Introduction to differential calculus (Chapter 13) 349DERIVATIVES 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
=::::::”4037 Cambridge
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Y:\HAESE\CAM4037\CamAdd_13\349CamAdd_13.cdr Tuesday, 7 January 2014 9:53:24 AM BRIAN