Chapter 34
Partial differentiation
34.1 Introduction to partial
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 volume V of a cylinder is given by V = πr 2 h. The volume
will change if either radius r or height h is changed.
The formula for volume may be stated mathematically
as V = f (r, h) which means ‘V is some function of r
and h’. Some other practical examples include:
(i) time of oscillation, t = 2π
l
g
i.e. t = f (l, g).
(ii) torque T = I α, i.e. T = f (I, α).
(iii) pressure of an ideal gas p =
mRT
V
i.e. p = f (T, V ).
(iv) resonant frequency f r =
1
2π
√
LC
i.e. f r = 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 a partial derivative of the function.
34.2 First order partial derivatives
A ‘curly dee’, ∂, is used to denote a differential coefficient in an expression containing more than one
variable.
Hence if V = πr 2 h then
∂V
∂r
means ‘the partial
derivative of V with respect to r, with h remaining
constant’. Thus,
∂V
∂r
= (πh)
d
dr
(r
2
) = (πh)(2r) = 2πrh.
Similarly,
∂V
∂h
means ‘the partial derivative of V with
respect to h, with r remaining 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, since n =1 when written in the form
∂ n V
∂r n .
First order partial derivatives are used when finding the
total differential, rates of change and errors for functions
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. If z = 5x 4 + 2x 3 y 2 − 3y find
(a)
∂z
∂x
and (b)
∂z
∂ y
.
(a) To find
∂z
∂x
, y is kept constant.
Since z = 5x 4 + (2y 2 )x 3 − (3y)
then,
∂z
∂x
=
d
dx
(5x
4
) + (2y
2
)
d
dx
(x
3
) − (3y)
d
dx
(1)
= 20x
3
+ (2y
2
)(3x
2
) − 0.
Hence
∂z
∂x
= 20x 3 + 6x 2 y 2 .
Partial differentiation
34.1 Introduction to partial
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 volume V of a cylinder is given by V = πr 2 h. The volume
will change if either radius r or height h is changed.
The formula for volume may be stated mathematically
as V = f (r, h) which means ‘V is some function of r
and h’. Some other practical examples include:
(i) time of oscillation, t = 2π
l
g
i.e. t = f (l, g).
(ii) torque T = I α, i.e. T = f (I, α).
(iii) pressure of an ideal gas p =
mRT
V
i.e. p = f (T, V ).
(iv) resonant frequency f r =
1
2π
√
LC
i.e. f r = 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 a partial derivative of the function.
34.2 First order partial derivatives
A ‘curly dee’, ∂, is used to denote a differential coefficient in an expression containing more than one
variable.
Hence if V = πr 2 h then
∂V
∂r
means ‘the partial
derivative of V with respect to r, with h remaining
constant’. Thus,
∂V
∂r
= (πh)
d
dr
(r
2
) = (πh)(2r) = 2πrh.
Similarly,
∂V
∂h
means ‘the partial derivative of V with
respect to h, with r remaining 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, since n =1 when written in the form
∂ n V
∂r n .
First order partial derivatives are used when finding the
total differential, rates of change and errors for functions
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. If z = 5x 4 + 2x 3 y 2 − 3y find
(a)
∂z
∂x
and (b)
∂z
∂ y
.
(a) To find
∂z
∂x
, y is kept constant.
Since z = 5x 4 + (2y 2 )x 3 − (3y)
then,
∂z
∂x
=
d
dx
(5x
4
) + (2y
2
)
d
dx
(x
3
) − (3y)
d
dx
(1)
= 20x
3
+ (2y
2
)(3x
2
) − 0.
Hence
∂z
∂x
= 20x 3 + 6x 2 y 2 .