3Partial differentiation
IA Differential Equations
3.3 Implicit differentiation
Consider the contour surface of a function
F
(
x, y, z
) given by
F
(
x, y, z
) = const.
This implicitly defines
z
=
z
(
x, y
). e.g. If
F
(
x, y, z
) =
xy
2
+
yz
2
+
z
5
x
= 5,
then we can have
x
=
5−yz
2
y
2
+z
5
. Even though
z
(
x, y
) cannot be found explicitly
(involves solving quintic equation), the derivatives of
z
(
x, y
) can still be found
by differentiating F (x, y, z) = const w.r.t. x holding y constant. e.g.
∂
∂x
(xy
2
+ yz
2
+ z
5
x) =
∂
∂x
5
y
2
+ 2yz
∂z
∂x
+ z
5
+ 5z
4
x
∂z
∂x
= 0
∂z
∂x
= −
y
2
+ z
5
2yz + 5z
4
x
In general, we can derive the following formula:
Theorem (Multi-variable implicit differentiation). Given an equation
F (x, y, z) = c
for some constant c, we have
∂z
∂x
y
= −
(∂F )/(∂x)
(∂F )/(∂z)
Proof.
dF =
∂F
∂x
dx +
∂F
∂y
dy +
∂F
∂z
dz
∂F
∂x
y
=
∂F
∂x
∂x
∂x
y
+
∂F
∂y
∂y
∂x
y
+
∂F
∂z
∂z
∂x
y
= 0
∂F
∂x
+
∂F
∂z
∂z
∂x
y
= 0
∂z
∂x
y
= −
(∂F )/(∂x)
(∂F )/(∂z)