Cambridge Additional Mathematics

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Introduction to differential calculus (Chapter 13) 349

DERIVATIVES OF COMPOSITE FUNCTIONS


The reason we are interested in writing complicated functions as composite functions is to make finding
derivatives easier.

Discovery 5 Differentiating composite functions


The purpose of this Discovery is to learn how to differentiate composite functions.

Based on the rule “if y=xn then dy
dx

=nxn¡^1 ”, we might suspect that if y=(2x+1)^2 then
dy
dx
= 2(2x+1)^1. But is this so?

What to do:
1 Expand y=(2x+1)^2 and hence find
dy
dx

. How does this compare with 2(2x+1)^1?


2 Expand y=(3x+1)^2 and hence find
dy
dx

. How does this compare with 2(3x+1)^1?


3 Expand y=(ax+1)^2 whereais a constant, and hence find
dy
dx

. How does this compare with
2(ax+1)^1?


4 Suppose y=u^2.

a Find
dy
du

.

b Now suppose u=ax+1,soy=(ax+1)^2.

i Find
du
dx

. ii Write
dy
du


fromain terms ofx.

iii Hence find
dy
du

£
du
dx

. iv Compare your answer to the result in 3.


c If y=u^2 whereuis a function ofx, what do you suspect
dy
dx
will be equal to?

5 Expand y=(x^2 +3x)^2 and hence find
dy
dx
.

Does your answer agree with the rule you suggested in4c?
6 Consider y=(2x+1)^3.

a Expand the brackets and hence find dy
dx

.

b If we let u=2x+1, then y=u^3.

i Find
du
dx

. ii Find
dy
du
, and write it in terms ofx.


iii Hence find
dy
du
£
du
dx

. iv Compare your answer to the result ina.


7 Copy and complete: “Ifyis a function ofu, anduis a function ofx, then
dy
dx
=::::::”

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