“JUST THE MATHS”
UNIT NUMBER
14.2
PARTIAL DIFFERENTIATION 2
(Partial derivatives of order higher than one)
by
A.J.Hobson
14.2.1 Standard notations and their meanings
14.2.2 Exercises
14.2.3 Answers to exercises
UNIT 14.2 - PARTIAL DIFFERENTIATION 2
PARTIAL DERIVATIVES OF THE SECOND AND HIGHER ORDERS
14.2.1 STANDARD NOTATIONS AND THEIR MEANINGS
In Unit 14.1, the partial derivatives encountered are known as partial derivatives of the first
order; that is, the dependent variable was differentiated only once with respect to each
independent variable.
But a partial derivative will, in general contain all of the independent variables, suggesting
that we may need to differentiate again with respect to any of those variables.
For example, in the case where a variable, z, is a function of two independent variables, x
and y, the possible partial derivatives of the second order are
(i)
∂2z
∂
, which means
2
∂x
∂x
∂z
;
∂x
∂2z
∂
, which means
2
∂y
∂y
∂z
;
∂y
!
(ii)
!
(iii)
∂2z
∂
, which means
∂x∂y
∂x
∂z
;
∂y
∂2z
∂
, which means
∂y∂x
∂y
∂z
.
∂x
!
(iv)
!
The last two can be shown to give the same result for all elementary functions likely to be
encountered in science and engineering.
Note:
Occasionally, it may be necessary to use partial derivatives of order higher than two, as
illustrated, for example, by
∂3z
∂ ∂
,
which
means
∂x∂y 2
∂x ∂y
"
1
∂z
∂y
!#
and
∂4z
∂
, which means
2
2
∂x ∂y
∂x
"
∂ ∂
∂x ∂y
∂z
∂y
!#!
.
EXAMPLES
Determine all the first and second order partial derivatives of the following functions:
1.
z = 7x3 − 5x2 y + 6y 3 .
Solution
∂z
∂x
= 21x2 − 10xy;
∂z
∂y
∂2z
∂x2
= 42x − 10y;
∂2z
∂y 2
∂2z
∂y∂x
= −5x2 + 18y 2 ;
= 36y;
∂2z
∂x∂y
= −10x;
= −10x.
2.
z = y sin x + x cos y.
Solution
∂z
∂x
∂2z
∂x2
∂z
∂y
= y cos x + cos y;
∂2z
∂y 2
= −y sin x;
∂2z
∂y∂x
= sin x − x sin y;
= −x cos y;
∂2z
∂x∂y
= cos x − sin y;
2
= cos x − sin y.
3.
z = exy (2x − y).
Solution
∂z
∂x
∂2z
∂x2
∂z
∂y
= exy [y(2x − y) + 2]
= exy [2xy − y 2 + 2];
∂2z
∂y 2
= exy [y(2xy − y 2 + 2) + 2y]
= exy [x(2x2 − xy − 1) − x]
= exy [2xy 2 − y 3 + 4y];
∂2z
∂y∂x
= exy [x(2x − y) − 1]
= exy [2x2 − xy − 1];
= exy [2x3 − x2 y − 2x];
∂2z
∂x∂y
= exy [x(2xy − y 2 + 2) + 2x − 2y]
= exy [y(2x2 − xy − 1) + 4x − y]
= exy [2x2 y − xy 2 + 4x − 2y];
= exy [2x2 y − xy 2 + 4x − 2y].
14.2.2 EXERCISES
1. Determine all the first and second order partial derivatives of the following functions:
(a)
z = 5x2 y 3 − 7x3 y 5 ;
(b)
z = x4 sin 3y.
2. Determine all the first and second order partial derivatives of the function
w ≡ z 2 exy + x cos(y 2 z).
3. If
!
x
,
z = (x + y) ln
y
show that
x2
2
∂2z
2∂ z
=
y
.
∂x2
∂y 2
3
4. If
z = f (x + ay) + F (x − ay),
show that
∂2z
1 ∂2z
= 2 2.
∂x2
a ∂y
14.2.3 ANSWERS TO EXERCISES
1. (a) The required partial derivatives are as follows:
∂z
∂x
= 10xy 3 − 21x2 y 5 ;
∂z
∂y
∂2z
∂x2
= 10y 3 − 42xy 5 ;
∂2z
∂y 2
∂2z
∂y∂x
= 15x2 y 2 − 35x3 y 4 ;
= 30x2 y − 140x3 y 3 ;
∂2z
∂x∂y
= 30xy 2 − 105x2 y 4 ;
= 30xy 2 − 105x2 y 4 .
(b) The required partial derivatives are as follows:
∂z
∂x
∂2z
∂x2
∂z
∂y
= 4x3 sin 3y;
∂2z
∂y 2
= 12x2 sin 3y;
∂2z
∂y∂x
= 3x4 cos 3y;
= −9x4 sin 3y;
∂2z
∂x∂y
= 12x3 cos 3y;
= 12x3 cos 3y.
2. The required partial derivatives are as follows:
∂w
= yz 2 exy + cos(y 2 z);
∂x
∂w
= z 2 xexy − 2xyz sin(y 2 z);
∂y
∂w
= 2zexy − xy 2 sin(y 2 z);
∂z
∂2w
∂2w
∂2w
2 2 xy
2 2 xy
2
2 2
2
=
y
z
e
;
=
z
x
e
−2xz
sin(y
z)+4xy
z
cos(y
z);
= 2exy −xy 4 cos(y 2 z);
∂x2
∂y 2
∂z 2
4
∂2w
∂2w
=
= z 2 exy + z 2 xyexy − 2yz sin(y 2 z);
∂x∂y
∂y∂x
∂2w
∂2w
=
= 2zxexy − 2xy sin(y 2 z) − 2xy 3 z cos(y 2 z);
∂y∂z
∂z∂y
∂2w
∂2w
=
= 2zyexy − y 2 sin(y 2 z).
∂z∂x
∂x∂z
3.
∂2z
1
y
∂2z
1
x
=
−
and
= − + 2.
2
2
2
∂x
x x
∂y
y y
4.
∂2z
∂2z
00
00
=
f
(x
+
ay)
+
F
(x
−
ay)
and
= a2 f 00 (x + ay) + a2 F 00 (x − ay).
∂x2
∂y 2
5