chapter4 differentiation

8/3/2019 Chapter4 Differentiation

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Differentiation

Concept of Differentiation.2

Derivative of Trigonometric Function..20

The Chain Rule...23 Implicit Differentiation..27

Tangent Line and Normal Line...30


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Differentiation at one point

a. Tangent LinesThe secant line connecting P and Q has slope

Concept of Differentiation

Introduction ( two topic in the same theme)

PQ x cmx c

=

( ) ( )limx c

f x f cmx c

=

Ifxc, then the secant line

through P and Q will approachthe tangent line at P. Thus theslope of the tangent line is c

f(c) P

x

f(x)Q

x-c

f(x)-f(c)

2Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology


3/40

b. Instantaneous Velocity

Let a particle travel around an axis and position of

particle at time tis s=f(t). If the particle has acoordinatef(c) at time c andf(c+h) at time c+ h.

chapter

4Differentiation

c

c+h

Elapsed time distancetraveled

s

f((c)

f((c+h)



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chapter

4Differentiation

Thus, the average velocity during the time interval[c,c+h] is

rata-rata

( ) ( )f c h f cv

h

+ =

If h0, we get instantaneous velocity at x = c:

Letx = c+ h, instantaneous velocity can be written as

limx c

f(x) f(c)v

x c

=

rata-rata0 0

( ) ( )lim limh h

f c h f cv v

h

+ = =



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From two cases: The slope of the tangent line and

instantaneous velocity has the sameformula.

chapter

4Differentiation

e n on : e rs er va ve o a x = c,denoted by is defined by

if its limit exist.

( )f c

( ) ( )( ) lim

x c

f x f cf c

x c

=



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Other notation :

Example: Let evaluate

( ), ( )

df cy c

dx

)x(f1

= )3('f

chapter

4Differentiation

3 3

1 13 33 lim lim

3 3x x

f(x) f( ) x f'( )x x

= =

3 3

3 ( 3)lim lim

3 ( 3) 3 ( 3)x x

x x

x x x x

= =

3

1 1lim

3 9x x

= =



7/40

Derivatives from the left at c, denoted by , isdefined by

Derivatives from the right at c, denoted by , is

( ) ( )( ) lim

x c

f x f cf c

x c

=

( )f c

( )f c+

Derivative from the right

and the left

e ne y

A functionf is said differentiable at c( exist) if

and

( ) ( )cf c f +

=

( )( )) (f c f cf c+

= =

( ) ( )( ) lim

x c

f x f cf c

x c++

=

( )f c



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Let

+


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b.1 1

( ) (1) 1 2 (1 2 1)(1) lim lim

1 1x x

f x f xf

x x

+ ++

+ + = =

1 1

2 2 1lim 2 lim 1

1 ( 1)( 1)x x

x x

x x x+ +

= = =

+

chapter

4Differentiation

Thus, f is differentiable at x = 1 .


(1) 1f =


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Theorem: If f is differentiable at point cthenf is

continuous at c.

Proof: We will prove

Since

)()(lim cfxfcx =

cxcxcfxf

cfxf

+= ,).()()(

)()(

Differentiable

and Continuity

Thus,

The converse, however, is false , a function may be

continuous at a point but not differentiable.

cx

+=

)()()(

)(lim)(lim cxcx

cfxfcfxf

cxcx

( ) ( )lim ( ) lim lim( ) ( ) '( ) 0 ( ) x c x c x c

f x f c f c x c f c f c f cx c

= + = + =



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12/40

0 0 0

( ) (0) 0(0) lim lim lim 10 x x x

f x f x xf x x x

= = = =

We will show thatf is not differentiable at x = 0.

chapter

4Differentiation

0 0 0(0) lim lim lim 1.

0 x x xf

x x x+ + ++

= = = =

(0) 1 1 (0)f f + = =Because

then f is not differentiable at x = 0.



13/40


14/40

f is continuous at x = 1 if f is continuous from theleft atx = 1 and f is continuous from the right or

.limlim1 xx ==

chapter

4Differentiation

a.f is continuous at x = 1 (necessary condition).

11 xx

+

2

1 1

lim lim

1

1

x xa x b ax

a b a

b a

+

= + =

= + =

=



15/40

2

'

11

2 2

( ) (1)(1) lim lim1 1

( 1) 1lim lim

xx f x f x b af

x x

x a a x

+ = =

+ = =

chapter

4Differentiation

b. Derivatives from the left = derivatives from theright at c(sufficient condition).

2)1()1(''

==+

aff

Thus, a = 2 and b = 1.

1 1

1 1

'

1 11

1 1( 1)( 1)

lim lim 1 21

( ) (1) 1(1) lim lim lim1 1 1

x x

x x

x xx

x xx x

xx

f x f ax a x f a a x x x+

+

+

= = + =

= = = =



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( ) ( )( ) lim ,

t x

f t f x f x x

t x

=

( )f x

Derivative Definition

The derivative with respect tox of the function

f(x), denoted by , defined by the formula

-

0

( ) ( )( ) lim ,h

f x h f x f x xh

+ =

( ), , , , ( ).x xdy df x y D y D f xdx dx

dx

dy


,

for which the limit exist.

Other Notation:

Notation is called as Leibniz Notation.


17/40

1( );

r

rd x rx r Rdx

=

( ) 0f x =

chapter

4Differentiation

By using derivative definition, we have formula:

1. If then .( ) f x k =

2.

( ( ) ( ))( ) ( )

d f x g x f x g x

dx

+ = +

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

d f x g xf x g x f x g xdx

= +

2

( ( ) / ( )) ( ) ( ) ( ) ( ), ( ) 0

( )

d f x g x f x g x f x g xg x

dx g x

=


3.

4.

5.


18/40

Proof for formula number 4:

Let

0 0

( ) ( ) ( ) ( ) ( ) ( )'( ) lim limh h

h x h h x f x h g x h f x g xh xh h

+ + + = =

chapter

4Differentiation

( ) ( ) ( ),h x f x g x=

hhlim0=

++

++=

h

xfhxfxg

h

xghxghxf

h

)()()(

)()()(lim

0

h

xfhxfxg

h

xghxghxf

hhhh

)()(lim)(lim

)()(lim)(lim

0000

++

++=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )f x g x g x f x f x g x f x g x = + = +



19/40

1. Find of 3 2( ) 3 4. f x x x= + +

Answer:2 2

( ) 3 3 2 0 3 6 f x x x x x = + + = +

2. Find of3 2

( ) ( 1)( 2 3). f x x x x= + + +

2 2 3 =

( )f x

Example

( )f x

2 3( ) .1

xf xx

+=

+

2 2 2 2

2 2 2 2 2 2

1 ( 1) 2 ( 3) 1 6 2 6 1( ) .

( 1) ( 1) ( 1)

x x x x x x x xf x

x x x

+ + + + = = =

+ + +

3. Find of

222296334234

++++++= xxxxxx

22985234

++++= xxxx


( )f x

Answer:


20/40

( ) si ( ) on c s f x x f x x ==

( ) co ( ) is s n f x x f x x = =

Derivative of

Trigonometric Function

a.

b.

=

Proof:

s in s in( ) lim

t x

t xf x

t x

=

02

sin( )2lim cos( ) lim

2( )

2

t xt x

t xt x

t x

+=

2 cos ( ) sin( )2 2lim

t x

t x t x

t x

+

=

cos 1 cosx x= =


.


21/40

0

0

cos( ) cos( ) lim

cos cos( ) sin sin( ) coslim

h

h

x h xf x

h x h x h x

h

+ =

=

2cos sin

hx

chapter

4Differentiation

b. Let ( ) cos f x x=

0 0

cos cos s n s n s n2lim lim sin

h h

x xx

h h h

= =

2

20

2

( / 2) 0 0

cos ( sin )sin( )2lim( sin )

( / 2) 4

sin( / 2) sin( )cos lim sin lim

/ 2 4

h

h h

hx h

h

xh h

h h hx x

h h

=

=

cos 0 si inn sx x x= =



22/40

The derivative of remaining trigonometric functions canbe obtained by using formula derivative of u/v

2 2

2

2 2

(tan ) (sin / cos ) cos sin 1sec

cos cos

d x d x x x xx

dx dx x x

+= = = =

chapter

4Differentiation

c.

2 2

2

2 2(cot ) (cos / sin ) sin cos 1 csc

sin sind x d x x x x x

dx dx x x = = = =

2

(sec ) (1/ cos ) sin sin 1

tan seccos cos cos

d x d x x xx xdx dx x x x= = = =

2

(csc ) (1/sin ) cos cos 1csc cot

sin sin sin

d x d x x xx x

dx dx x x x

= = = =


d.

e.

f.


23/40

dydu

dydx

dudx

=

dy

du

du

dx

dy1sin

2+= x

The Chain Rule

Lety = f(u) and u=g(x). If and exists, then

dx1

2+= xu

xdx

du2=

uy sin=

udu

dycos=

2 2

cos( 1)2 2 cos( 1)

dy dy du x x x xdx du dx= = + = +


Let so that

Because of

and then


24/40

dy dy du dv

dx du dv dx=

Lety = f(u) and u=g(v), v = h(x), ifdx

dv

dv

du

du

dy,, exists, then

Example :

chapter

4Differentiation

dxdy 4 3sin ( 5)y x= +

53

+= xv 23x

dx

dv=Answer: Let

)5cos(cos3

+== xvdv

du

4

uy =

3 3 3

4 4sin ( 5)

dy

u xdu = = +


sinu v=

1. Find if


25/40

2 2 2( ) ( ( )) 1

d f x f x x

dx = +

chapter

4Differentiation

2. Find if

so that2 3 3 3

12 sin ( 5) cos( 5)dy dy du dv

x x xdx du dv dx

= = + +

2 2 2 2

2

( ( )) 1 ( ) 2 1

1( )2

d f x x f x x x

dx

xf xx

= + = +

+ =


Answer:


26/40

The nth derivative off(x) is the derivative of (n-1)th off(x).

The first derivative :( )

( )df x

f xdx

=

2

( ) ( 1)( ) ( ( ))

n nd f x f x

dx

=

The Higher Derivative

The second derivative :

The nth derivative :

Example: Find if

Answer:

2( )f x dx =

( )( )( )

n

n

n

d f xf x

dx=

xxy sin4 3 +=

212 cos y x x = + 24 sin y x x =

y



27/40

Function defined explicitly

22

cos

y x x

y x

= + +

=

Implicit Differentiation

Differentiate both sides with respect tox.

3 2 210x y x y+ + =

2 2sin( ) 1 xy x y+ = +



28/40

Answer:3 2 2

10x y x y+ + =

3 2 210x y x y+ + = 2 2sin( ) 1 xy x y+ = +

By implicit differentiation find dy/dx ifchapter

4Differentiation

a. b.

a.

3 2 2

3 2 2

2 2 3

3 2 2

2 2

3

( ) (10)

( ) ( ) ( ) (10)

(3 2 ') 2 0

(2 1) 2 3

2 3

2 1

x x

x x x x

D x y x y D

D x y D x D y D

x y x y y x y

x y y x x y

x x yy

x y

+ + =

+ + =

+ + + =

+ =

=

+



29/40

2 2

2 2

sin( ) 1

(sin( ) ) ( 1)cos( )( ) 2 2 0

cos 2 2 cos

x x

xy x y

D xy x D y xy y xy x yy

x x x x

+ = +

+ = +

+ + = +

=

b.

chapter

4Differentiation

2 cos( )

cos( ) 2

x y xyy

x xy y

=



30/40

Tangent line at point (x0,y0)

at point (x0,y0)0 0( ),y y m x x m y = =

Tangent Line

and Normal Line

Normal Line and tangent line perpendicular eachother.

Normal line at point (x0,y0)

at point (x0,y0)0 01

( ), y y x x m ym

= =30

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology


31/40

2 23 4 (2) 3 2 4 2 4 y x x y = = =

6223

+= xxy

Answer:

1. Find the tangent line and normal line at point (2,6) of

Example

24 = xy

)2(46 = xy

1 1 16 ( 2) 6

4 4 2

1 13.

4 2

y x y x

y x

= = +

= +

,

The normal line at point (2,6) :



32/40

2. Find the tangent line and normal line at point whichx = 1

0622

=xyyx

Answer:

If x = 1 is substituted in equation we found

chapter

4Differentiation

)0()6(22

xx DxyyxD =

062

= yy ( 3)( 2) 0y y + = ,y = 3 andy = - 2.

Thus, we get points (1,3) and (1,-2)

By implicit differentiation

2 22 2 ( ) 0 0 xy x yy y xy + + =



33/40

0''2222

=+ xyyyyxxy

222')2( xyyyxyx =

2

2

2

2

y xyy

x y x

=

At point (1,3)

(1,3)

3 2 1 9 15| 3y

= = =

chapter

4Differentiation

3

1

3

1)1(

3

13 == xxy

The tangent line

33)1(33 +== xxy

63 =+ yx

The normal line

83 = yx33



34/40

At point (1,-2)

(1, 2)

2 2.1.4 10

| 22.1.( 2) 1 5y

= = =

The tangent line

chapter

4Differentiation

22)1(22 ==+ xxy

42 =yx

The normal line

2

1

2

1)1(

2

12 +==+ xxy

32=+

yx34



35/40

f xa x x

( );

=+


36/40

Find first derivative of

)12()1()(3

+++= xxxxf

1)(

3 22/1++=

xxxf4

5

Problem Set 2

1

1)(

+=

x

xxf

1)(2

=x

xxf

1

1)(

2

2

+

=

x

xxf

6

7

8



37/40

10(2 3)x=

Find dy/dx

y

x x

x x=

+

+

2

2

2 5

2 39

10

Problem Set 3

y x= sin3

( )xxy = 24 4cos2

1

1

+=

x

xy

11

12

13

1437


2

sin tan( 1) y x x= +


38/40

( )y x= sin 2 1

( )y x= 2 3 4

x=

15 Find the second derivative of

a.

b.

Problem Set 4

x+

1( )y x= cos2

( ) 0f c =

f x x x x( ) = + 3 2

3 45 6

g x ax bx c( ) = + +2

(1) 5, (1) 3g g= = (1) 4g =

16 Find the value of c so that if

17 Find the values of a, b and c if

and .

.

d.



39/40

y

( ) y xy+ =sin 1

x x y y3 2 23 0 + =

Find by implicit differentiation.

18

19

Problem Set 5

xyxyx =+)sin(2

20

21

tan( ) 2 0 xy y =



40/40

sin( ) 1 y xy+ =

22 Let

Find tangent line and normal line at point (,1).

(0) 2, (0) 0 , (0) 3 f g g = = = ( ) (0).f g 23 If , find

Problem Set 6

4040

chapter4 differentiation

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