AM015 LECTURE & TUTORIAL NOTES

AM015/7. Limits

CHAPTER 7: LIMITS

LECTURE 1 OF 5
At the end of lesson, student should be able to :

7.1 a) State limit of a function f(x) as x approaches a given value a, lim f (x)  L
xa

7.1 b) State the basic properties of limit

7.1 LIMITS
Definition of Limit
A function f(x) is said to approach a constant L as a limit when x approaches a as below,

lim f (x)  L

xa

We are interested not in the value of f (x) when x  a but in the behavior of f(x) as x

comes closer and closer to a.

Means that as x gets closer to a, but x  a , f(x) gets closer to L, read as “the limit of f as

x approaches a is L”.

Notation means x approaches a from the right
means x approaches a from the left
x  a means x approaches a from both sides
x  a

x a

The limit of a function can be evaluated by using different method

(1) Intuitive method
(2) By using properties

(1) Intuitive Method

Computation of a limit (intuitive approach)
(a) A table of values of x versus f(x) is drawn up
(b) The values of f(x) are observed to see whether they approach a particular

value.
Step 1 : Select several values of x close enough to a.

Step 2 : Observe the pattern of corresponding f  x values

Step 3 : Then guess the value of f  x when x is approaching a.

Page 1 of 15

AM015/7. Limits

Example 1

Find lim (x2 1) =
x0

Solution

xapproaching0 from the left xapproaching 0 from the right
0 0.001 0.25 0.5
x -1 -0.5 -0.25 -0.001 1
f(x)= x2+1 y

x

(2) By using properties
Properties of limit

Properties of Limits Example
(1) If f (x) = c , where c is a constant
lim 5 
lim f (x)  lim c  c
x3

xa xa

(2) If f (x) = x, then lim x  a lim x 
xa
x3
(3) If f (x) = xn, where n is a positive integer (n
lim x3 
> 0), then lim x n  a n
x a x2

4 lim f (x)  g(x)  lim f (x)  lim g(x) lim(x5 10) 
xa xa xa
x1

5 lim f (x).g(x)  lim f (x).lim g(x) lim(x2  5)(x 1) 
xa xa xa
x4

Page 2 of 15

AM015/7. Limits

(6) If c is a constant, then lim 5x3 

lim c. f (x)  c.lim f (x) x10

xa xa

7 lim f (x) lim f (x) provided lim x
xa g(x)  xa x2 10
x5
lim g(x)
xa

lim g(x)  0

x a

(8) If f is a polynomial function, then lim(3x2  4x 10)

lim f (x)  f (a) x2

xa lim x  4

(substitution) x0

(9) lim n f (x)  n lim f (x) where n is a
xa xa

positive integer and f(a)>0

Note: By substituting a in the lim f(x)  L , then L is the exact value of f(x) when x = a and
xa

the approximate value of f(x) when x approaches a, i.e. the limit is equal to exact
value.

Limit of the Rational Function

1) For a rational function, the limit can be found by substitution when the denominator is not

zero. If f(x) and g(x) are polynomials and c is any number, then

lim f (x)  lim f (x) f (c) provided g(c)  0

xc

xc g(x) lim g(x) g(c)
xc

Example 2

Find the limits (if exist)

a lim 1 b lim 3x  4
x2
x1 x3 x  2

EXERCISE

Find the limits of b 2x2  x  3
x3  4
a  lim x2 x4 lim
x2  2x 1
x2 x3

Page 3 of 15

QA016/7. Limits

LECTURE 2 OF 5

At the end of lesson, student should be able to :

lim f (x)
7.1 c) Find xc
lim g(x) when lim f (x)  0 and lim g(x)  0
xa xa

xc

*Use the following methods: i. factorisation; and ii. multiplication of conjugate

lim f (x) 0

2) If xc (indeterminate form), this function is undefined when x = c, but its limit may

lim g(x) 0
xc

exist, then we use:

(i) factorization method

(ii) multiplication of conjugates method

(i) Finding limit using factorization method
Step 1: Factorize the numerator and denominator and
Step 2: Simplify common / like factor

Example 1 blim x2  x  6
Find the limits (if exist)
x3 3  x
a lim x2  9

x3 x  3

 c  lim x3  8  d  lim e2x 1
x2  4 ex 1
x2 x0

Example 2

If f (t) = 2t2 + 1, determine the value of lim f (t)  f (2) if exists
t2 t  2

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QA016/7. Limits

(ii) Finding limit using multiplication of conjugates method for functions related to surd
Step 1: Multiply both numerator and denominator by the conjugate of numerator or

denominator.
Step 2: Simplify the fraction.

Example 3 blim x  2
Find the limits
x2 x  2  2
a lim 2  x 1

x1 x 1

clim x 1

x1 2( x 1)

EXERCISE 2. Find the limits
1. Find the limits (if exist)
 a  lim  x  3
a lim 4x2  x  3 x3
x 3
x1 x 1
b lim1 x 1
blim 2x2  5x  2
x0 x
x2 x  2

c lim (3  t)2  9

t0 t

Page 5 of 15

QA016/7. Limits

LECTURE 3 OF 5

At the end of lesson, student should be able to :

7.1 d) Find one-sided limits. i. lim f (x)  L ; and ii. lim f (x)  M
xa xa

7.1 e) Determine the existence of the limit of a function. * lim f (x)  lim f (x)
xa xa

**Exclude: End point limit such as lim x, x  0 , where lim x  0 but lim x does
x0 x0 x0

not exist.

One - Sided Limits f (x)  x2  9
Consider the graph below x3

y

6

3

3x

left hand limit Right hand limit
lim f (x)  6 and lim f (x)  6
x3 x3

One sided limit
Left hand limit Right hand limit

lim f (x)  M lim f (x)  L

xa xa

Example 1 b lim(x  2) 
Find x3

a lim(x  2) 
x3

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QA016/7. Limits

Existence of Limits

lim f (x)  lim f (x)
Limit exist if and only if xa xa

lim f (x)  L , lim f (x)  L
xa
xa

lim f (x)  L
xa

Example 2

If f (x)  3x2 1, find the left-hand and right hand limit of f at x= 4 and hence determine

lim3x2 1

x4

Estimating a limit from a graph

Example 3
Use the graph below to estimate each limit at x= -1, 0, 2, 4

y

2

1

02 4 x

Page 7 of 15

QA016/7. Limits

Example 4

y 
2

1
x

1 2 34

The above diagram shows the graph of the function f. Find

lim f (x)

(a) x1

lim f (x)

(b) x2

lim f (x)

(c) x3

Limit for Piecewise Function

Example 5

 4  x 2 ;x 1 lim h(x).
(a) Given h( x) 2  x 2 . Find x1
;x 1

Page 8 of 15

QA016/7. Limits

f (x)  2 ; x 1
5 ; x  1. Find lim f (x) .
(b) Given
x1

Example 6

ax  3 ;x2
If f (x)  bx2  c ;2 x4
;x4
8

Determine the value of a, b and c given that lim f (x)  5 , lim f (x) and lim f (x) exist.
x 2 x2 x4

Page 9 of 15

Limit for Absolute Values QA016/7. Limits

Example 7 b lim 3 x 1
Find the limit (if exist)
x1 x 1
alim x 3
x 3

clim (x  3)(x  2)

x2 x  2

EXERCISE b lim  x2 1 
x2
1. Find

a lim  x2 1 
x2

2. Given

f  x  2x x4 lim f (x)
2x . Find
 3 x4
x4

f ( x)  3x  3 ;xk
 ;xk
3. If  x2 +5

Determine the values of k if lim f (x) exist.
xk

Page 10 of 15

QA016/7. Limits

LECTURE 4 OF 5

At the end of lesson, student should be able to :

7.1 f) Find infinite limits. * i. lim f (x)   and ii. lim f (x)   *Sketch of the
xa xa

graphs may be necessary to explain (i) and (ii). * lim f (x)   and lim f (x)  
xa xa

are undefined limits.

7.1 g) Find limits at infinity i. lim f (x)  L and ii. lim f (x)  M.
x x

Infinite Limits

Consider the graph of f x  1 , as x approaches 0, x2 also approaches 0, and 1 becomes
x2 x2

very large.

x f x

1 1
 0.5 4
 0.1 100
 0.01 10 000
 0.001 1 000 000

f  x  1 From the graph our informal definition is
x2
i)as x  a , f(x) increases without bound, then we
a write lim f (x)  
a
xa

ii) as x  a , f(x) increases without bound, then
we write lim f (x)  

xa

iii) Therefore lim f (x)   (undefined)
xa

From the graph our informal definition is

i)as x  a , f(x) decreases without bound, then we
write lim f (x)  

xa

ii) as x  a , f(x) decreases without bound, then
we write lim f (x)  

xa

iii) Therefore lim f (x)   (undefined)
xa

Page 11 of 15

a QA016/7. Limits

Example 1 From the graph our informal definition is
Find one sided limit
i)as x  a , f(x) increases without bound, then we
1 write lim f (x)  
(a) lim
xa
x2 x  2
ii) as x  a , f(x) decreases without bound, then
Example 2 we write lim f (x)  
Find limit if exist
xa
(a) lim 1
x1 x 1 iii) Therefore lim f (x) does not exist.
xa

1
(b) lim

x2 x  2

(b) lim 2 1 22
x
x2

Limits At Infinity
y

y =L
x

Page 12 of 15

QA016/7. Limits

1. (a) If x moves increasingly far away from the origin in the positive direction (i.e. x   )
and f(x) gets closer to L, then we write lim f (x)  L

x

(b) If x moves increasingly far away from the origin in the negative direction (i.e. x   )
and f(x) gets closer to L, then we write lim f (x)  L

x

(c) In case 1(a) and 1(b), L is known as the limit to infinity. L must be a real number.

2. If lim f (x)   or lim f (x)   , we say that the limit does not exist.
x x

Limit at infinity for lim  f (x)  g(x)
x

Example 3 (b) lim 1 x2
Find the limit x

a lim(ex  2)
x

(c) lim (2  x)
x

EXERCISE

1. Find the limits (if exist)

(a) lim x  3 (b) lim x2  x  2
x3 x  3 x2  2x 3
x3

2. Find

 a x2 3 b
lim (  ) lim x2  x 10 
x  x2 x

Page 13 of 15

QA016/7. Limits

LECTURE 5 OF 5

At the end of lesson, student should be able to :

7.1 h) Determine lim f (x) when lim f (x) and lim g(x) are undefined. *Emphasise i.
x g(x) x x

lim  1   0 and ii. lim  1   0 for n>0. *Exclude: Vertical and horizontal asymptotes.
x xn  x xn 

Limit at Infinity for Rational Function

lim 1 0 lim 1 0
xn xn
1. For n >0 , x and x

Example 4

Find lim f (x) and lim f (x) .
x x

(a) f (x)  1 (b) f (x)  3 5
x2 x2 x3

2. If lim f (x)   , then we will have to divide the numerator and denominator by the
g(x) 
x

highest power of x of the denominator.

Note 

:

is an indeterminate form.

Example 5

Find

 a  lim x2 1 b lim  x2 
2x2 1 x  
x  3x 1 

Page 14 of 15

Example 6 QA016/7. Limits

Find limit if (exist) b lim x2  2

a lim x 1 x 3x  6
x x2 1

3. For the function, if lim  f (x)  g(x)     (indeterminate form)
x
i) Multiply by the conjugate
ii) Divide the numerator and denominator by the highest power of x of the denominator.

 Example 7
lim 4x2  x  2x 

Find x

EXERCISE

Find

 a  lim 2x3  x2  3 b lim x
x3  x2 x  3x2 1
x

 c  lim x2 1 d lim x2  x  6
2x2 1 3 x
x x

e lim x 1 f  lim x2  2
x x2 1
x 3x  6

Page 15 of 15

AM015/ 7.Limit

TUTORIAL 1 OF 6

Learning outcomes:

7.1 a) State limit of a function f(x) as x approaches a given value a, lim f  x  L
xa

7.1 b) State the basic properties of limit

1. Evaluate the following limit, if exist

(a) lim12 (b) lim 3x  2
x2
x5

(c) x2 1 (d) 3x2  2
x2 1 x2  4
lim lim

x2 x 2

(e) lim y3 (f) lim x2  7x 10
3y  x5 x  3
y3 y3 9

(g) lim h (h) lim  y  35
h4 h  4 y4

   2. If lim f x  2 and lim g x  64 , find
x1 x1

(a) lim gx  4 lim fx
x1 x1

fx
 (b) lim

x1 g x

(c)  limf x g x   f x 2 

x1

 f x  5
 3. If lim
x4 x  2  1 , find lim f x
x4

ANSWER

1. (b) 8 (c) 5 2 (e) 0 (f) 0 (g) 1 (h) 1
3 3
(a) 12 (d)

3

2. (b) 1 (c) 20
(a) 56 32

3. 7

Page 1 of 9

AM015/ 7.Limit

TUTORIAL 2 OF 6

Learning outcomes:

lim f  x lim f  x  0and lim g x  0
7.1 c) Find xa when
xa xa
lim g  x
xa

*Use methods: i. factorisation

ii. multiplication of conjugate

1. Evaluate the following limit, if exist

(a) lim x2  4x  5 (b) lim 2x2  x  1
x1 x  1 x2  1
x1

(c) lim x3  4x2  2x  8 (d) lim x3  27
x4 x4 x3 x  3

(e) lim x2 (f) lim p  9
x3 8 p9 p  3
x2

(g) lim x  7 (h) lim 2  x
x7 3x  4  5 x4 4  x

(i) lim x2  25 (j) lim x  1  1
x5 2x  1  3 x2 x  2

(k) lim t 2  81 (l) x  2  2x
t9 3  t x2  2x
lim

x2

1. (a) 6 (b) 3 ANSWER (d) 27 (e) 1
(f) 6 2 (c) 18 (i) 30 12

(k) 108 (g) 10 (h) 1 (j) 1
3 4 2

(l)  1
8

Page 2 of 9

AM015/ 7.Limit

TUTORIAL 3 OF 6

Learning outcomes:

7.1 d) Find one-sided limits. i. lim f  x  L ; ii. lim f  x  M .
xa xa

7.1 e) Determine the existence of the limit of a function. *

lim f x  lim f x
xa xa

3x2  6 , x2

.

x2
 1. x 
Given that f  8  x, Evaluate lim f x , if exists.

x2

x  4 x  4
 2.  4  x
Given f (x) x  4 . Evaluate lim f x , if exists.
x4

3. Determine the limit (if exist) at the given x

(a) f x  x  2 at x  2
(b) h x  3 2x  4 at x  2

(c) f  x   x  3 x  2 , at x  2

(x  2)

x2  2 , 1  x  2
4. Given the function as, f (x)  1 2x , 2  x  4 determine the limits at (if exist) ;

 x ,x4

(a) x = 2 (b) x = 4 (c) x = 9

 2  x, x  1

5. Given f  x    x2 1 1 x  2 Evaluate each of the following limits, if exists.
 2x 1 2x4.

 1x8 x4
 4

State the reason if the limit does not exist.

(a) lim fx (b) lim fx (c) lim fx (d) lim fx
x1 x2 x3 x4

ANSWER

1. 6

2. does not exist

3. (a) 0 (b) 3 (c) does not exist

4. (a) does not exist (b) does not exist (c) 3

5. (a) does not exist (b) 5 (c) 7 (d) 9
Page 3 of 9

AM015/ 7.Limit

TUTORIAL 4 OF 6

Learning outcomes:

7.1 e) Determine the existence of the limit of a function. * lim f  x  lim f  x
xa xa

 2x 4 xk
kx  5k, xk
 1.
Given f (x)  . Find the value of k>0 if lim f x exist.

xk

2. Determine the values of A and B if the limit of f(x) exist at the points x = -1 and x = 4

4 x  1
f(x)  Ax  B
1 x  4

Ax2  Bx  1 x  4

 x3  c x0
Given f (x)  ax2  b
3. 0  x  2 determine the value of c if lim f (x)  2 . Find
x0
 2x  2 x2


a and b if lim f (x) and lim f (x) exist.
x0 x2

 x ,x 1

4. Given f  x   3 , x  1 . Determine the limit at x 1 (if exist)

2x 1 , x 1

5. Evaluate limit at x 1 for the function f  x   2x 1 , x 1
 , x 1
 0

9  x2 x3
x3
6. A function is defined as f  x  x6 3
162x  54x2
 x3  27 x3

Determine the limit (if exist) at x=3

1. 1 c=2 ANSWER
3. a = 2 , b = -2 2. A= 3,B=-7
5. 1 4. 1
6. -6

Page 4 of 9

AM015/ 7.Limit

TUTORIAL 5 OF 6

Learning outcomes:

7.1 f) Find infinite limits. * i. lim f  x   and ii. lim f  x   .
xa xa

   7.1 g) Find limits at infinity i. lim f x  L and ii. lim f x  M
x x

1. Find the limits (if exist):

2  1 2
 1 x 1
3 x2
 alim  x   9 (b) lim x 

x3 x1

(c) lim 3  x2 (d) lim x  4
x4 x 2  16 x1 x2  2x  3

2. For the graph of the function given below, find the limit (if exist)

f(x)

2

1

012 3 4x

(a) lim f x (b ) lim f  x (c) lim f  x

x 1 x2 x 3

3. Use the graph below to determine the indicated limits

y

4

-4 -2 1 4x

-1

-2

a) lim f (x) b) lim f (x) c)lim f (x) d)lim f (x) e) lim f (x)
x4 x2 x1 x4 x
Page 5 of 9

AM015/ 7.Limit

 x , x 1

 x 1, 1  x  3
4. The function f(x) is defined as f (x)  
8  2x, 3  x  6
 4, x  6

 Find (i) lim f x (ii) lim fx (iii) lim fx
x1 x x3

ANSWER

1. (a)  (b)  (c)  (d) 

2. (a) 1 (b) does not exist (c)  (undefined)

3. (a)  (undefined) (b) does not exist (c) does not exist (d) 2 (e) 2

4. (a) does not exist (b) -4 (c) 2

Page 6 of 9

AM015/ 7.Limit

TUTORIAL 6 OF 6

Learning outcomes:

   7.1 g) Find limits at infinity i. lim f x  L and ii. lim f x  M
x x

   lim f x
 7.1 h) Determine x , lim f x and lim g x are undefined.
lim g x x x

x

1. Find the limits ( if exist):

(a) lim 2t 2 (b) lim 16x2  4 (c) lim x2  4
2 x 4x2  3 x x  2
t  t 6
(e) lim 2x3  x2  1
(d) lim (x  1)2 x 2x3  1 (f) lim 5x3  4x  9
x2  x 1 7  3x3
x 2 (h) lim x2 1 x
x 4x 1
(g) lim 8x 1 (i) lim x2 1
x x2  9 (k) lim 4x3  5x2  1 x 4x 1
x 9x3  x
(j) lim 2x 1  (l) lim x x  2  x
x 4  x2 x

 x3 , x1
 x2  9


Function f is defined as  10  px, 1x3
 2. f x   x2 3x5

 x x5

 2x2  1

(a) Evaluate,

(i) lim fx
x5

(ii) lim fx
x3

(iii) lim fx
x

 (b) Find the value of p if lim f x exists
x3

ANSWER

1. (c)  (d) 1 (e) 1 (f)  5 (g) 8 (h) 1 (i)  1
(a) 2 (b) 4 2 3 44

(j) 2 (k) 2 (l) 1
3

2. (a) (i) 25 (ii)  1 (iii) 0 b) 1
63

Page 7 of 9

AM015/ 7.Limit

EXTRA EXERCISES

SESSION 2017/2018

1. Evaluate each of the following limits, if exists

(a) lim 2x 6 3
x2  2x 
x3

(b) lim 3x3 11x2  x  1
4x  2x3
x

 ax  7, x  k

2. Given f  x   3,  k x 1
 x2 1, 1 x k

3x 1, k  x  5
 16, x5

(a) Find the values of k if lim f  x exist
xk

(b) Hence, by using the value of k >0, find the value of a if lim f  x exist.
xk

SESSION 2018/2019

1. Find lim x  x2  5x
x

 x  4, x  1
2. Given f  x   A, 1 x  B where A and B are constants.
 x  1,
xB

(a) Find the values of A and B if the limit of f x exist at x  1 and x  B

 (b) Hence, find lim f x
x6

SESSION 2019/2020

 3x x3

 2x2  7x  3

 m, x  3
n
1. A function is defined as f x    x 3x5
7

 x7 5x7

 x 7

 p  2 x 7

Where m, n and p are constants.

 (a) Find lim f x
x3

     (b) Determine the values of m and n such that lim f x  f 3  lim f x
x3 x3

   (c) Find the values of p such that lim f x  lim f x
x7 x7

Page 8 of 9

AM015/ 7.Limit

SESSION 2020/2021

1. Determine lim x  9
x9 6  x  3

2. Find lim 2x1 5x
x
7x2  3

 x2  2x 8 x4
 x4
 4x
3. The function is defined as f  x   m

 6x  n x4



Where m and n are constants

If lim f  x  f 4 and lim f  x  f 4 , find the values of m and n.
x4 x4

ANSWERS

SESSION 1 (b)  3 2. (a) 3 4
2017/2018 2
1. (a) (b)
SESSION
2018/2019 2 3
1.  5
SESSION 2. (a) A  5, B  4 (b) 7
2019/2020 2
(b) n 8 m1 (c) p  2  2 7
SESSION 1. (a)  1 35 5
2020/2021 5
2. 2 6 3. n  30 m  6
1.  10
7

Page 9 of 9

AM015/8. Differentiation

CHAPTER 8: DIFFERENTIATION

LECTURE 1 of 5

LEARNING OUTCOMES:

At the end of this topic, student should be able to;
8.1 a) Find the derivative of a function f(x) using the first principle
8.2 a) Apply the rules of differentiation:

i) Basic rules
ii) Sum rules

8.1 DERIVATIVE OF A FUNCTION

In calculus, a branch of mathematics, the derivative of a function is a measurement of how the
function changes when the values of its input change. The process of finding derivative of a
function is called differentiation.

Let say, we have two points, P(x, f (x)) and Q(x  h, f (x  h)) which lies on the curve y  f (x) .
h is use to denote a small increment of length in the direction of the x-axis.

Thus the gradient of the chord is given by Recall: m  y2  y1
x2  x1
P  QN
PN

 f (x  h)  f (x)
xhx

 f (x  h)  f (x)
h

Now, if we move Q nearer to P , say to point Q1 , Q2 ,…, h  0 , then the gradient of the chords
P, PQ1, PQ2 ,…will give better and better approximations for the gradient of the tangent at P
and therefore, the gradient of the curve at P .

Page 1 of 15

AM015/8. Differentiation

By definition,

f ' ( x)  lim  f (x  h)  f ( x) 
 h 
h0

= the gradient of the chord PQ as Q moves nearer to P

=the gradient of the tangent line at P

=the gradient of curve y  f (x) at P

=the derivative of a function f (x) with respect to x

The process of finding derivative y  f (x) using definition f '(x)  lim  f (x  h)  f (x)  is
 h 
h0

called as differentiation using the first principle.

 dy means y differentiated with respect to x .
dx

 The 'd' is the ‘operator’, operating on some function of x . e.g. d  x2   2x

dx dx

 Though we choose to use a fractional form of representation, dy is a limit and it is not a
dx

fraction, i.e. dy does not mean dy  dx
dx

 The common ways to denote the derivative of a function y  f x are as follows:

i. f xread as f prime x

ii. dy read as dydx
dx

iii. y read as y prime

 The differentiation of a function should be with respect to the independent variable for
example, if

i. y  f xthen dy  f x

dx

ii. y  htthen dy  h(t) and so on.

dt

Example 1

Find the derivatives of the following functions using the first principle.
(a) f x  5x  4 (b) y  2x 1x  5

Page 2 of 15

AM015/8. Differentiation

Example 2

If f (x)  4x  2x2 , find f '(x) from first principle. Hence calculate f '(2) and f '(2) .

8.2 RULES OF DIFFERENTIATION

a) Rules of Differentiation
i) Basic Rules
1) The Derivative of a Constant Function (Constant Rule)

If k is a real number, the derivative of

f (x)  k is f '(x) = 0

In other words, the derivative of a constant function is zero.

Example 3 (b) y  3 (c) f (x)  e3
Find the derivatives of

(a) f x  8

2) The Derivative of a function of form xn (Power Rule)

If n is a real number, the derivative of

f x  xn is f x  nxn1

3) The Derivative of a Function of the form kxn (Constant Multiple Rule)

If k and n is a real number, the derivative of
f (x)  kxn is f '(x)  knxn1

Example 4 (b) f x  1
Find the derivatives of
x2
3

(a) y  x 2

Page 3 of 15

AM015/8. Differentiation

ii) The Derivative of a Sum or Difference of Functions (Sum or Difference Rule)

If u(x) and v(x) are differentiable functions, the derivative of
f (x)  u(x)  v(x) is f (x)  u(x)  v(x)
f (x)  u(x)  v(x) is f (x)  u(x)  v(x)

Example 5 (b) g(x)  x3  1 x2  4
Find the derivatives of 2 4x

(a) f (x)  2x2  5 x

Exercise

1. Find the derivatives of the following functions using the first principle.

(a) y  53x ans: -3

(b) y  x2  2x ans: 2x  2

2. Find the derivatives of ans: 0

(a) f (x)   ans: 6

(b) mx  12 3

x x2

2

(c) y  8 x  2 5 1 ans: 8 4  20 x 3
3x2 3 3x3 3
 4x3

Page 4 of 15

AM015/8. Differentiation

LECTURE 2 of 5

LEARNING OUTCOMES:

At the end of this topic, student should be able to;
8.2 a) Apply the rules of differentiation:

iii) Chain rules
iv) Product rules
v) Quotient rules

8.2 RULES OF DIFFERENTIATION

iii) The Chain Rule

If y  f u is a differentiable function of u and u  g(x) is a differentiable function
of x , therefore if y  f gxis a differentiable function of x , then

dy  dy  du
dx du dx

or, equivalently, d  f gx  f gx.gx

dx

Example 1

Find dy if y  u3  3u2 1 and u  x2  2
dx

In general, any composite function of the form y   f (x) n , involving some function f (x)

raised to a rational power n , is called the General Power Rule, and it is a special case of the

Chain Rule.

Let y  un , where u  f (x) ,
Then dy  nun1 , and du  f '(x) ,

du dx

Using the Chain Rule, dy  dy  du
dx du dx

 So dy  nun1  f '(x), substitute u  f (x) , therefore
dx

General Power Rule

 d  f (x) n n f (x)n1  f '(x)

dx

Page 5 of 15

AM015/8. Differentiation

Example 2 (b) y  2
Find the derivatives of
1
 (a) y  2x4  9x  6 4  3x2 1 5

iv) The Derivative of a Product (Product Rule)

If u(x) and v(x) are differentiable functions, the derivative of
f (x)  u(x)v(x) is f (x)  u(x)v(x)  v(x)u(x)

The Product Rule can be written in function notation as
(u v)'  u v'  vu '

Example 3

Differentiate each of the following functions

 (a) y  x 1 13x2 (b) p(x)  (3x2 1)(7  2x)3

(c) f (x)   1 1(x 1)
x 

Page 6 of 15

AM015/8. Differentiation

v) The Derivative of a Quotient (Quotient Rule)

If u(x) and v(x) are differentiable functions, the derivative of

f (x)  u(x) is f '(x)  v(x)u'(x)  u(x)v'(x)
v(x)
v(x)2

The Quotient Rule can be written in function notation as

u   v.u  u.v
 v  v2

Example 4 (b) y  (x2 1)
Differentiate each of the following functions, (x4 1)

(a) y  1 x 1x 1

 x  33
(c) f (x)  4  3x2

Exersice
Find the derivatives of

 x  3  x2 1 3  ans: x2 1 2 6x3 13x2  72x  7
 (a)f   x  4 
(x)  x  42

(b) f (x)  x1  x 2 1  x2  2x 1

ans:  x 12

Page 7 of 15

AM015/8. Differentiation

LECTURE 3 of 5

LEARNING OUTCOMES:

At the end of this topic, student should be able to;

8.2 b) Perform Second Order Differentiation

8.3 a) Find the differentiation of:
i) ex
ii) ef(x)

b) Second Order Derivatives

Consider the function y  f (x) .Differentiate y with respect to x , the first derivative of f ,

that is dy  f (x) is obtained. Differentiate dy with respect to x again, the second derivative
dx dx

of f , that is d  dy   d2y  f (x) obtained.
dx  dx  dx2

First derivative y f x dy or d  f (x)
f  x or
dx dx

Second derivative y d  dy   d2y d2  f (x)
dx  dx  dx2 dx2

Example 1

Find the first and second order derivatives of p(x)  2x4  9x3  5x2  7.

Example 2

Given y  ax2 b, show that x2 d2y  2y
x dx2

Page 8 of 15

AM015/8. Differentiation

8.3 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(a) Differentiation of Exponential Functions

Let y  ex  loge y  x Change index form to logarithmic form
ln y  x
Differentiate both sides with respect to y
1  dx
y dy

dy  1
dx  dx 

 dy 

But  1
1
 y 

y

 ex

Hence d ex   ex

dx
The Chain Rule can be used to differentiate exponential functions of the form of y  e f x ,

where f x is some functions of x .

Let y  eu , where u  f x.

Then dy  eu and du  f x.

du dx

Using the Chain Rule,

dy  dy  du
dx du dx

 eu  f (x)

 e f (x)  f (x)

 f (x)e f (x)

Therefore, (Derivative of the power) x
(Original exponential function)
d e f x   f xef x
dx

Derivative of Exponential Functions

If y  ex then dy  ex
dx

If y  e f (x) then dy  f (x)e f (x)
dx

Page 9 of 15

AM015/8. Differentiation

Example 3

Differentiate each of the following with respect to x

(a) y  e3x4 (b) y  5e3x

Example 4

Find the derivatives of the following functions

(a) f (x)  e3x 2x 1 (b) f ( x)  e2x
1 ex

Example 5

Given that y  2e2x  3e3x . Show that d2y  dy  6 y  0
dx2 dx

Exercise ans: 24(2x 1)

1) Find f (x) for f (x)  (2x 1)3

 (4x 1) dy and d2y dy  2(2x2  x  6)
(x2  3) dx dx2 . dx x2  3 2
2) Given y , find  ans:

d2y  2(4 x3  3x2  36x  6)
dx2 x2  3 3
 ans:

3) Find the derivatives of the following functions  ans: 2x2e2x 2  x2

 (a) f (x)  e2x 2  x2

(b) f x   2 1 ex ans:  ex 2x  2  x2
 x
x2

Page 10 of 15

AM015/8. Differentiation

LECTURE 4 of 5

LEARNING OUTCOMES:

At the end of this topic, student should be able to;
8.2 b) Perform Second Order Differentiation
8.3 a) Find the differentiation of:

iii) ln x
vi) ln f(x)

8.3 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(a) Differentiation of Logarithmic Functions

The natural logarithmic function ln x is differentiable for all x  0.
y  ln x

dy  1
dx x

Natural logarithmic functions of the form y  ln f x, where f x is some functions of x , can

be differentiated using the Chain Rule.

Let y  lnu where u  f x.

Then dy  1 and du  f x.

du u dx

Using the Chain Rule Derivative of Logarithmic Functions

dy  dy  du If y  ln x then dy  1
dx du dx dx x

 1  f (x) If y  ln f (x) then dy  f (x)
u dx f (x)

 1  f (x)
f (x)

 f (x)
f (x)

Example 1

Differentiate the following with respect to x .

(a) y  ln 5x (b) y  ln3x 14 (c) g(x)  ln3 1 2x

Page 11 of 15

AM015/8. Differentiation

Example 2
Differentiate the following functions

(a) f x  ln  x  3  2x  53  (b) f x  ln  x 1 
  1 x 

Example 3 (b) y  ln x
Find the derivatives of y ex

(a) y  x ln x

Example 4

If y  ln  x  , prove that d2y  (4 x  1)  dy 2 0
 2x 1  dx2  dx 

Exercise ans: 2ex  1  ln x 
1) Differentiate the following functions  x 

(a) f (x)  ex ln x2 ans: 1 ln(x 1)
(x 1)2
(b) f (x)  ln(x 1)
x 1 Ans: 2x

 2) Differentiate y  log10 x2 1 with respect to x . (ln10)(x2 1)

Page 12 of 15

AM015/8. Differentiation

LECTURE 5 of 5

LEARNING OUTCOMES:

At the end of this topic, student should be able to;
8.4 a) Solve the first derivatives implicitly

8.4 IMPLICIT DIFFERENTIATIONS

(a) Implicit Function

So far differentiation has been done involving functions of the form y  f (x) , which is said to
define y explicitly as a function of x .

Examples: y  3x2  7x 1, y  ex 3lnx2 1

An implicit function is one in which a relationship between x and y is given without having y

as an explicit or clearly defined function of x . y is defined implicitly as a function of x .

Examples: y2  3yx3  x 1, y x  y2  6

An implicit function involving y and x can be differentiated with respect to x as it stands.

To differentiate y with respect to x : d y  dy
dx dx

To differentiate yn with respect to x : d yn  nyn1 dy
dx dx

Proof:

y  f (x)

yn   f (x)n

d yn  dy  f (x)n Use Chain Rule

dx dx

 n f (x) n1 f (x)

 nyn1 dy
dx

We can generalize this as follows:

To differentiate a function of y with respect to x , we differentiate with respect to y and then

multiply by dy .
dx

Page 13 of 15

AM015/8. Differentiation

(b) Implicit Differentiation
Suppose we want to differentiate the implicit function

y2  4x  6x2  3y 1. Differentiate both sides of the
equation with respect to
d  y2  d 4x  d 6x2  d 3y
dx dx dx dx

2y dy  4  12x  3 dy 2. Collect all terms with on one side of the equation
dx dx

2y dy  3 dy  12x  4
dx dx

2y  3 dy 12x  4 3. Factor out

dx

dy  12x  4 4. Solve for
dx 2y  3

Example 1 (b) x3  y3  2xy  8

Find dy in term of x and y if

dx
(a) x2  y2  8

Example 2 (b) x2 y  y2 ln x  x

Find dy in term of x and y if

dx

(a) yey  ex  x

Example 3

Given ye2x  2x 1. Prove that dy  2e2x  2y .
dx

Page 14 of 15

Exercise AM015/8. Differentiation

1) Given that y3  x2 y  x  3y2  0, ans: 2xy 1
(a) find dy in terms of x and y 3y2  x2  6y
dx
(b) evaluate dy when y  3 ans:  1 , 82
dx 99

2) Find dy for the function ln(xy)  2exy ans:  y
dx x

3) If y  x2 y  4 , show that  1  x2  dy  2xy  y .
x  x  dx x2

Page 15 of 15

AM015/ 8: Differentiation

CHAPTER 8: DIFFERENTIATION

TUTORIAL 1 of 7

1. By using first principles, find dy for the following functions.
dx

(a) y = 3x +1 (b) f (x) = 4x2 − 7x + 5

(c) y = (2x − 3)(x + 2) ( )(d) f x = 3x2 − 2x + 1

2. Find the derivative in each case (b)f (x) = x4 − 2x2 + 3

(a) y = 6x4 + 4x2 + 3x + 5 (d)f (x) = x4 − 4x3

(c) f (x) = 3x5 − 6x4 (f) y = 1 − 1
x4
2 x + x3

(e) f (x) = x3 + x +1

3. Differentiate each function with respect to x.

(a) x4 + 3x2 − 5x (b) (x +1)(x − 3) x2 − 5x + 3

(c)

x3

(d) y = x2 +1 (e) y = 2 x(x3 + x2 +1)
x

Answer: (c) 4x+1 (d) 6x-2
1 (a) 3 (b) 8x-7
(b) 4x3 − 4x (c)15x4 − 24x3
2 ( a)24x3 + 8x + 3
(e) 2 + 1 (f) −4 − 1 + 1
(d) 4x3 −12x2 2x x5 2x
1 2

3x3 3x3

3 (a) 4x3 + 6x − 5 (b) 2x − 2 (c) − 1 + 10 − 9
x2 x3 x4

(d) 3 x− 1 53 1
2
3 (e) 7x2 + 5x2 + x

2x2

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AM015/ 8: Differentiation

TUTORIAL 2 of 7

1. Find dy for the following function
dx

(a) 1 (b) y = (2x5 − x2 )8 (c) y = 2 (d) y = 6x2 −1
5x + 2 (3x3 − 6)4

2. Use product rule to differentiate the following. (d) y = x3 x2 + 2

(a) y = x x −1 (b) y = x(5 − x)3 (c) y = (x2 +1)2 (1+ x)3

3. Differentiate the following by using quotient rule

2 2

a) y = x b) y = 6x3 (c) y =  3x + 2
x2 +1 x3 + 2x2  2−x

(d) y = x (e) y = (x2 + 4)3
1− x 3x3 + 4

4. Show that: (b) d  3x 3  = 3(3 − x2 )
dx  x2 +  (x2 + 3)2
( )a) d ( x − 9)(x + 2)4 = (x + 2)3(9x − 70)
dx 2 x − 9

Answers: (b)16x(2x5 − x2 )7 (5x3 −1) −72x2 6x
1 (a) − 5 (c) (3x3 − 6)5 (d)
(5x + 2)2 6x2 −1

2 3x − 2 (b) (5 − 4x)(5 − x)2 (c) (x2 +1)(1+ x)2[7x2 + 4x + 3]

(a)

2 x −1

2x2 (2x2 + 3)
(d)

x2 + 2

35 16(3x + 2)
−2x3 (7x + 8)
1 (c) (2 − x) 3 1
(b) (d) 3
(a) 3 (x3 + 2x2)2 2 x (1− x) 2

(x2 +1)2

3x(x2 + 4)2 (3x3 + 8 −12x)
(e) (3x3 + 4)2

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AM015/ 8: Differentiation

TUTORIAL 3 of 7

1. Find f '(x) and f "(x) for each of the following functions

(a) y = 3x3 + 2x2 + 4x + 7 (b) y = 4(x2 + 6) (c) f (x) = x2
x 2 + x2

2. If y = 3x + 2 , prove that x2 d2y + x dy − y = 0 .
x dx2 dx

3. Given y = Ax + Bxex with A and B are constants.

Show that x2 y"− (x2 + 2x) y '+ (x + 2) y = 0 .
dy

4. Find for the following functions.

dx

(a) y = e3x (b) y = 5e2x2 (c) y = eln x2

(d) y = ex + 1 (e) y = 1 (f) y = eax2 +bx
ex ex

5. Given that y = 3xe−2x , find dy and d2y.
dx dx 2

( )6. Given y = Ax + B ex , find the constants A and B if y = 3 and dy = 5 when x = 0.
dx

Answers: (b) 8 x− 3  ; 81+ 6  x(4 + x2) ; 2(4 − x2 )
1 x2  x3  c) 3
5
(a) 9x2 + 4x + 4 ; 18x + 4
(2 + x2)2 (2 + x2)2
4 (a) 3e3x (b) 20xe2x2
(c) 2x (d) ex − 1 (e) −1 (f)
(2ax + b)eax2 +b ex 2 ex
5 3e−2x (1− 2x) ; −12e−2x (1− x)

6 A = 2,B = 3

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AM015/ 8: Differentiation

TUTORIAL 4 of 7

1.Differentiate each function with respect to x.

(a) ln(4x + 1) (b) ln 3 − x2 (c) ln x 4 (d) (ln x)2 (e) ln 3x +1
 x +1 
(i) x + ln x
(f) ln 1 + x ln x ( )(h) ln ln x2 x − ln x
1− x (g) x2

2. Differentiate the following.

(a) y = xex2 (b) y = e−x ln x (c) y = (xln x) − x ( )(d) y = e2x ln 2x

(e) y = e x (f) y = ex (g) y = (x2 + 2x)ex2 +2x (h) y = (2x2 − 3)e−4x
1− e2x 1+ ln x

dy ( )(b) y = ln x2ex2−5x+1

3. Find for

dx

(a) y = (2x + 3)3 ln (4x + 5)

Answer: (b) − x 4 2 ln x (e)
14 3− x2 (c) (d)
(a)
4x +1 x x

2

(x +1)(3x +1)

1 1 − 2 ln x 1 2 − 2ln x
(g) (h) (i) (x − ln x)2
(f) (1− x)(1+ x) x ln x
x3

( )2
(a) e x2 (2x 2 + 1) (b) e−x  1 − ln x  (c) ln x (d) e2x  1 + 2 ln 2x  (e) ex e2x +1
x   x  (1 − e2x )2

ex[x(1+ ln x) −1] (g) 2(x + 1)3 e x2+2x (h) −4e−4x (2x − 3)(x +1)
(f) x(1+ ln x)2

3 2 (2x 3)2  2 (2x + 3) 3ln (4x 5) (b) 2 + 2x − 5
x
(a) +  + +
 4x + 5 

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AM015/ 8: Differentiation

TUTORIAL 5 of 7

1. Given that ln x + y = exy , find dy .

dx

( )2. Find dy for e4x x2y + 1 = 1
dx

3. Find dy when:
dx

(a) x2 y3 = 8 (b) y2 = ln(x2 + y 2 ) (c) y 2 x + e2y x2 + 2 y = 0

(d) x2 + e xy + y2 = 1 (e) x2 ln y + y 2 ln x = 2

4. By using the laws of logarithm, find dy ex+ y = (2x −1)2
if x +1 ( x − 3)5 .
dx

5. If y − x2 y = 4, show that  1 − x2  dy = 2xy + y .
x  x  dx x2

6. If y2 = 1 , prove that y3 + dy = 0.
2x −1 dx

( ) ( )7. Given 
ln x + 1 , show that dy = 1  1 ln x + 1 
y= −
dx x  x + 1 2x 
x

Answers: ( )2.
dy −4 x2y +1 − 2xy
2 y(x + y)exy −1 dx = x2
1. 1− 2x(x + y)exy

3. (a) − 2 y x −( y2 + 2xe2 y ) − 2x − yexy
3x (b) (c) 2xy + 2x2e2 y + 2 (d)
xexy + 2 y
(x2 + y 2 −1) y

(−2x2 ln y − y 2 ) y
(e)

(x2 + 2 y 2 ln x)x

4. 4 − 1 − 5 −1 (5, 6, 7 ) Shown
2x −1 2(x +1) x − 3

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AM015/ 8: Differentiation

TUTORIAL 6 of 7

1. Find dy for the following functions. Give your answers in the most simplified for

dx

(a) y = (2x2 − 3)e−4x (b) y = 3x ln  x2 + 1  (c) y = x2
 5x + 2  3x + 5

2. Find dy for 2x2 y + y2 = 2ey by using implicit differentiation.
dx

3. Find the slope of the curve y = 5x2 − 3 at the point = 2 by using the first principle of

differentiation.

4. If y = Ax2 + Bx , show that x2 d 2 y − x dy + y = 0
2 dx2 dx

5. Differentiate the following with respect to x. Give your answer in the simplest form.

a) f (x) = xe3x (b) f (x) = (3x + 2)(x + 4)3
x+3

6. Evaluate dy for the curve 3y2 + xy − 3x2 = 1 at point (1,1)
dx

7. Find f '(x) for f (x) = 4x2 − 7x + 5 by using the first principle.

8. Given y = Ax + Bxex with A and B are constants. Determine the values of A and B
if dy = 0 and d 2 y = 2 at x = 0 .

dx dx2

Answers:

1. (a) dy = −4e−4x (2x − 3)(x + 1)
dx

(b) dy = 6x2 − 15x + 3ln  x2 +1  (c) dy = x(3x +10)
dx x2 +1 5x + 2  5x +2  dx (3x + 5)2

2. dy = − 2xy 3. 20
dx x2 + y − ey

4. shown 5. (a) 3e3x[x2 + 3x +1]
(x + 3)2

(b) 3 (x + 4)(2x + 3)
3x + 2

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AM015/ 8: Differentiation

6. dy = 11 7. f '(x) = 8x − 7
dx 13
8. A = -1, B = 1

TUTORIAL 7 of 7

1. Find f '(2) for f (x) = 2x2 − x by using the first principle.

2. Differentiate the following with respect to x. Give your answer in the simplest form.

a) f (x) = ( x − 9)(x + 2)4 b) f (x) = (2x + 4)3 ( )c) f (x) = ln (2x + 3)3 e3x
x−3

3. a) Find dy for y3 = 2x2 y + xe4y − 2 by using implicit differentiation. Hence, evaluate dy at
dx dx

y =0.

b) Prove that if y = ( Ax + B) e−2x then d2y + 4 dy + 4y = 0 .
dx2 dx

y = (A + Bx)e3x , find dy d2y d2y − 6 dy + 9 y = 0.
dx dx2 . Hence, show that dx2 dx
4. Given and

5. Find f '(x) for f (x) = −2x2 + x +1by using the first principle. Hence, find f '(2).

6. Find dy for y2 + 3x2 y2 − 7xy +1 = 0 by using implicit differentiation.
dx

Hence, determine all values of dy when y = 1.
dx

7. Differentiate the following with respect to x. Give your answer in the simple form.

a) y = (x2 − 3) e5x

b) y = ln  x −1 3
 5x2 + 2 

7 of 11

Answers: AM015/ 8: Differentiation
1 f '(2) = 7
3(2x + 5)
2 ( x + 2)3 (9x − 70) 2(2x + 4)2 (2x −11)
b) ( x − 3)2 c)
a) 1
2x +3
2(x −9)2
b) proved
3 dy = 4xy + e4y dy = − 1 at y = 0
dx 3y2 − 2x2 − 4xe4 y dx 16
a) ,

4 dy = 3e3x ( A + Bx) + Be3x d2y = 6Be3x + 9e3x ( A +
dx , dx2
Bx)

, shown.

5 f '(x) = −4x +1, f '(2) = −7.

6 dy = 7 y − 6xy2
dx 2 y + 6x2 y − 7x

at  1 ,1 , dy = 15,
 3 dx

at (2,1), dy = −5 .

dx 12

( )7
a) dy = e5x (5x2 + x −15)
dx x2 − 3
1
2

b) dy = (3 10x − 5x2 + 2)
dx ( x −1)(5x2 + 2)

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AM015/ 8: Differentiation

EXTRA EXERCISE

1. Use suitable rules of differentiation to find the derivative of the following functions.
Give your answer in simplest form.

(a) f (x) = x3e7x (b) h(x) = x ln x − eln x

2. (a) Apply implicit differentiation to find dy for the equation (x − y)2 = xy − 2 . Hence,
dx

solve for x when dy = 0 . Give your answer in exact value.
dx

(b) Determine the values of A, B and C for y = Ax3 + B(x −1)2 + Cx , if dy = 2 and
dx

d2y =1 at point ( 2,1) .
dx2

3. Find f ' (x) for f (x) = 3x2 − 2 by using the first principle . hence or otherwise, find
the value of f ' (x) at x=1.

4. Given y = Ae−x ln x , where A is a constant.
(a) Show that x2 d 2 y − yx2 + Ae−x (2x +1) = 0
dx2
(b) Find the value of A if dy = 1 at x=1. Give your answer in exact value.
dx

5. If 4y − 3x2 − 5xy2 = 8 , find dy in terms of x and y. Hence, evaluate dy when x=0.
dx dx

[5M]

6. (a) Given y = (x x3 and dy = Ax2(x + B) . Find the values of A and B [3M]
+ 1)2 dx (x +1)3 [4M]

(b) Given 1 , find dy
in the simplest form.
y = 4x(3x3 − 2)2 dx

7. (a) Given y = (4 − 5x)e−3x , find d2y d2y
dx2
in the simplest form. Hence, evaluate

dx2

when x=0.

[6M]

(b) Find dy in terms of x and y, given that ln(2 y) = y ln x , where y>0 and x>0. Give
dx

dy [6M]
your answer in the simplest form. Hence, evaluate when x=1.

dx

9 of 11

AM015/ 8: Differentiation

8. (a) Use first principle to find the derivative of f (x) = 5x2 − 2x + 7 [4M]

(b) Differentiate 2x − 3 with respect to x. Give your answer in the simplest form.
4x2 −9

[5M]

9. (a) Find the second derivative of y = 1 − 1 [3M]
x x2 [6M]

(b) Given y = xe−2x , calculate the value of x when d2y =0
dx2

10. (a) Using implicit differentiation , find dy for x ln( y +1) = e2y−3x [6M]
dx

(b) Given y = (mx + n)e2x , where m and n are constants. Find the values of m and m

if y = −3 and dy = 8 when x=0. [6M]
dx

Answers: (b) ln x

1 (a) x2e7x (7x + 3)

2 3y − 2x ; x =  3 10 (b) A = − 3 , B = 5,C = 1
4
(a)
2y − 3x 5

3 f '(x) = 6x; f '(1) = 6

4 (a) shown (b) e

5 dy = 6x + 5y2
dx 4 −10xy , 5

6 2(15x 3−4) y2 1
(a) A=1, B=3 (b)
3x3 − 2 (b) ,

7 x(1− y ln x) 4

(a) 3e−3x (22 −15x) , 66

8 (a) 10x-2

(b) 6(2x − 3) @ 6@6
(2x + 3) (4x2 − 9) (2x + 3)3 (2x − 3)
1

(4x2 − 9)2

9 (a) 3 −5 − 6 x −4

4 x2

(b) x=1

10 −3( y +1)e2y−3x − ( y +1) ln( y +1)
(a) x − 2( y +1)e2 y−3x

(b) n = -3, m=14

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AM015/ 8: Differentiation
11 of 11