2.2.3 n 2.2.4 continue

ADDITIONAL
MATHEMATICS
F5 KSSM5UO
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 21/02/2021 2.2 The first derivative
 Sunday
2.2.3 Determine the first derivative of composite function
 02:30 – 04:00 pm

2.2.4 Determine the first derivative of a funtion involving product and quotient
of algebraic expressions

First Derivative of Composite Function 2.2 : The First Derivative

 To differentiate the function = + , we expand the function into
= + + before it is differentiate term by term to get = + .



 However, what if we want to differentiate the function = + ?

 Then 2 + 3 4 will be too difficult to expand unless we consider the function
as a composite function consisting of two simple functions.

Let’s explore the following method…

From Discovery activity 4,
We found that there are other methods to differentiate functions like = 2 + 3 2.

However, the method used in step 3 is much easier to get the derivative of an expression which
is in the form + , where ≠ , that is difficult to expand.

For function = = 2 + 3 2. In this case, is a function of and is a
Let = ℎ = 2 + 3. function of . Hence, we say that = is a
Then composite function with = ( ) and = .

= = 2

To differentiate a function like this, we will
introduce a simple method known as a

chain rule, which is


= ×

In general, the first derivative of a composite function is as
follows :

If = and = ℎ , then differentiating respect to will give

′ = ′ × ℎ′( )

That is, = ×



(a) = − With chain rule

then,
= ×
= 6 -4 and = 7 6
= 7 6 6 − 4
= 7 3 2 − 4 6 6 − 4
= 7 6 − 4 3 2 − 4 6
= 42 − 28 3 2 − 4 6

−
= −

(b) =
+

Then and = − − − = −

=



With chain rule
= ×

3
= − u4 2


= − +

(c) = +

With chain rule,
= ×

1 Then,
= − 12
and = 1 12−1 = − 1
2 = 12 2 2
12

=−
2 6 2 + 8

6
= − 6 2 + 8

 Self- exercise 2.4

First derivative of a function involving product and quotient of
algebraic expressions

From Discovery Activity 5 result,
 It is shown that there are more than one method of differentiating

functions involving two algebraic expressions multiplied together like
the function = + − .

 However, in cases where expansion of algebraic expressions is difficult
such as + − , the product rule illustrated in step 3 is often used to differentiate
such functions.

 In general, the formula to find the first derivative of functions involving
the product of two algebraic expressions which is also known as the product rule is as follows

If and are functions of , then

= +

From Discovery Activity 6,

It is shown that apart from using the product rule in differentiating a function involving

the quotient of two algebraic expressions such as = − a quotient rule

illustrated in step 3 can be also used.

 In general,
the quotient rule is stated as follows:

If and are functions of , and ≠ 0, then

= −

2

Example 8 Differentiate each of the following expressions with respect to .

Hence,
= +
(a) + −
Given y = 2 + 1 − 3 4 = 2 + 1 × 4 − 3 3 + − 3 4 × 2
= 4 2 + 1 − 3 3 + 2 − 3 4
let = 2 + 1 = 2 − 3 3 2 2 + 1 + − 3
and = − 3 4

we get = 2



and = 4 − 3 4−1 − 3 = 2 − 3 3 3 2 − 3 + 2



= 4 − 3 3(1)

= 4 − 3 3

Example 8 Differentiate each of the following expressions with respect to .

Given y = 3 + 2 4 − 1 (b) 3 + 2 4 − 1
let = 3 + 2 = +
Hence ,

1 = 3 + 2 × 2 + 4 − 1 × 3

and = 4 − 1 = 4 − 1 2 4 −1

we get = 3 = 2 3 +2 + 3 4 − 1

4 −1

and = 1 4 − 1 12−1 4 − 1
2
2 3 +2 +3 4 −1
= 4 −1
1 −21
=2 4 − 1 4

2 = 18 +1
=
4 −1
4 − 1

Example 9 Given = + , find

(a) the expression for (b) the gradient of tangent at =

When = 6 ,

Solution
Let = and = + 3

Then, ′ + ′ 3 6 + 2
= 2 6 + 3

= + 3 + + 3

1 24
= + + 3(1) =6

2 + 3 =4

+ 2 + 3 Hence, the gradient of the tangent at = 6 is 4.
=

2 + 3

3 + 2
= 2 + 3

Example 10 (a) Given = 2 +1 , find
2−3

Solution 2 2 − 6 − 4 2 + 2
Let = 2 + 1 and = − = 2 − 3 2

Then, and −2 2 − 2 − 6
= 2 = 2 − 3 2

= 2 −2 2 + + 3
= 2 − 3 2
Therefore, −
=

2

2 − 3 2 − 2 + 1 2
= 2 − 3 2

Example 10 (b) Given = , show that = 2 −1
4 −1 4 −1 3

′ − ′ Solution
2

4 − 1 − 4 − 1
=
4 − 1 2 4 −1 4 −1 −2 ÷ 4 −1

4 − 1 − 2 4 − 1 4 − 1 − 2 4 −1 1

= 4 − 1 4 − 1 4 − 1 4 − 1 − 2 1
4 − 1 4 − 1

4 − 1 × 4 − 1

4 − 1 4 − 1 − 2 2 − 1 2 − 1
= = = 4 − 1 4 − 1

4 − 1 4 − 1 4 − 1 4 − 1
4 − 1 − 2

=
4 − 1 4 − 1

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