(vi)y=e−^2 x

This is a function of a function. Here we will practise a short hand

approach to the function of a function rule.

dy

dx

=

d(e−^2 x)

d(− 2 x)

×

d(− 2 x)

dx

=e−^2 x(− 2 )=− 2 e−^2 x

(vii)y=

√

1 −x^2 =( 1 −x^2 )

1

2

Again, function of a function – ‘differentiate the ( )

1

(^2) and multiply
by the derivative of the ( )’.
dy
dx

1
2
( 1 −x^2 )−
1
(^2) ×(− 2 x)=−
x
( 1 −x^2 )
1
2
=−
x
√
1 −x^2
8.2.5 Implicit differentiation
➤
229 244➤
Ifyis defined implicitly in terms ofxby an equation of the form (91
➤
)
f(x,y)= 0
then
dy
dx
may often be found by differentiating throughout using the function of a function
rule. This is best illustrated by the example given below:
Example
If
x^2 +y^2 = 1
then differentiating through with respect toxgives
2 x+
d
dx
(y^2 )=
d
dx
( 1 )
so, using the function of a function rule on the left-hand side
2 x+ 2 y
dy
dx
= 0
Hence
dy
dx
=−
x
y
Solution to review question 8.1.5
A.Differentiating through with respect tox, using the function of function
and product rules we have
d
dx
(x^2 + 2 xy+ 2 y^2 )=
d
dx
( 1 )