Higher Engineering Mathematics

G

Differential calculus

30

Differentiation of implicit functions

30.1 Implicit functions

When an equation can be written in the formy=f(x)
it is said to be anexplicit functionofx. Examples
of explicit functions include

y= 2 x^3 − 3 x+4, y= 2 xlnx

and y=

3ex

cosx

In these examplesy may be differentiated with
respect toxby using standard derivatives, the prod-
uct rule and the quotient rule of differentiation
respectively.
Sometimes with equations involving, say,yandx,
it is impossible to makeythe subject of the formula.
The equation is then called animplicit function
and examples of such functions include
y^3 + 2 x^2 =y^2 −xand siny=x^2 + 2 xy.

30.2 Differentiating implicit functions

It is possible todifferentiate an implicit function
by using thefunction of a function rule, which may
be stated as

du

dx

=

du

dy

×

dy

dx

Thus, to differentiatey^3 with respect tox, the sub-

stitution u=y^3 is made, from which,

du

dy

= 3 y^2.

Hence,

d

dx

(y^3 )=(3y^2 )×

dy

dx

, by the function of a

function rule.
A simple rule for differentiating an implicit func-
tion is summarised as:

d

dx

[f(y)]=

d

dy

[f(y)]×

dy

dx

(1)

Problem 1. Differentiate the following func-

tions with respect tox:

(a) 2y^4 (b) sin 3t.

(a) Letu= 2 y^4 , then, by the function of a function

rule:

du

dx

=

du

dy

×

dy

dx

=

d

dy

(2y^4 )×

dy

dx

= 8 y^3

dy

dx

(b) Letu=sin 3t, then, by the function of a function

rule:

du

dx

=

du

dt

×

dt

dx

=

d

dt

(sin 3t)×

dt

dx

=3 cos 3t

dt

dx

Problem 2. Differentiate the following func-

tions with respect tox:

(a)4ln5y (b)

1

5

e^3 θ−^2

(a) Letu=4ln5y, then, by the function of a func-

tion rule:

du

dx

=

du

dy

×

dy

dx

=

d

dy

(4 ln 5y)×

dy

dx

=

4

y

dy

dx

(b) Letu=

1

5

e^3 θ−^2 , then, by the function of a func-

tion rule:

du

dx

=

du

dθ

×

dθ

dx

=

d

dθ

(

1

5

e^3 θ−^2

)

×

dθ

dx

=

3

5

e^3 θ−^2

dθ

dx