“JUST THE MATHS”
UNIT NUMBER
14.6
PARTIAL DIFFERENTIATION 6
(Implicit functions)
by
A.J.Hobson
14.6.1
14.6.2
14.6.3
14.6.4
Functions of two variables
Functions of three variables
Exercises
Answers to exercises
UNIT 14.6 - PARTIAL DIFFERENTIATION 6
IMPLICIT FUNCTIONS
14.6.1 FUNCTIONS OF TWO VARIABLES
The chain rule, encountered earlier, has a convenient application to implicit relationships of
the form,
f (x, y) = constant,
between two independent variables, x and y.
It provides a means of determining the total derivative of y with respect to x.
Explanation
Taking x as the single independent variable, we may interpret f (x, y) as a function of x and
y in which both x and y are functions of x.
Differentiating both sides of the relationship, f (x, y) = constant, with respect to x gives
∂f dx ∂f dy
.
+
.
= 0.
∂x dx ∂y dx
In other words,
∂f dy
∂f
+
.
= 0.
∂x ∂y dx
Hence,
∂f
dy
∂x
= − ∂f
.
dx
∂y
1
EXAMPLES
1. If
f (x, y) ≡ x3 + 4x2 y − 3xy + y 2 = 0,
determine an expression for
Solution
dy
.
dx
∂f
∂f
= 3x2 + 8xy − 3y and
= 4x2 − 3x + 2y.
∂x
∂y
Hence,
dy
3x2 + 8xy − 3y
=− 2
.
dx
4x − 3x + 2y
2. If
f (x, y) ≡ x sin(2x − 3y) + y cos(2x − 3y),
determine and expression for
Solution
dy
.
dx
∂f
= sin(2x − 3y) + 2x cos(2x − 3y) − 2y sin(2x − 3y)
∂x
and
∂f
= −3x cos(2x − 3y) + cos(2x − 3y) + 3y sin(2x − 3y).
∂y
Hence,
dy
sin(2x − 3y) + 2x cos(2x − 3y) − 2y sin(2x − 3y)
=
.
dx
3x cos(2x − 3y) − cos(2x − 3y) − 3y sin(2x − 3y)
14.6.2 FUNCTIONS OF THREE VARIABLES
For relationships of the form,
f (x, y, z) = constant,
let us suppose that x and y are independent of each other.
2
Then, regarding f (x, y, z) as a function of x, y and z, where x, y and z are all functions of
x and y, the chain rule gives
∂f ∂x ∂f ∂y ∂f ∂z
+
.
+
.
= 0.
∂x ∂x ∂y ∂x ∂z ∂x
But,
∂x
∂y
= 1 and
= 0.
∂x
∂x
Hence,
∂f
∂f ∂z
+
.
= 0,
∂x ∂z ∂x
giving
∂f
∂z
∂x
;
= − ∂f
∂x
∂z
and, similarly,
∂f
∂z
∂y
.
= − ∂f
∂y
∂z
EXAMPLES
1. If
f (x, y, z) ≡ z 2 xy + zy 2 x + x2 + y 2 = 5,
determine expressions for
dz
dx
and
dz
.
dy
Solution
∂f
= z 2 y + zy 2 + 2x,
∂x
3
∂f
= z 2 x + 2zyx + 2y
∂y
and
∂f
= 2zxy + y 2 x.
∂z
Hence,
∂z
z 2 y + zy 2 + 2x
=−
∂x
2zxy + y 2 x
and
∂z
z 2 x + 2zyx + 2y
=−
.
∂y
2zxy + y 2 x
2. If
f (x, y, z) ≡ xey
determine expressions for
dz
dx
and
2 +2z
,
dz
.
dy
Solution
∂f
2
= ey +2z ,
∂x
∂f
2
= 2yxey +2z ,
∂y
and
∂f
2
= 2xey +2z .
∂z
Hence,
2
∂z
ey +2z
1
=−
2 +2z = −
y
∂x
2xe
2x
and
2
∂z
2yxey +2z
=−
= −y.
∂y
2xey2 +2z
4
14.6.3 EXERCISES
1. Use partial differentiation to determine expressions for
dy
dx
in the following cases:
(a)
x3 + y 3 − 2x2 y = 0;
(b)
ex cos y = ey sin x;
(c)
sin2 x − 5 sin x cos y + tan y = 0.
2. If
x2 y + y 2 z + z 2 x = 10,
where x and y are independent, determine expressions for
∂z
∂z
and
.
∂x
∂y
3. If
xyz − 2 sin(x2 + y + z) + cos(xy + z 2 ) = 0,
where x and y are independent, determine expressions for
∂z
∂z
and
.
∂x
∂y
4. If
r2 sin θ = (r cos θ − 1)z,
where r and θ are independent, determine expressions for
∂z
∂z
and
.
∂r
∂θ
5
14.6.4 ANSWERS TO EXERCISES
1. (a)
dy
4xy − 3x2
= 2
;
dx
3y − 2x2
(b)
dy
ex cos y − ey cos x
= x
;
dx
x sin y + ey sin x
(c)
dy
5 cos x cos y − 2 sin x cos x
=
.
dx
5 sin xsiny + sec2 y
2.
∂z
2xy + z 2
=− 2
∂x
y + 2zx
and
∂z
x2 + 2yz
=− 2
.
∂y
y + 2zx
3.
∂z
yz − 4x cos(x2 + y + z) − y sin(xy + z 2 )
=−
∂x
xy − 2 cos(x2 + y + z) − 2z sin(xy + z 2 )
and
∂z
xz − 2 cos(x2 + y + z) − x sin(xy + z 2 )
=−
.
∂y
xy − 2 cos(x2 + y + z) − 2z sin(xy + z 2 )
4.
∂z
2r sin θ − z cos θ
=
∂r
r cos θ − 1
and
∂z
r2 cos θ + rz sin θ
=
.
∂θ
r cos θ − 1
6