Understanding Engineering Mathematics

This is, of course, an incorrect application of the rule for the deriva-

tive ofxn. We will come back to 2xlater (Section 8.2.5) – it needs

some more rules of differentiation, which takes us to the next section.

8.2.4 Rules of differentiation

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229 244➤

In order to differentiate more complicated functions than those in the standard derivatives
table we need to be able to differentiate sums, differences, products and quotients. Also we
sometimes need to differentiate a composite function or ‘function of a function’ (97
➤
).
These rules of differentiation can be proved by using the limit definition of differentiation
(see Chapter 14). Here we will concentrate on making them at least plausible, and on
getting used to using them.
The derivative of a sum (difference) is the sum (difference) of the derivatives.

d

dx

(f(x)±g(x))=

df(x)

dx

±

dg(x)

dx

We say differentiation is alinear operation. An extension of this is:

d

dx

(kf(x))=k

df(x)

dx

fork=constant

which simply says that to differentiate a constant multiple of a function you only have to
take out the constant from the derivative.
It would be nice if the derivative of a product was a product – unfortunately it is not.
To obtain the correct result, suppose we want to differentiatey=uvwhereu=u(x),
v=v(x)are functions ofx.
Following our recipe of Section 8.2.2 letxincrease tox+δx,say.Thenuwill increase
tou+δuandvtov+δv,say.Soyincreases by:

δy=(u+δu)(v+δv)−uv

=uδv+vδu+δuδv

Now we neglectδuδvbecause it is of second degree (i.e. like (δx)^2 notδx)so we have

δyuδv+vδu

So, dividing through byδx

δy

δx

=u

δv

δx

+v

δu

δx

If we now letδx→0wegettheproduct rule:

dy

dx

=

d(uv)

dx

=u

dv

dx

+v

du

dx

A nice physical way to look at this is to think of how much the area of a rectangular plate
increases when you heat it. Suppose the plate has sidesuandv. Due to linear expansion