the derivative. definition example (1) find the derivative of f(x) = 4 at any point x

48
The Derivative

Upload: kristopher-bryan

Post on 27-Dec-2015

220 views

Category:

Documents


0 download

TRANSCRIPT

The Derivative

Definition

0

0

0

0

,

)()(

:lim

lim0

xatfofderivativethecalledisexistsif

xt

xftf

itThe

fdomxLet

xt

Example (1)Find the derivative of f(x) = 4 at any point x

0

44

)()()(

0lim

lim

lim

xt

xt

xt

xt

xt

xftfxf

Example (2)Find the derivative of f(x) = 4x at any point x

4

)(4

44

)()()(

4lim

lim

lim

lim

xt

xt

xt

xt

xt

xt

xt

xt

xt

xftfxf

Example (3)Find the derivative of f(x) = x2 at any point x

x

xx

xt

xtxt

xt

xt

xt

xftfxf

xtxt

xt

xt

xt

2

))((

)()()(

)lim(

lim

lim

lim22

Example (4)Find the derivative of f(x) = x3 at any point x

2

22

22

22

33

3

)(

))((

)()()(

lim

lim

lim

lim

x

xxxx

xtxt

xt

xtxtxt

xt

xt

xt

xftfxf

xt

xt

xt

xt

Example (5)Find the derivative of f(x) = x4 at any point x

3

2

22

22

22

2222

44

4

)2(2

))((

))((

))()((

))((

)()()(

lim

lim

lim

lim

lim

x

xx

xxxx

xtxt

xt

xtxtxt

xt

xtxt

xt

xt

xt

xftfxf

xt

xt

xt

xt

xt

Example (6)Find the derivative of f(x) = 3x3 + 5x2 - 2x + 7 at any point

2109

2)2(5)3(3

)4(&)3((

2)(5)(3

)(2)(5)(3

72537253

)()()(

2

2

2233

2233

2233

2323

2limlim5lim3

lim

lim

lim

lim

xx

xx

examplespreviousfromxt

xt

xt

xt

xt

xt

xt

xt

xt

xtxtxt

xt

xxxttt

xt

xftfxf

xtxtxt

xt

xt

xt

xt

QuestionsFind from the definition the derivative of each of the following functions:

1 .f(x) = √x

2 .f(x) = 1/x

3 .f(x) = 1/x2

4 .f(x) = 5 / (2x + 3)

Power Rule

Let:

f(x) = xn , where n is a real number other than zero

Then:

f'(x) = n xn-1

If f(x) = constant , then f'(x) = 0

Algebra of Derivatives

)(

)()()()()()(.4

)()()()()()(.3

)()()(.2

)()()()(.1

:

tanint)()(

2 xg

xgxfxfxgx

g

f

xfxgxgxfxfg

xfcxcf

xgxfxgf

Then

tconsacandxpoaatexistxgandxfthatAssume

Example (1)

)()(.5

)()(.4

)()(.2

)(,)(.1

:

85)(

&374)( 23

xg

f

xf

g

xfg

xgxf

followingtheofeachfind

xxg

xxxfLet

Solution

2

232

223

223

223

22

23

)85(

)5)(374()1412)(85()()(.4

)374(

)1412)(85()5)(374()()(.3

)1412)(85()5)(374()()(.2

5)(

1412)2(7)3(4)(.1

:

85)(

&374)(

x

xxxxxx

g

f

xx

xxxxxx

f

g

xxxxxxfg

xg

xxxxxf

followingtheofeachfind

xxg

xxxfLet

The Chain RuleThe derivative of composite function

for the case f(x) = gn(x)

Let:f(x) = gn(x)Then:f' (x) = ngn-1(x) . g'(x)

Example:Let f(x) = (3x8 - 5x + 3 )20

Then f(x) = 20 (3x8 - 5x + 3 )19 (24x7 - 5)

Examples (1)

36

7

5 9

9

7

5 9

9

5 99 59

5 99 599

9

45

32)(.3

32)(.2

1111)(.1

:

xx

xxxf

xxxf

xxxxxx

xxxf

functionsfollowingtheofeachofderivativetheFind

379 3)1()(.7 xxxxxf

73

3)(.6

12

2)(.5

12)(.412)(.4

9

24

4 58

4 584 58

xx

xxxf

xxxf

xxxfxxxf

514

914

54

94

910

98

59

95

59

95

91

91

5

9

9

5

5

99

5

9

1

9

199)(

11

11)(.1

108

99

5 99 5

5 9

9 5

9

99

9

xxx

xxxxxxf

xxxxxxxx

xxx

xx

xx

xxf

514

59

59

5

2718327)(

32

32)(.2

86

9

79

7

5 9

9

xxxxxf

xx

xxxf

514

59

59

59

5

271832745

43045332)(

4532

45

32)(.3

86

9316

252167

9

3167

9

36

7

5 9

9

xxxxxx

xxxxxxxf

xxxx

xx

xxxf

116124

5)(

12

12)(.4

78

8

4 58

41

45

xxxxf

xx

xxxf

116122

5)(

122

12

2)(.5

78

8

4 58

49

45

xxxxf

xx

xxxf

29

82439

9

24

73

)39)(3()212)(73()(

73

3)(.6

xx

xxxxxxxxf

xx

xxxf

]3

139)1()1(7[

]3)1([2

1)(

]3)1([

3)1()(.7

3

28769

2

1

3

179

2

1

3

179

379

xxxxx

xxxxxf

xxxx

xxxxxf

Example (2)

)()(.4

)(

)()(.3

)()()(.2

)()()(.1

:)3(:

1)3(,2)3(,5)3(,6)3(

2

2

xhxxxf

xh

xgxf

xgxxxf

xhxgxf

casesfollowingtheofeachforfFind

hhggLet

Solution

)()()(

1)3(,2)3(,5)3(,6)3(

:

xhxgxf

and

hhgg

haveWe

16)5(2)1(6

)3()3()3()3()3(

)()()()()(.1

ghhgf

xgxhxhxgxf

)()1()(

,5)3(,6)3(

:

2 xgxxf

and

gg

haveWe

603030

)16()6()5()39()3(

)12()()()()(

)()()(.22

2

f

xxgxgxxxf

xgxxxf

Example (3)

28)7(4)2())1(

)2()1()2(

)2())2(()2(

)())(()(

))(()(

:

)2(

7)2(,1)2(&4)1(

))(()(

hg

hgf

hhgf

xhxhgxf

xhgxf

Soluion

fFind

hhg

xhgxf

Homework

exx

xxxf

xxxxf

xxxxf

xx

xxxf

xxxf

xxxf

functionsfollowingtheofeachofderivativetheFind

33

52)(.6

123

2)(.5

123)(.4

42

1810)(.3

1810)(.2

1810)(.1

:

2

24

4 525

4 525

2

35

4

9 5

5 2

100

9 5

5 2

9 5

5 2

Answers of Questions(1)

Find from the definition the derivative of

each of the following functions:

1 .f(x) = √x

2 .f(x) = 1/x

3 .f(x) = 1/x2

4 .f(x) = 5 / (2x + 3)

1

0

2

11

1

))((

)()()(

limlim

limlim

xallforabledifferentiisf

xxx

xtxtxt

xt

xt

xt

xt

xftfxf

xtxt

xtxt

2

0

1

1

11)()(

)(

2

lim

lim

limlim

xallforabledifferentiisf

x

tx

xttx

tx

xtxt

xt

xftfxf

xt

xt

xtxt

3

0

22

)(

)(

))((

11)()(

)(

322

22

22

22

22

22

22

lim

lim

lim

limlim

xallforabledifferentiisf

xxx

xxx

xxxt

tx

xtxt

txtx

xtxt

tx

xtxt

xt

xftfxf

xt

xt

xt

xtxt

4.

2

3.

)32(

10

)32)(32(

110

)32)(32(

110

)32)(32(

1

)32)(32)((

)32)(32(

22

)32)(32(

)32()32(

32

5

32

5)()(

)(

2

lim10

lim10

lim5lim5

limlim

xallfordiffisf

xxx

xtxt

xtxt

tx

xt

xt

tx

xt

xt

tx

xtxt

xt

xftfxf

xt

xt

xtxt

xtxt

Differentiability & Continuity1 .If a function is differentiable at a point, then it is continuous at that point.

Thus if a function is not differentiable at a point, then it cannot be continuous at that point.

But the converse is not true. A function can be continuous at a point without being differentiable at that point.

2 .A point of the graph at which the graph of the function has a vertical tangent or a sharp corner is a point where the function is not differentiable regardless of continuity

Examples(1) Sharp Corner

2;

2;8

3

)(

:

xx

xxf

Let

This function (Graph it!) is continuous at the point x=2, since the limit and value of the function at that point are equal ( Show that!) but it is not differentiable at that point, since the right derivative

of f at x=2 is not equal to the left derivative a that point.

existnodoesf

Thus

tt

ftf

atfofderivativeLeft

thatShowt

t

t

ftf

atfofderivativeLeft

xxx

xx

)2(

'

002

88

2

)2()(

2

!122

8

2

)2()(

2

limlimlim

limlim

222

3

22

Examples(2) Vertical Tangent

When both right and left derivatives are +∞ or both are - ∞

3)(

:

xxf

Let

This function (Graph it!) is continuous at the point x=0, since the limit and value of the function at that point are equal ( Show that!) but it is not differentiable at that point, since the right derivative (and also the left derivatives) of f do not exist at x=2 ( both are +∞)

3 20

3

0

3

00

3 20

3

0

3

00

1limlim

0

0lim

0

)0()(lim

0

1limlim

0

0lim

0

)0()(lim

0

tt

t

t

t

t

ftf

xatfofderivativeleftThe

tt

t

t

t

t

ftf

xatfofderivativerightThe

tttt

tttt

Examples(3) Cusp

When the one of the one-sided derivative is +∞ and the other is- ∞

3 2)(

:

xxf

Let

This function (Graph it!) is continuous at the point x=0, since the limit and value of the function at that point are equal ( Show that!) but it is not differentiable at that point, since at x=2 the right

derivative does not exist ( is + ∞) and also the left derivatives does not exist and is(∞-

30

3 2

0

3 2

00

30

3 2

0

3 2

00

1limlim

0

0lim

0

)0()(lim

0

1limlim

0

0lim

0

)0()(lim

0

tt

t

t

t

t

ftf

xatfofderivativeleftThe

tt

t

t

t

t

ftf

xatfofderivativerightThe

tttt

tttt

Limits Involving Trigonometric Functions

All trigonometric functions are continuous a each point of their domains, which is R for the sine & cosine functions(→The limit of sinx and cosx at any real number a are

sina and cosa respectively)

, R-{ π/2, - π/2, 3π/2, - 3π/2,…………} for the Tangent and the Secant functions

and R-{0, π, - π, 3π, - 3π ,…………} for the Cotangent and the Cosecant functions.

Important Identity

111

cos

1sin

cos

1sincossin

tan

).1(.2

1tan

.2

1sin

.1

limlim

limlimlim

lim

lim

00

000

0

0

xx

x

xx

x

xxx

x

x

fromroofP

x

x

x

x

xx

xxx

x

x

Examples (1)

filesPDFothertwoonexamplesmoreSee

xxx

xx

xxx

xx

xxx

xxx

xx

xx

x

x

x

x

x

x

xx

xx

x

xx

xx

x

3

22

30

)1(2)1(20

35

tan2

55sin

20

35

tan2

5sin4

35

tan25sin4

35

tan25sin4.2

7

5)1(

7

5

5

5sin5

5

5sin

7

5sin.1

limlimlimlim

lim

limlim

lim75

lim71

lim

0

6

0

006

0

70

70

00

0

Example (2)

fofcontinuitytheeInvestigat

xf

Let

xx

x

x

xx

x

0;sin

0;0

0;1

sin

2

4

)(

:

f is continuous for all x other than zero. To check, whether it is continues, as well at x=0, we need to that its limit at x=0 is equal to f(0), which is given as zero.

Solution

0

)0(0)(

)(01

sin)(

0)0(1

sin

sinsin)(

lim

limlim

limlim

limlimlim

0

4

00

02

2

0

2

2

0

2

00

xatcontinuousisf

fxf

theoremsqueezethebyx

xxf

xx

x

xx

x

x

xxf

x

xx

xx

xxx

Example (3)

0

)(

:

0;1sec8

0;3sin1 22

xatcontinuousisfwhetherCheck

xf

Let

xx

xxx

For f to be continuous at x=0, we need its limit at x= 0 to exist and to equal the value at that point, which is 9. Since its right limit at x=0 is equal also 9, it remains that its left limit at that point be equal to that value.

Solution

9)1)(1(93

3sin

3

3sin9

93

3sin

3

3sin3sin

1)(

:Pr

.9)0()(

:

91)1(8)1sec8()(

91)1(810sec8)0(

limlim

limlimlim

lim

limlim

00

02

00

0

00

2

x

x

x

x

x

x

x

xx

xxf

oof

fxf

needWe

xxf

f

xx

xxx

x

xx

Example (4)

membersucheachofformulatheWrite

functioncontinuousaconstituemembersitsofwhichFind

xf

functionsoffamilyfollowingtheConsider

xcx

x

xcxcx

.

)(2;

)2cos(

2;)sin(1

22

All members are continuous for all x other than 2. For a member to be continuous at x=2, we need the limit of the member function at x= 2 to exist and to equal the value at that point, which is 1/2c. Since the right limit at x=2 of any member is equal o 1/2c, it remains that its left limit at that point be equal to that value.

Solution

2222

1

2

1:

2

1)(

)sin(

)/()sin(

)sin(1

)(

)2()(

:

)2(2

1

2

0cos)2cos()(

2

1

2

0cos)2cos()2(

limlim

limlimlim

lim

limlim

22

222

22

2

22

ccccc

needWe

ccx

cx

cx

cxcx

cxcx

cxxf

fxf

needWe

fcccx

xxf

cccx

xf

xx

xxx

x

xx

2;2

)2cos(

2;)2sin(4

12

)(

:

xx

x

xxx

xf

membercontinuoustheofformulaThe