UNIT 02 - Differentiation

7/27/2019 UNIT 02 - Differentiation

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TOPIC 2

DIFFERENTIATION

1

PREPARED BY MOHD IZWANNI B. SARIMAN

2.0 INTRODUCTION

Differentiation is used widely in solving problems especially in Engineering, Medical,

Commerce, Agriculture, and many other fields. When you find the rate of change of a

quantity with respect to another quantity, you are actually doing differentiation. Another

name for differentiation is the derivative of the function.

If a person is running up a hill, his vertical height will change with respect to his horizontal

height. This rate of change of vertical height with respect to horizontal height is called itsgradient.

heighthorizontalofchange

heightverticalofchangeGradient

Generally, to find the differentiation of y with respect to x, just write the symboldx

dyto

represent it.

Differentiation concepts are applied in displacement and velocity . Velocity, v, is the

differentiation of displacement s, unto time, t, and is written as ds/dt, whilst acceleration,

a, is the differentiation of velocity, v, unto time, t, and is written as dv/dt.

UNIT 2 : DIFFERENTIATION

If y is a function of x, then, y = f(x),

The differentiation of y with respect to x,dx

dy, can be written as f '(x) .

Therefore,dx

dy= f '(x)


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2.1 IDENTIFY BASIC DIFFERENTIATION FORMULAE.

2.1.1THE BASIC RULE OF DIFFERENTIATION

Differentiation of simple algebraic functions is quite easy. Look at the samples below.

If y = x1

,dx

dy= 1x

0= 1.

If y = x2

,dx

dy= 2x

1= 2x.

If y = x3

,dx

dy= 3x

2.

From the pattern shown above, if y = x4

, then,dx

dy= 4x

3.

The first Rule of Differentiation is:

i. Multiply the index of x with the constant for x.

ii. Decrease the index of x by 1 unit.

Thus:

, for all values of n.

, where a is a constant.


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Example 2.1

Finddx

dyfor the cases below :

a)y = x6

dx

dy= 6x6 1 = - 2x5.

b) y = - 2x5

dx

dy= 5 x ( 2 ) x5 1= 10x4

c) y =x

5

y =x

5= 5x 1

dx

dy= 1 (5x 1 1 ) = 5x2 or

2

5

x

c) y = x 1/3

3

2

3

1 x

dx

dy

d) y = 4y = 4 = 4 x 0

dx

dy= 0 (4) x 0 1 = 0

f)

g)

h)

i)

j)


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k)

l)

m)

2.1.2 DERIVATIVE OF A CONSTANT FUNCTION

Example 2.2

Find the derivative for each of the following functions :

a)

b)

c)


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2.1.3 DERIVATIVE OF AN IDENTITY FUNCTION

, where a is a constant

Example 2.3


a)

b)

c)

d)

e)

2.1.4DERIVATIVE OF ADDITION AND SUBTRACTION OF FUNCTIONS.


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Example 2.4


a. y = 2x + 6x

dx

dy(4)(2)x

3+ 1(6)x

0

= 8x3

+ 6

b. y =x

xx3

23

dx

dy3(x)

22 (x)

0+ 3( -1)x

-2

= 3x2

- 2 -3x2

c. y = 45 21

x

dx

dy 45

2

1 12

1

x

= 42

52

2

2

1

x

= 23

2

5 x =

2

3

2

5

x

d. y = 7 - 9x

dx

dy727 x

2=27 x

2

2.1.3APPLY THE DIFFERENT TECHNIQUES OF DIFFERENTIATION

(i) DERIVATIVE OF A COMPOSITE FUNCTION.

The Rules you have learned in the earlier part of this unit however may not be able to solve

compound functions like y = (1+x)4

or21 xy .

So, we can say for , where k is more than 2 , fraction number andnegative number, we can solve the equation by using :

a) Chain Rule

b ) Extended Power Rule (E.P.R)

1 23

4


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a) Chain Rule.

The Chain Rule states that if y = f (u), where u is a function of x, then

Example 2.5

Finddxdy for the following functions.

a) y = ( 2x + 3 )Let u = 2x + 3 ,

then y = u4

2dx

du, 34u

du

dy

dudy

dxdu

dxdy

)4(2 3udx

dy

b) 42 xy Let u = 42 x ,

then y = 21

u

2

dx

du

x ,

2

1

2

1

udu

dy

du

dy

dx

du

dx

dy

)2

1(2 2

1

uxdx

dy

()


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b) ALTERNATIVE WAY (DIFFERENTIATION OF A COMPOSITE FUNCTION) / EXTENDED

POWER RULE

The examples shown above can also be solved using an alternative way to use the Chain

Rule. Where the solution :-

Using same question as above but use altenative way :-

)()()( 1 baxdx

dbaxkbax

dx

d nknkn

a. 4)32( xdx

d= 14)32(4

x )32( xdx

d= 2

11 )(

knn baxkanx

= 2)32(4 3 x

= 3)32(8 x

)()()( 1 baxdx

dbaxkbax

dx

d nknkn

b. 42 xdx

d= 2/12 )4( x

dx

d= )4()4(

2

1 221

2

xdx

dx

11 )( knn baxkanx

11)(

knn baxkanxdx

dy

)()()( 1 baxdx

dbaxkbax

dx

d nknkn


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= xx 2)4(2

1 2/12

=

42 xx

Example 2.6

Find the derivative for each of the following functions by using Extended Power Rule :

a. y = ( 2x + 3 ) )32()32(4 3 x

dx

dx

dx

dy

= )2()32(4 3x = 8(2x + 3)

3

b. y = ( 2 + x )

)2()2(3 222 xdx

dx

dx

dy

= 3( 2 + x2

)2

( 2x )

= 6x (2 +x2

)2

= 6x ( 4 + 4x2

+ x4

)

= 24 x + 24x3

+6x5

(ii) DERIVATIVE OF PRODUCT OF FUNCTIONS

If y is formed by multiplying the functions u and v, i.e. y = uv, then

(PRODUCT RULE)

dx

duv

dx

dvu

dx

dy


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Example 2.7

Find

dx

dyfor the following functions

a) y = x4 ( 3x + 5 )Let u = x4 and v = ( 3x + 5 )6

Therefore, 34xdx

du and )3()53(6 5 x

dx

dv

Therefore, )4()53()53)(18(

3654

xxxxdx

dy

= )4()53()53(18 3654 xxxx

= )53(29)53(2 53 xxxx

= 1069)53(2 53 xxxx

dx

dy= 1015)53(2 53 xxx

b) y = ( x 4 )2

( 2x3

3x + 5 ) . y = ( x

4 )2

( 2x3

3x + 5 )

Let u = ( x4 )2

and v = 2x33x + 5

Therefore, )1)(4(2 xdx

duand 36 2 x

dx

dv

Therefore, )4(2)532()36()4( 322 xxxxxdx

dy

)]532(2)36)(4[()4( 32 xxxxx

dx

dy)2292410()4(

23 xxxx

c) y = . y = Let u = 2x and v = (3x + 1)

1/2

dx

duv

dx

dvu

dx

dy

(3x +5)5 Greatest

Common Factor.


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Therefore, 2dx

duand 2

1

1)-(3x2

3

dx

dv

Therefore, 21)+(3x1)-(3x2

32 2

1

2

1

xdx

dy

1)+(3x21)-(3x3 21

2

1

xdx

dy

1)+(3x231)-(3x 21

xdx

dy

291)-(3x 21

xdx

dy

13

29

x

x

dx

dy


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(iii) DERIVATIVE OF QUOTIENT OF FUNCTIONS

Ify =

v

u, where u and vare functions ofx, then

(QUOTIENT RULE)

Example 2.8

Differentiate the following:

a) y =10)1( x

x

Let u = x and v = ( x + 1)10

Then, 1dx

duand 9)1(10 x

dx

dv

Using the Product Rule2v

dx

dvu

dx

duv

dx

dy

,

Then,210

910

))1((

10)1()1()1(

x

xxx

dx

dy

210

910

))1(()1(10)1(

xxxx => ]

)1(10)1([)1(20

9

xxxx

dx

dy=

11)1(

)91(

x

x

2v

dx

dvu

dx

duv

dx

dy


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b) y =1

)12( 43

x

x

Let u = ( 2x 3 + 1 )4 and v = x

1

Then, )6()12(4 233 xxdx

du and 1

dx

dv

Using the Product Rule2v

dx

dvu

dx

duv

dx

dy

Then,2

43332

)1(

)1()12()12)(24)(1(

x

xxxx

dx

dy

2

323

)1(])12()1(24[)12(

xxxxx

=2

3233

)1(

]122424[)12(

x

xxxx

dx

dy=

2

3233

)1(

]122422[)12(

x

xxxx


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2.2 APPLY THE DIFFERENTIATION FORMULA FOR TRIGONOMETRIC, LOGARITHMIC

AND EXPONENTIAL FUNCTIONS.

2.2.1 APPLY THE FORMULA TO SOLVE RELATED PROBLEMS(a) DERIVATIVE OF TRIGONOMETRIC FUNCTIONS

So far, you have learned how to differentiate algebraic functions. However, there are

functions in other forms that require different formulas to solve. You are going to learn

how to find the derivative of trigonometric functions. Basically, you need to remember the

following formula:

xx

dx

dcossin

xecxecx

dx

dcotcoscos

xxdx

dsincos

xxx

dx

dtansecsec

xxdx

d2sectan

xecx

dx

d 2coscot


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Example 2.9

Find

dx

dyfor each of the following functions:

a. y = sin 3xBy using the Chain Rule,

du

dy

dx

du

dx

dy.

Let u = 3x then y = sin u

3

dx

duand u

du

dycos

udx

dycos3

xdx

dy3cos3

Or by using the extended trigonometric rule

:

b. y = cos ( 3x +1 )

Let u = 3x + 1 , then y = cos u

Then, 3dx

du, udu

dysin

)sin(3 udx

dy => )13(sin3 x

dx

dy


:

c. y = tan ( x 2 + 2 )Let u = x

2+ 2 , then y = tan u

xdx

du2 u

du

dy2sec


:


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uxdx

dy2sec2 =>

)2(sec2 22 xxdx

dy

Note: From the answers in the example above, you will find that the following are true:

By making use of the Chain Rule, the derivative of trigonometric functions can be done

using the following properties:

)cos()sin( baxabaxdx

d

)sin()cos( baxabaxdx

d

)(sec)tan( 2 baxabaxdx

d

xxnxdx

dnn cossinsin 1

xxnxdx

dnn sincoscos 1

xxnxdx

d nn 21 sectantan

and

)cos()(sin)(sin 1 baxbaxanbaxdx

dnn

)sin()(cos)(cos 1 baxbaxanbaxdx

dnn

)(sec)(tan)(tan 21 baxbaxanbaxdx

dnn


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Example 2.10

Find

dx

dyfor each of the following functions.

a) Let , When

,

and

By using Chain Rule :

,

TRIGONOMETRY IDENTITIES

1. 1cossin 22 2. 22 tan1sec 3. 22 cot1cos ec

4. cossin22sin 5. 22 sincos2cos

2

sin21

1cos2 2

6.

2

tan1

tan22tan

7. sincossina Rb 8. sincossina Rb 9. cossincosa Rb

10. cossincosa Rb


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b)

Let , When , and By using Chain Rule :

,

c) Let , When , and By using Chain Rule :

,


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d) 3

Let , When , and

, (cancel out 2 and 8)

(b) DERIVATIVE OF LOGARITHMIC FUNCTIONS

Lets move on to logarithmic functions. To find their derivatives, you will need another setof formulas. Study the following formula:

xx

dx

d 1ln

Before we want to solve the logarithmic function questions, we need to know some natural

logarithm rules and properties. Then, apply the rules and properties to the questions and

solve the question by using differentiation.

, where a and b is a constant


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Rule name Rule

Product rule ln(x y) = ln(x) + ln(y)

Quotient rule ln(x / y) = ln(x) - ln(y)

Power rule ln(xy) = y ln(x)

Example 2.11

Find the derivative of the following logarithmic functions:

a) y = ln x4 => y = 4 ln xUsing the formula,

Therefore,

)1

(4xdx

dy =>

x

4

b) y = ln ( 2x2 + 5 )3 (Power Rule)y = 3 ln ( 2x

2+ 5 )

Therefore, )52

4(3

2

x

x

dx

dy

52

122

x

x

c) y = ln x31 = 21

31ln x (Power Rule)

y = )31ln(2

1x

Therefore, )31

3(

2

1

xdx

dy

)31(2

3

x

d) y = lnx

2 (Quotient Rule)

y = ln 2 ln x

Therefore,xdx

dy 10

x

1

e)

(Lets try this in class) f)

(Lets try this in class)

xx

dx

d 1ln


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(c) DERIVATIVE OF EXPONENTIAL FUNCTIONS

An exponential function is a function in the form y = ex

. To find its derivative, you will have

to remember the following rule:

Given y = ex

, thenxx ee

dx

d

andaxax aee

dx

d

baxbax aeedx

d

Example 2.12

Differentiate these functions:

a) y = e3xxe

dxdy 33

b) y = e1 - 2xxe

dxdy 212

c) y = e 2x +1 d)


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122

xedx

dy

e)

(Lets try this in class) f)

(Lets try this in class)

2.2.2 USE VARIOUS DIFFERENTIATION TECHNIQUES TO SOLVE PROBLEMS

We also can use other method such as product rule and quotient rule to solve the

trigonometric, logarithmic and exponential functions.

Example 2.13 :

a) By using Product Rule :

, ,

b) By using Product Rule :


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,

,

[ ]

c) By using Quotient Rule :

,

, ,

d)

By using Product Rule :


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,

,

e) By using Chain Rule:

Let , When , and , f)

By using Extended Power Rule :

g) By using Product Rule :


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,

,

h)

By using Quotient Rule :

, 2 ,

()

,


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2.3 UNDERSTAND PARAMETRIC EQUATION.

If a curve is defined parametrically by two equations, x=f(t) and y=g(t), where f and g are

functions of parametert, the derivative can be obtained by applying the chain rule.

For example, ifx = 2at2

and y = 4at, find

In this case, = 4at and so =

Also .

Hence:

Example 2.14 :(Lets try this in class)

Find in terms of t

a) b)

Ans :

Ans :


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c) d)

e) f) Ans : Ans :

Ans : Ans :


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2.4 SECOND ORDER DIFFERENTIATION

We can find the second derivative by using various technique of differentiation.

Finding the second derivative for parametric equation is a little trickier.

We use the fact that:

For example, to find the second derivative in the above example, therefore:


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Activity 2(a)

1. Find the derivatives (differentiation) of the cases below:

a. 4x b. 2x5c.

x

6d. x

8

e. 10 f. 5x1/5

2. Differentiate these functions with respect to x:

a. 4x5

b.6

5

2

x

c.34

1

xd.

2

3

x

e.25

12

x e. 5

24

5

3x


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Activi ty 2(b)

Finddx

du,

du

dyand

dx

dyfor

a. 8)2

( xx

y b. y = (5xx3

)5

c. xy 2 d. )3( xxy

e.3

12

xy f.

2

3 2x

xy

Feedback for Activity 2(b)

a. 122 x , 8u

7

, )12

()2

(8 27

xxx

b. 53x2, 5u

4, 5(5xx

3)4

( 53x2

)

c. 1, 21

2

1 u ,

x

22

1

d . 2x + 3, 21

2

1 u , )32()3(

2

12

1

2

xxx

e. 2x , 23

2

1 u

, 23

2 )3(

xx

f. 3x2

- 4x 3, 2

1

2

1 u ,

23

32

22

43

xx

xx


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Activity 2(c)

Differentiate the following

a. y = x ( 4 x )8b. y = ( 3x + 1 )3 ( x + 2 )5c. y = xx 5)12(

d. y = xx 6)12( 2

e. m = (n3 )( 52n )6

Feedback for Activi ty2(c)

a. (4x )7 (49x )b. (3x + 1 )2 ( x + 2 )4(24x + 23)

c.x

xx

2

)122()12( 4

d. x

xx

62

)1049)(12(

e. (52x )5

(4114x)


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DIFFERENTIATION


Activi ty 2(d)

Find the derivatives of the following functions

a.x2

3

b.232

1

t

c.

1

2

z

z

d.1

12

x

x

e.4

32

x

x

f.

1

12

k

k

Feedback for Activity 2(d)

a.2

)2(

3

x

b.22)32(

6

t

t

c.2)1(

2

z

d. 22

2

)1(

12

x

xx

e.2)4(

11

x

f.3

)1(2

52

k

k


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Activity 2(e)

Find the derivative of the following trigonometric functions

a. sin 5x

b. cos4

x

c. tan 3x

d. 2 tan x + 3 sin 2x

e cos ( 2x5 )

f. sin x + tan ( 2x + 1)g. sin ( 34x

3)

h. cos ( 3x2

+ 2 )

i. tan ( 2x2

+ 3x + 1 )

2. Differentiate these trigonometric functions with respect to x

a. sin 3 4xb. cos 6 2xc. tan 2 3xd. cos 2 ( x 2 + 1 )e. sin 2 ( 2x + 1 )f. cos ( x 3)


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Feedback for Activi ty2(e)

1. a. 5 cos 5x

b.4

sin4

1 x

a. 3 sec 2 3xb. 2 sec 2 x + 6 cos 2xc. 2 sin ( 2x5 )d. cos x + 2 sec

2

( 2x + 1 )

e. 12 x2 cos ( 3 4x3 )f. - 6x sin ( 3x2 + 2 )i. ( 4x + 3 ) sec

2(2x

2+ 3x + 1 )

2. a. 12 sin2

4x cos 4x

b. 12 cos5

2x sin 2x

c. 6 tan 3x sec2 3xd. 4x cos (x

2+ 1 ) sin ( x

2+ 1 )

e. 4 sin ( 2x + 1 )cos ( 2x + 1)f. 3x 2 sin ( x3 )


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Activi ty 2(f)

1. Differentiate the following functionsa.ln 3xb.ln ( 12x )c.

4)35(

3ln

x

d.

)34(

9ln

x

e. ln (x1 )(x + 2)5

Feedback for Activity 2(f)

1. a.x

1

b.x21

2

c.x35

12

d.x34

3

e.)2)(1(

)12(3

xx

x


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Activi ty 2(g)

1. Differentiate these functionsa. e 2xb. e x/3c. e 3x + 1d. e 12xe. 2e3x + 8e 2xf. e

xe

x

Feedback for Activity2(g)

1. a. 2e2x

b. 33

1x

e

c. 3e3x + 1

d. -2e1 2x

f. 6e3x

-16e2x

g. ex

+ ex


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SELF ASSESSMENT 2(a)

1. Finddx

dyfor each of the following

a. y = 10x5b. y = 3

4

1x

c. 5x 3

d. y = - 6x-2

2. Differentiate these functionsa. y = 2x

2x + 1

b. y = ( x4 )2

c. s = t3

( t + 4 )

d.x

xxy

23

e. z = ( k + 1 )( 2k3)

f. p = )3

2(2

qqq

3. Finddx

dyfor each of the following

a. y = ( 7x3 )5b. p = 5 ( q3 )

3

c.5)23(

4

xy

d. y = ( 3x25 )

7

e.

)12(

1

x

y

f.x

xy

33


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SOLUTIONS TO SELF ASSESSMENT 2(a)

1. a. 50x4b. 2

4

3x

c. 15x 4

d. 12x

2. a. 14 xdx

dy

b. )4(2 x

dq

dp

c. )3(4 2 ttdt

ds

d. xdx

dy2

e. 14 kdk

dz

f. 36 2 q

dq

dp

3. a. 35 ( 7x 3 )4

b. 15(q3 )2

c. 40(32x )-6

d. 42x(3x2

5)6

e.3)12(

1

x

f .2

32

xx


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SELF ASSESSMENT 2(b)

1. Use the Product Rule to find the derivatives of these functions

a. 5)1( xx

b. 1xx

c. 5)3)(2( xx

d. 132 xx

e.722 )3( xx

f. )3()52( 22 xx

2. Use the Quotient Rule to find the derivatives of these functions

a.2)1( x

x

b.22 )3( k

k

c.1

3

l

l

d.1

1

p

p

e.)1(

512

2

m

m

f.

1

52

z

z


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SOLUTIONS TO SELF ASSESSMENT 2(b)

1. a. ( x1 )4

( 6x1)

b.12

23

x

x

c. ( x3 )4

( 6x + 7 )

d.1329

xx

e. 2x( x23 )

6( 8x

2- 3 )

f. 2 (2x5 ) ( 4x25x + 6 )

2 a.3)1(

1

x

x

b.3

2

)1(

)1(3

k

k

c.2)1(2

23

l

l

d.2)1(

2

p

e.22 )1(

14

m

m

f.3)1(2

)32(3

z

z


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SELF ASSESSMENT 2(c)

1. Differentiate these functions

a. cot xb.

x

xsin

c. tan 2 ( 5x + 3 )d. sin 2 x + 2 cos 2( x- 1)

2. Find the derivative for

a. ln 7xb. 1ln 2 x c.

2)34(

2ln

x

d. ln (2x3)(x + 5)4

3. Differentiate with respect to x

a. e x + 1b. ex + e-xc. ( ex + e-x )2d.

x

x

e

e

2

1


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TOPIC 2

DIFFERENTIATION

44


SOLUTIONS TO SELF ASSESSMENT 2(c)

1. a. cosec2

x

b.2

sincos

x

xxx

c. Let y = tan 2 ( 5x + 3 )dy/dx = 5(2) tan (5x + 3 ) sec

2(5x + 3 )

= 10 tan ( 5x + 3 ) sec2

( 5x + 3 ) .

d. sin 2x10 cos4 (x1 ) sin ( x 1)

2. a.x

1

b.12 x

x

c.34

8

x

e. Let y = ln (2x 3)(x + 5)4Therefore, y = ln ( 2x 3) + 4 ln (x + 5 )

Then, dy/dx = )5

1(4

)32(

)2(1

xx

=)5)(32(

)32(4)5(2

xx

xx

)5)(32(

128102

xx

xx

=)5)(32(

)15(2

xx

x


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TOPIC 2

DIFFERENTIATION

45

3. a. ex + 1

b. ex

- e-x

c. Let y = ( exe

x)

2

Therefore y = e

x

. e

x

+ e

x

. e

- x

e

x

e

x

e

x

e

x

= e2x

+ e2x

2dy/dx = 2e

2x+(-2 ) e

2x

= 2 ( e2x

e-2x

)

d.2)2(

22x

xx

e

ee

UNIT 02 - Differentiation

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Transcript of UNIT 02 - Differentiation