Understanding Engineering Mathematics

uandvwill increase tou+δuandv+δurespectively, withδuandδvvery small. The
area therefore increases to

(u+δu)(v+δv)=uv+vδu+uδv+δuδv

So the increase in area will be

vδu+uδv+δuδv

Sinceδu,δvare very small,δuδvisveryvery small, so we can neglect it and the increase
in area is approximately

vδu+uδv

which is the form of the product rule.
The correspondingquotient rule is

d

dx

(u

v

)

=

(

v

du

dx

−u

dv

dx

)/

v^2

We will see why below.
We can handle many functions by using the standard derivatives and the sum, product,
quotient rules. We could not handle something like cos(ex)however. This is an example
of afunction(cosine)of a function(exponential) or a composition of functions (97

➤
).
In general we write:

y=f(u) whereu=g(x)

e.g. y=cosu whereu=ex

To differentiate such a functionwith respect toxwe use thefunction of a functionor
thechain rule:

dy

dx

=

dy

du

du

dx

So

dy

dx

=−sinu×ex=−exsin(ex)

Think of this ‘rule’ of change ofuw.r.t.x=rate of change ofyw.r.t.utimes rate of
change ofuw.r.t.x. Note that the chain rule suggests a useful result if we take the special
casey=x:

dy

du

du

dx

=

dx

du

du

dx

=

dx

dx

= 1

so

du

dx

= 1

/

dx

du

or

dy

dx

= 1

/

dx

dy

This result is useful in implicit differentiation for example (see Section 8.2.5). While it

shouldnotencourage you to think of

dy

dx

as afractionthere are occasions when it is