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BA201 ENGINEERING MATHEMATICS 2 2012 28 CHAPTER 2 DIFFERENTIATION 2.1 FIRST ORDER DIFFERENTIATION What is Differentiation? Differentiation is all about finding rates of change of one quantity compared to another. We need differentiation when the rate of change is not constant. Notation for the Derivative IMPORTANT: The derivative (also called differentiation) can be written in several ways. This can cause some confusion when we first learn about differentiation. The following are equivalent ways of writing the first derivative of y = f(x): ' ' or ( ) or dy f x y dx 2.1.1 RULES OF DIFFERENTIATION A. Derivative of Power Function n y ax So 1 n dy nax dx Examples: 1. Find the derivative of y = -7x 6 Note: We can do this in one step: We can write: OR y' = -42x 5
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Page 1: CHAPTER 2 DIFFERENTIATION - · PDF fileBA201 ENGINEERING MATHEMATICS 2 2012 28 CHAPTER 2 DIFFERENTIATION 2.1 FIRST ORDER DIFFERENTIATION What is Differentiation? Differentiation is

BA201 ENGINEERING MATHEMATICS 2 2012

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CHAPTER 2 DIFFERENTIATION

2.1 FIRST ORDER DIFFERENTIATION

What is Differentiation?

Differentiation is all about finding rates of change of one quantity compared to another. We need differentiation when the rate of change is not constant.

Notation for the Derivative

IMPORTANT: The derivative (also called differentiation) can be written in several ways. This can cause some confusion when we first learn about differentiation.

The following are equivalent ways of writing the first derivative of y = f(x):

' ' or ( ) or dy

f x ydx

2.1.1 RULES OF DIFFERENTIATION

A. Derivative of Power Function

ny ax

So 1ndynax

dx

Examples:

1. Find the derivative of y = -7x6

Note: We can do this in one step:

We can write: OR y' = -42x5

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Example 1 Find the derivative for each of the following function.

a) 3y x b) 52y x c) xy 4 d) 2

1y

x

e) 2y x f)

2

3y x g) 3

2

3y

x h)

5 3

2y

x

i) 1

yx

j) 3

5y

x

k) y x l)

2

3

3

2

y

x

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B. Derivative of a Constant Function

y a

So 0dy

dx

Example Find the derivative for each of the following functions.

a) 1y b) y c) 40y d) 1

3y

2.1.2 THE DERIVATIVE OF SUMMATION AND SUBSTRACTION

If ( ) and ( ) are differentiable functions, the derivative of

y f x g x y f x g x

and

' 'dy

f x g xdx

' 'dy

f x g xdx

Examples:

1. Find the derivative of y = 3x5 - 1

y = 3x5 − 1

Now,

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And since we can write:

So,

2. Find the derivative of

Now, taking each term in turn:

(using )

(using )

(since -x = -(x1) and so the derivative will be -(x0) = -1)

(since )

So

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Example

Find the following derivatives;

a) 42 6y x x

b) 3 32y x x

x c) 33 4s t t d) 2 2 3p q q

e) 8 41 1

34 2

y x x

f)

4 29 5y x x x

f)

1 4 2y x

x g) 3 21 2y x x

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Exercise

Find the derivative of the following function;

i. 4 2f x x x

ii. 116 9f x x

iii. 8 5y x x

iv. 3 1y x

x

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2.1.3 THE DERIVATIVES OF COMPOSITE FUNCTION

Chain Rule

If y f u , where, u is a function of x, so:

dy dy du

dx du dx

This means we need to

1. Recognise u (always choose the inner-most expression, usually the part inside brackets, or under the square root sign).

2. Then we need to re-express y in terms of u. 3. Then we differentiate y (with respect to u), then we re-express everything in

terms of x.

4. The next step is to find du

dx.

5. Then we multiply dy

du and

du

dx.

Example 1:

Differentiate each the following function with respect to x.

i. y = (x2+ 3)5

In this case, we let u = x2 + 3 and then y = u5.

We see that:

u is a function of x and y is a function of u.

For the chain rule, we firstly need to find and .

So

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ii.

In this case, we let u = 4x2 − x and then .

Once again,

u is a function of x and y is a function of u.

Using the chain rule, we firstly need to find:

and

So

i. 3

4y x

ii. 1

4 9y

x

iii. 2 4y x

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The Extended Power Rule

An extension of the chain rule is the Power Rule for differentiating. We are finding the derivative of un (a power of a function):

k

ny ax b

1

11

kn n

kn n

dy dyk ax b ax b

dx dx

dykan ax b

dx

Example:

1.

In the case of we have a power of a function.

If we let u = 2x3 - 1 then y = u4.

So now

y is written as a power of u; and u is a function of x [ u = f(x) ].

To find the derivative of such an expression, we can use our new rule:

where u = 2x3 - 1 and n = 4.

So

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We could, of course, use the chain rule, as before: dx

du

du

dy

dx

dy*

a) 3

4y x

b) 3

5 4y x c) 2 4y x

d) 8

3 6 7y x

e)

5

12

xy

f)

1

7

xy

g) 5

1 3y

x

h)

5

2

9 4y

x

i)

7 23 24 3 2y x x x

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2.1.4 DERIVATIVE OF A PRODUCT FUNCTION

If u and v are two functions of x, then the derivative of the product uv is given by...

In words, this can be remembered as:

"The derivative of a product of two functions is the first times the derivative of the second, plus the second times the derivative of the first."

Example:

If we have a product like

y = (2x2 + 6x)(2x3 + 5x2)

we can find the derivative without multiplying out the expression on the right.

We use the substitutions u = 2x2 + 6x and v = 2x3 + 5x2.

We can then use the PRODUCT RULE:

We first find: and

Then we can write:

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Exercise:

a) 2 34 2 3 5y x x x

b) 2

1 1y x x c) 1 2y x x

d) 10 3y x x

e) 4

3 28 1y x x f) 5 6

5 1 5y x x

2.1.5 DERIVATIVE OF A QUOTIENT FUNCTION

(A quotient is just a fraction.)

If u and v are two functions of x, then the derivative of the quotient u/v is given by...

In words, this can be remembered as:

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"The derivative of a quotient equals bottom times derivative of top minus top times derivative of the bottom, divided by bottom squared."

Example:

1. We wish to find the derivative of the expression:

Solution:

We recognise that it is in the form: .

We can use the substitutions:

u = 2x3 and v = 4 − x

Using the quotient rule, we first need to find:

And

Then

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2. Find if .

Solution

We can use the substitutions:

u = 4x2 and v = x3 + 3

Using the quotient rule, we first need:

and

Then

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a)

2

353 4

xy

x

b) 28 3

2 3

xy

x

c)

432 1

1

xy

x

d) 2

1

2 3y

t

e) 2

5

2 2g s

s s

f)

22 4 3

2 3

x xg x

x

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Challenge

Find the derivative of

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2.1.6 DERIVATIVE OF LOGARITHMIC FUNCTION

If, lny x

1dy

dx x

lny u 1dy

dx u

lnd a

ax bdx ax b

Example:

Differentiate each of the following functions;

i. 2lny x

12

12 1

2

dy dx

dx x dx

x

x

ii. 4lny x

iii. 3

2ln 2 5y x

iv. ln 1 3y x

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Exercise:

1. 4

5lny

x

2. 2

3ln 3 7y x 3. 2ln 3 4y x x

4. ln1

xy

x

5.

4

3ln

5 3y

x

6.

5ln 1 2y x x

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2.1.7 DERIVATIVE OF EXPONENTIAL FUNCTION

If, xy e

So, xdye

dx

Example:

Differentiate each of the following functions;

i. 4xy e

4

4

4

4

4

4

x

x

x

dy de x

dx dx

e

e

ii. 23xy e

2

2

2

3 2

3

3

3

6

6

x

x

x

dy de x

dx dx

e x

xe

iii. 2

3x

y e

2

3

2

3

2

3

2

3

x

x

dy de x

dx dx

e

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Exercise:

i. 5

1 xy e ii. 3

1

2 3 xy

e

iii.

xey

x

iv. 2xy xe v. vi. 2 lnxy e x

31 xy e

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2.1.8 DERIVATIVE OF TRIGONOMETRY FUNCTIONS

If, cosdy

xdx

cosy x sindy

xdx

tany x 2secdy

xdx

Example:

Differentiate each of the following with respect to x;

i.

ii. 3cos2y x

3 sin 2 2

3 sin 2 2

6sin 2

dy dx x

dx dx

x

x

iii. tan 6y x

iv. sin 1y x

v.

vi. 3cos 3y x

vii. 52cos 2 1y x

siny x

2siny x

2cos

2cos 1

2cos

dy dx x

dx dx

x

x

2siny x

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Exercise 1:

i. 2sin 2 4y x x ii. 2

cos 2 1y x iii. tan 2y x

iv. 2

1

siny

x v. 43tan

4

xy vi. 4 1

sin2

y x

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Exercise 2:

i. cosy x x

ii. 2 sin 2y x x iii. sin

1 sin

xy

x

iv. sin 2 cos3y x x

v. tan x

yx

vi. 3 sin 2xy e x

vii. 3

3x

xy

e

viii. lny x x ix.

ln 2xy

x

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2.1.9 PARAMETRIC DIFFERENTIATION

The implicit of relationship of x and y can be expressed in a simpler form by using a third variable, known as the parameter.

Example:

Find dy

dxin terms of the parameter for

1. 2 3,x t y t t

2

2

x t

dxt

dt

3

23 1

y t t

dyt

dx

dx

dt

dt

dy

dx

dy*

t

t2

1*13 2

t

t

2

13 2

2. 22 , 4 4 4x t y t t

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3. , sintx e y t

Exercise:

Find dy

dx in terms of the parameter for

i. 33 2, 1x t y t

ii. 2 35cos , 7sinx t y t iii. 2

1, 1x y t

t

iv. 2 3 33 3 , 3x t t y t t

v. cos2 , 2 sin 2x a y a a vi. 3sin , tx t y e

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2.1.10 SECOND DERIVATIVE

The second derivative is what you get when you differentiate the derivative. Remember

that the derivative of y with respect to x is written dy

dx. The second derivative is written

2

2

d y

dx, pronounced "dee two y by d x squared".

Example:

Find dy

dx and

2

2

d y

dx if

a) 3 32y x x

x

13 32 xxx

22 316 xxdx

dy

2

2 316

xx

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Exercise:

Find dy

dx and

2

2

dx

ydif :

i. 2 345 11f t t t t

ii.2

12y

x

iii. 3 22 21 74 86y x x x

iv. 2

2 1p q

v. 35 2y x x vi. 2 1 3f x x x

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POLITEKNIK KOTA BHARU JABATAN MATEMATIK, SAINS DAN KOMPUTER

BA 201 ENGINEERING MATHEMATICS 2

PAST YEAR FINAL EXAMINATION QUESTIONS 1) Using the suitable method

differentiate the following variables.

a) ( ) ( )

b)

c)

d) ( )

e) √

2) Differentiate the equation below.

a)

b) ( )

c)

d)

e) ( )

f) ( )

3) Derive the equation below:

a) ( )( )

b) √

c) ( )

d) ( )

e)

4) Using the suitable method

differentiate the following variables

a)

b) ( )

c) ( )

d) ( )

e)

f)

( )

5) Differentiate the equation below.

a) (

)

b) √

c)

d)

e)

6) Derive the equation below:

a)

b) ( )

c) ( )

d)

e)

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7) Find

for the following equations

a)

b) ( )( )

c)

d)

e) ( ) ( )

f)

8) Find

for the following equations

a)

b) ( )

c) √

d)

e)

9) Using the right method, differentiate

the functions given.

a) (

)

b) ( )

c)

d) ( )

e)

10) Derive the equation below:

a) √

b)

c)

d) ( )( )

e) ( )

f) ( )

11) Find the dy

dxfor the parametric

functions given below in terms of t.

a) 4 333 5,

4y t x t t

b) 2 25 , lny t x t

c) 2 32 3, 4x t y t t

12) Find the second derivatives for the

function

a) 2

5 2 3 1y x x

b) 3

3 5f x x

c) 2

24 2 3y x x