

u x x u xy v x x v x

yx x

0

( ) ( )( ) ( )

limx

u x u u xv x v v x

x

0

[ ( ) ] ( ) ( )[ ( ) ]lim

[ ( ) ] ( )x

u x u v x u x v x v

v x v v x x

0

( ) ( )lim

[ ( ) ] ( )x

uv x u x v

v x v v x x

0

( ) ( )lim

[ ( ) ] ( )x

u vv x u xx xv x v v x

2

( ) ( ) ( ) ( ) ( 0 as 0)

( )

u x v x u x v xv x

v x

2

This establishes the formula

( ) ( ) ( ) ( ) ( ) ( ( ) 0)

( ) ( )

u x u x v x u x v xv x

v x v x

Note

This completes the proof of theorem 1.

and functions theif that asserts theoremThis 1 xu

)0)( ( )(

)(

)(

1)2(

2

xv

xv

xv

xv

s.derivative the

of (quotient)product not the is (quotient)product

theof derivative heout that t it turnsfor ,surprising

bemay you But s.derivative theof )difference(or

sum theis )difference(or sum theof derivative the

s,other wordIn quotient. andproduct ,difference

sum, their do so then ,at sderivative have xxv

(3) Theorem 1 can extend to any finite number of derivable functions.

The following examples use the formulas for the derivative of the sum, the difference, the product and the quotient.

xxxfxf sin)( if Find Example 1

Solution

)(sinsin)()( xxxxxf

xxxx

cossin2

1

By the formula (1) of theorem 1

.cot)( if )( Find xxfxf

xx

xfsincos

)( Since

x

xxxxxf

2sin

)(sincossin)(cos)(

x

xxxx2sin

coscossinsin xx

22

cscsin

1

xx 2csc)(cot isThat

xx 2sec)(tan Similarly

Example 2

Solution By the formula (4) of theorem 1

Example 3

Solution By the formula (4) of theorem 1

xxfxf csc)( if Find

xxf

sin1

)( Since

x

xxf

2sin

)(sin)(

xx

x

xcotcsc

sin

cos2

xxx cotcsc)(csc isthat

xxx tansec)(sec Similarly

x

xxxfxf

sin1

cos)( if )( Find

2)sin1(

)sin1(cos)sin1()cos()(

x

xxxxxxxf

2)sin1(

coscos)sin1)(sin(cos

x

xxxxxxx

Example 4

Solution By the formulas of theorem 1

2)sin1(

)1(sinsincoscos

x

xxxxx

)0( findthen ),()(

,)0(),,0(on continuous is )( If

fxxgxf

agUxg

Using the definition of derivatives, we have at x=0

xfxf

fx

)0()(lim)0(

0

x

xxgx

0)(lim

0

)(lim0xg

x

ag )0(

Example 5

Solution

2.The Rules for Finding Derivatives of Inverse Functions

Theorem 2

and at derivable is of

function inverse then the,0 if , interval

open an on derivable and monotonic be )(Let

1 xyfxxfy

yfI

yfx

)(

1)(1

yfxf

dydxdx

dy 1or

Proof

,0 as ,0)()(

is that ,continuous monotonic

is ,on continuous monotonic isit

then,on derivable and monotonic is Since

11

1

xxfxxfy

xfyI

Iyfx

.0 if )()( and 11 xxfxxf

.0 if 0)()( isThat 11 xxfxxfy

x

xfxxfxf

x

)()(lim)(

11

0

1

)()(lim

0 yfyyfy

y

yyfyyfy

)()(

1lim

0

)(1yf

Example 6

Solution

xyy arcsin if Find

]2

,2

[ as sin ,arcsin Since

yyxxy

yydydx

cos)(sin

22 1sin1 xy

21

11

xdy

dxdx

dy

21

1)(arcsin is,that

xx

21

1)(arccos Similarly,

xx

Example 7

Solution

xyy cotarc if Find

,cotarc Since xy

thus,),,0( as cot yyx

yydydx 2csc)(cot )1()cot1( 22 xy

21

11 Therefore

xdydxdx

dy

21

1)cotarc( is,that

xx

21

1)(arctan Similarly,

xx

3.The Derivative of a Composite Function (The Chain Rule)

This section presents the most important technique for

finding the derivative of a function. It turns out that we

can easily compute the derivative of a composite

function if we know the derivatives of the functions

from which it is composed.

Theorem 2 (The chain rule)

a is then , offunction derivable a is

)( and offunction derivable a is If

xgfyx

xguuufy

Proof

0

dlim

d x

y y

x x

dx

du

du

dy

dx

dyxguf

dx

dy

x

is, that ),()(

and offunction derivable

0lim , 0 if is small enoughx

y uu x

u x

0 0lim lim , since both limit existx x

y u

u x

Note

andfunction derivable a

is )]}([{ then functions, derivable

are )(),(),( if instance,For

functions. moreor threeofn compositio the

as upbuilt function a toextends rulechain The 1

xhgfy

xhvvguufy

0 0lim lim , since 0 as 0ux x

y uu x

u x

d d

d d

y u

u x

dx

dv

dv

du

du

dy

dx

dyxhvguf

dx

dy is, that ),()()(

The following examples apply the chain rule

Example 8

Solutionx

yy1

arctan if Find

Then .1

,arctan as expressed becan 1

arctan

xu

uyx

y

dx

du

du

dyy

xu

1

1

12

22

1

11

1

x

x

21

1

x

Example 9

Solution

21

2cos if Find

x

xyy

2

2

1

2 and

cos as expressed becan 1

2cos

x

xu

uyx

xy

dx

du

du

dyy )

1

2(sin

2

x

xu

22

2

2 )1(

22)1(2

1

2sin

x

xxx

x

x

222

2

1

2sin

)1(

)1(2

x

x

x

x

Example 10

Solution

xeyy cosln if Find

)cos(ln xey )(coscos

1 xx

ee

))(sin(cos

1 xxx

eee

xx ee tan

Example 11

Solution

3 3sin21 if Find xyy

)sin21(3 3 xy

)sin21()sin21(31 33

23

xx

])(sinsin320[)sin21(31 23

23

xxx

)cossin6()sin21(31 23

23 xxx

xxx cossin)sin21(2 23

23

Example 12

Solution

xeyy1

sin2

if Find

e xy

1sin2

xe x1

sin21

sin2

xxe x1

sin1

sin21

sin2

xxxe x11

cos1

sin21

sin2

2

1sin 11

cos1

sin22

xxxe x

xxe x

2sin

1 1sin

2

2

Example 13

Solution

x

xxyy

1

1 if Find

xx

xy11

xx

xxx

11

11

xx

xx

xxx

11

11

21

11

2)1(

)1()1(11

21

11

x

xxxx

xxx

xx

x

xxx

11

)1(11

2

Example 14

Solution

nxxx

xyy n cossin

1

1arctan if Find

)cos(sin11

arctan

nxxxx

y n

11

11

1

12 xx

xx

)(cossincos)(sin nxxnxx nn

22

2

)1(

)1()1(

)1(2

)1(

x

xx

x

xnxxxn n coscossin 1

nnxxn )sin(sin

nxxnnxxxnx

nn sinsincoscossin1

1 12

Example 15

Solution

function. derivable a is

where,)]([sin)(sin if Find 22 xfxfxfyy

)]([sin)(sin 22 xfxfdxd

y

)]([sin)(sin)]([sin)(sin2

1 22

22xfxf

xfxf

)]([sin)(sin2

)](sin[)](sin[2)(sin)(sin222 xfxf

xfxfxfxf

)]([sin)(sin

)()](cos[)](sin[cos)(sin)(sin22 xfxf

xfxfxfxxfxf

As these examples suggest, the chain rule is one of the most frequently used tools in the computation of derivatives.

4.The Derivatives of Elementary Functions

a. The formulas of derivatives for the fundamental elementary function

0))(1( C 1))(2( xx

xx cos))(sin3( xx sin))(cos4(

xx 2sec))(tan5( xx 2csc))(cot6(

xxx tansec))(sec7( xxx cotcsc))(csc8(

aaa xx ln))(9( xx ee ))(10(

axxa ln

1))(log11(

xx

1))(ln12(

21

1))(arcsin13(

xx

21

1))(arccos14(

xx

21

1))(arctan15(

xx

21

1)cotarc)(16(

xx

xx

21

))(17( 2

11)18(

xx

b. The rules of operations for derivatives

)()()()()1( xvxuxvxu

)()()()()()()2( xvxuxvxuxvxu

)()()3( xuCxCu

)0)( ( )(

)()()()()()(

)4(2

xv

xv

xvxuxvxuxvxu

dx

du

du

dy

dx

dy)5(

dy

dxdx

dy 1 6

Example 16

Solution

)ln( if Find 22 xaxydy

dx

)(1 22

22

xax

xaxdxdy

)(2

11

1 22

2222xa

xaxax

22

22

22

1

xa

xxa

xax

22

1

xa

22 xady

dx

2222 2

21

1

xa

x

xax

Example 17

Solution

xx xxydx

dy 33 33 if Find

exxxxy ln33 33 Since

)ln(03ln33ln2 xxx

dx

dye

xxx

xxxx e

xxx 1ln3ln33 ln2

)1(ln3ln33 ln2 xx exxx

Example 18

Solution

)4)(3(

)2)(1(ln if Find

xx

xxy

dx

dy

)4ln()3ln()2ln()1ln( xxxxy

4

1

3

1

2

1

1

1

xxxxdx

dy

We developed a formula called the chain rule for differentiating composite functions. The chain rule is the most commonly rule for differentiating functions.