Chapter 34

Partial differentiation

72.1 Introduction to inequalities

derivatives

In engineering, it sometimes happens that the variation
of one quantity depends on changes taking place in
two, or more, other quantities. For example, the vol-
umeVof a cylinder is given byV=πr^2 h.Thevolume
will change if either radiusror heighthis changed.
The formula for volume may be stated mathematically
asV=f(r,h)which means ‘Vis some function ofr
andh’. Some other practical examples include:

(i) time of oscillation,t= 2 π

√

l

g

i.e.t=f(l,g).

(ii) torqueT=Iα,i.e.T=f(I,α).

(iii) pressure of an ideal gasp=

mRT

V

i.e.p=f(T,V).

(iv) resonant frequencyfr=

1

2 π

√

LC

i.e.fr=f(L,C), and so on.

When differentiating a function having two variables,
one variable is kept constant and the differential
coefficient of the other variable is found with respect
to that variable. The differential coefficient obtained is
called apartial derivativeof the function.

34.2 First order partial derivatives

A ‘curly dee’,∂, is used to denote a differential coef-
ficient in an expression containing more than one
variable.

Hence ifV=πr^2 h then

∂V

∂r

means ‘the partial

derivative of V with respect tor, withhremaining

constant’. Thus,

∂V

∂r

=(πh)

d

dr

(r^2 )=(πh)( 2 r)= 2 πrh.

Similarly,

∂V

∂h

means ‘the partial derivative ofVwith

respect toh, withrremaining constant’. Thus,

∂V

∂h

=(πr^2 )

d

dh

(h)=(πr^2 )( 1 )=πr^2.

∂V

∂r

and

∂V

∂h

are examples of first order partial

derivatives,sincen=1 when written in the form

∂nV

∂rn

.

First order partial derivatives are used when finding the

total differential,rates ofchangeand errors forfunctions

of two or more variables (see Chapter 35), when finding

maxima, minima and saddle points for functions of two

variables (see Chapter 36), and with partial differential

equations (see Chapter 53).

Problem 1. Ifz= 5 x^4 + 2 x^3 y^2 − 3 yfind

(a)

∂z

∂x

and (b)

∂z

∂y

.

(a) To find

∂z

∂x

,yis kept constant.

Sincez= 5 x^4 +( 2 y^2 )x^3 −( 3 y)

then,

∂z

∂x

=

d

dx

( 5 x^4 )+( 2 y^2 )

d

dx

(x^3 )−( 3 y)

d

dx

( 1 )

= 20 x^3 +( 2 y^2 )( 3 x^2 )− 0.

Hence

∂z

∂x

= 20 x^3 + 6 x^2 y^2.