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DIFFERENTIATION
Name
........................................................................................
Differentiation
CHAPTER 9 : DIFFERENTIATION:
9.1: To determine the limit of a function
Don’t write the substitution.
If 0
0
or
is found in the solution then the function have to be
simplify by factorization.
Remember that if a 0 then 0
a
= and a
0
= 0 .
Example of factorization :
(a) (b)
x
baxbax
x
baxbxax 22
EXAMPLE:
Evaluate the following limits.
1. x
xx
62lim
3
2.
12
2lim
2
2
1
xx
xx
x 3.
xxxx
1lim
0
Question Step Solution/Answer
1. 1. Substitute (without write it down)
the value of x and write the answer.
8
2.
1. Substitute (without write it down)
the value of x and write the answer.
0
0
2. Simplify the function, hence
calculate the limit.
3.
1. Substitute (without write it down)
the value of x and write the answer.
2
1
x
Differentiation
EXERCISE 1:
Evaluate the following limits
1. 2
1lim
0 xx 3.
65
4lim
2
2
2
xx
x
x 5.
x
xx
x 2
21
lim
2. xxx
2lim0
4. xxx
1lim
0 6. xxx
x
63lim 2
0
9.2: To find dx
dy by first principle:
1. Using y and x .
2. Solving simultaneous equation .
3. Determine the expression for x
y
.
4. Using .x
ylim
0x
Differentiation
EXERCISE 1: Find the first derivative of xy 25 by using the first principle.
Step Solution
1. Let the function given be the first equation.
2. Let the equation using xandy be the
second equation.
3. Second equation minus first equation and express
y in terms of .x
4. Determine .x
y
5. Determine dx
dy by using .lim
0 x
y
x
EXAMPLE:
Find the first derivative of 74 xy by using the first principle.
Step Solution
1. Let the function given be the first equation.
74 xy ….. (i)
2. Let the equation using xandy be the second
equation.
)(...7)(4 iixxyy
3.Second equation minus first equation and express y
in terms of .x
xy 4
4. Determine .x
y
4
5. Determine dx
dy by using .lim
0 x
y
x
4
Differentiation
9.3: Differentiation of a function y = axn, differentiation of a sum and difference of algebraic
functions.
EXERCISE 2:
Find the first derivative of the following functions using the first
principle.
1. 54 xy
2. y2
2
1x
3. x
y3
2
1. Given y = xn then
dx
dy = n x
n 1
2. Given y = axn then
dx
dy = an x
n 1
3. Given y = k then dx
dy= 0 , where k is a constant.
4. Given y = kx then dx
dy= k , where k is a constant.
5. If f(x) = p(x) q(x) , then f / (x)= p
/ (x) + q
/ (x) where f
/ (x), p
/ (x) and q
/ (x) are first derivatives.
Differentiation
EXAMPLE:
Find dx
dy for the following functions :
(a) y = x4
step solution
1. make sure x is in index form before differentiate x4
2. differentiate y with respect to x
dx
dy
3. differentiate x4 with respect to x by Multiply 4 and index Minus 1 4x
3 4. complete answer
dx
dy= 4x
3
(b) y = x
6
step solution
1. make sure x is in index form before differentiate
2. differentiate y with respect to x
3. differentiate x
4 with respect to x by Multiply 4 and index Minus 1
4. complete answer
Differentiation
EXERCISE: Find dx
dy for the following functions:
1 y = x3
2 y = 5x
3 y = 7
4 y = 6x2
5 y = 12x
6 y = 25
x 5
7 y = 4x 3 8 y =
x
6
9 y = 4x
8 10 y = 8x4
3
11 y = 2x3
2 12 y = 3
x
3
13 y = 3 x4 14 y = 400
15 y = x
7
1
16 y = 2x3
1
Differentiation
EXAMPLE 1: Differentiate each of the following functions :
(a) y = 3x4 + 2x
3 – 4
step solution
1. make sure x is in index form before differentiate x4, x
3
2. differentiate y with respect to x
dx
dy
3. differentiate all the terms on the right hand side with respect to x 12x3 + 6x
2
4. complete answer
dx
dy=12x
3 + 6x
2
(b) f(x) = 2x
3
x
3
step solution
1. make sure x is in index form before differentiate 3x 1
, 3x 2
2. differentiate f(x) with respect to x f /(x)
3. differentiate all the terms on the right hand side with respect to x 3x
2 (2)3x
3
4. complete answer f
/(x)= 3x
2 (2)3x
3
Differentiation
EXERCISE: Differentiate each of the following functions :
1 y = x6 – 3x
3 + 6 2
y = 9x4x2
1 24
3 f (x) = 3x
2
x
2 4 f(x) = 1 +
2x
1
x
1
5 y = (x – 1) (2x + 5) 6 y = 2(2 3 )y x
7 v =
t
9t2 2
8 s =
2t
)t32)(1t(
Differentiation
9.4: Differentiation of a Composite function .
EXAMPLE:
Differentiate y = (2x3 + 5)
6 with respect to x:
steps solution
1. recognise the composite
function
k ( ax + b)n
(2x3 + 5)
6
2. differentiate y with respect to x
dx
dy
3. differentiate composite function
bracketinfunctionatedifferenti
bracketinfunctionbackcopy
kn1n
6 (2x
3 + 5)
5 (6x
2)
4. complete answer
dx
dy=6 (2x
3 + 5)
5 (6x
2)
Given y = k ( ax + b)n
Then dx
dy = k n ( ax + b)
n – 1 (a)
EXERCISE
Differentiate each of the following functions :
1 y = (4x +7)3 2 y = (x
3 + 6)
4
3 y = (6 4x2)
3 4
f (x) = 6
3 9x3
1
5 f (x) = (x 2 + 5x 3 )
4 6 y =
1x5
2
7 y = 4x
32
8 s = 2 4
5
( 2)t
9 v =
32 )6t4t(
1
10 y = 23 x3x65
Differentiation
9.5: Differentiation of the Product of Algebraic functions.
EXAMPLE: Find dx
dy for y = (3x
2 2)( x
2 +5x +4).
step solution
1. recognise the product function vuy
u = (3x2 2)
v = ( x2 +5x +4).
2. differentiate y with respect to x
dx
dy
3. differentiate product function
)uatedifferenti()vcopy(vatedifferentiucopy
(3x2 2) (2x + 5) + ( x2
+ 5x + 4) (6x )
4. complete answer dx
dy = (3x
2 2) (2x +5) + ( x2
+5x +4) (6x)
5. Simplify the answer dx
dy = 12x
3 + 45x
2 + 20x 10
Exercise : Finddy
dx
1. y = 2x3 ( 2 – x
5) 2. y = (4x +7) (3x
2 – 5)
y = (x3 + 6)
(1– 4x +2x
2 )
3. y = (x3 + 6)
(1– 4x +2x
2 ) 4. y = x
2 (x + 1)
4
5. y = x2 (4-3x )
2 6. y = (2x + 1) (x + 3)
3
Method – The Product Rule
If vuy , then find the first derivative of u and v.
Then substitute into the formula: dx
duv
dx
dvu
dx
dy
Differentiation
9.6: Differentiation of the Quotient of Algebraic functions.
EXAMPLE:
Steps solutions
1.
Determine whether product or quotient
function involved .
y = v
u quotient function.
2. Use the quotient rule for , y = v
u
2v
dx
dvu
dx
duv
dx
dy
3.
Determine u and v , and find dx
duand
dx
dv
by differentiating
u and v with respect to x .
u = x3 , v = 2 – x
4
23xdx
du 34x
dx
dv
4.
Substitute into the formula.
24
3324
2
432
x
xxxx
dx
dy
5 Simplify your answer.
24
662
2
436
x
xxx
dx
dy
24
62
2
6
x
xx
dx
dy
Method – The Quotient Rule
If
xv
xuy
, then find the first derivative of u and v.
Then substitute into the formula: 2v
dx
dvu
dx
duv
dx
dy
Given that 4
3
2 x
xy
, find dx
dy
. Guide: Don’t expand the denominator
Differentiation
EXERCISE:
1. y = 2
2
x
x 2. y =
3x
x5
3. y = 2
2
3
7
x
x
4. y = 1x
)x3( 2
5. f (x) = 3)1x(
5x2
6. f (x) =
2x5
)x31( 4
Differentiation
9.7: GRADIENT OF TANGENT AND NORMAL
EXAMPLE:.
Given that P( 1 , 4 ) is a point on a curve y = 6 + 5x -7x2
Find
(a) the gradient of tangent and the gradient of normal at point P.
(b) the equation of tangent and the equation of normal at point P.
1. Determine the gradient of tangent at a point by substitution the x-
coordinate into dx
dy .
2. Determine the gradient of normal by referring to the gradient of the
tangent.
3. Determine the equation of tangent and normal at a point of a curve by
using the formula
)( 11 xxmyy
Steps Solution
1. Find .dx
dy
x145dx
dy
2. Find the gradient of tangent by Substitution the value of x. 9)1(145dx
dy
3. Find the gradient of normal by
using formula.
m1 m 2 = 1 9
1mnormal
4. Find the equation of tangent.
y y1 = m1 (x x1 )
y = - 9x + 13
5. Find the equation of normal.
y y1 = m2 (x x1 )
4
15x
9
1y
Differentiation
EXERCISE
1. Given that P(1,7) is a point on curve y = 3x2 – x + 5 . Find
(a) the gradient of tangent and the gradient of normal at point P.
(b) the equation of tangent and the equation of normal at point P
2. Given that P(2,4) is a point on curve y = (2x – 3)(x + 2) .
Find
(a) the gradient of tangent and the gradient of normal at point P.
(b) the equation of tangent and the equation of normal at point P.
Differentiation
9.8: TURNING POINTS - MINIMUM OR MAXIMUM.
EXAMPLE: Find the turning point of the curve y = 3x2 – x
3 + 5 and determine its maximum and
minimum point.
Step Solution
1. To determine the turning point
i Find dx
dy .
i) y = axn ii) y = kx
dx
dy = a
1nnx dx
dy= k
dx
dy= 6x – 3x
2
ii. Let dx
dy = 0 .
and solve for x .
6x – 3x2 = 0
3x(2 – x) = 0
x = 0 , x = 2
iii. To determine the y-coordinate for each value of
x that satisfy the above equation .
y = 0 – 0 + 5 = 5
y = 3(2)2 - 2
3 + 5 = 9
iv. State the turning point (0, 5) and (2, 9)
2. To determine the maximum or minimum
point
i. Find 2
2
dx
yd
2
2
dx
yd = 6 – 6x
ii. Substitute the value of x to get the value of 2
2
dx
yd x = 0,
2
2
dx
yd= 6 – 0 = 6
2
2
dx
yd> 0 min
x = 2, 2
2
dx
yd=6-6(2) = -6 < 0 max
iii. State the coordinate Minimum point (0. 5)
Maximum point (2, 9)
1. To find the turning points
Find dx
dy .
Lets dx
dy = 0 , solve the equation to find x.
Find the value of y.
State in ordered pairs (x,y) .
2. To determine whether the turning point is a
maximum or minimum point.
Find 2
2
dx
yd.
Substitute the value of x to the product of the differentiation .
If 2
2
dx
yd< 0, maximum point.
If 2
2
dx
yd> 0, minimum point.
3. The difference between maximum/minimum point with
the maximum/minimum value
Point : in ordered pairs ( x, y)
Value : refers only to the value of y.
Differentiation
EXERCISE
Find (a) the turning point (b) determine the maximum or minimum point (c) max/min value , from
the following curve equations.
1. y = x2 +2x
+ 3
2. y = 4 + 8x -x2
3. y = 2x3+6x
2 -1 4. y = x
3 -12x +1
Differentiation
9.9 Rate of change
,
y
area
volume
dydifferentiate y with respect to x
dx
dxrate of change of x
dt
dyrate of change of
dt
dArate of change of
dt
dVrate of change of
dt
Formulae frequently used:
.
.
.
dy dy dx
dt dx dt
dA dA dr
dt dr dt
dV dV dr
dt dr dt
EXAMPLE: Given y=x2 + 4x +3 and the rate of change of x is 0.5 unit s
-1 when x= 2, find the rate of
change of y.
Step Solution
1. Using formula .
dy dy dx
dt dx dt
2. Given
0.5
dx
dt
dyfind
dx
2 4
2, 2(2) 4 8
dyx
dx
dyx
dx
3. Substitute into formula 1 .
dy dy dx
dt dx dt
= 8(0.5) =4
Differentiation
Exercise
1.Given y=x
2 - 6x +2 and x decreases at the rate
of 0.5 unit s-1
when x= 3, find the rate of change
of y.
2. Given y=x3 - 4x +5 and x increases at the rate
of 2 unit s-1
when x= 4, find the rate of change of
y.
4. The radius of a circle increases at the
rate of 2 cms-1
when its radius is 10
cm. Find the rate of change in its area
5. The radius of a sphere decreases at
the rate of 2 cms-1
when its radius is 8
cm. Find the rate of change in its
volume
Differentiation
9.10: Small change And Approximation.
The increment in y, xdx
dyy
x = the small change in x
If x is decreasing then x is negative .
The differences between the small change of y with the
approximate value of y
The small change of y is y
The approximate value of y = y origin + y
y origin = the value of y before x changes
EXAMPLE: Given that y = x2 + 5x, use differentiation to find the small change in y
when x increases from 3 to 3.01.
Step Solution
Keyword
- find y
1. Using formula
2. Determine the
Value of x
Value of x
Find dx
dy
- substitute the value of x
3. Substitute into the formula (1)
Differentiation
1. Given that y = 2x3 +6x
2 , find the small
change of y when x increase from 2 to 2.01
3. Find the approximate increase in the area of a
circle when its radius increases from 5 to 5.03
2. Given that y = 6t ³ + t², find the small change
of y when t increase from 2 to 2.01.
4. The radius of a cylinder is fixed at 4 cm. If the
height of the cylinder decreases by 0.25, find
the approximate decrease in the volume of the
cylinder.
