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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Y:\HAESE\CAM4037\CamAdd_13\349CamAdd_13.cdr Tuesday, 7 January 2014 9:53:24 AM BRIAN