12.5 The Chain Rule
Chain Rule for functions of a single variable: If y = f (x) and x = g(t) then
dy dx
dy
=
dt
dx dt
For functions of more than one variable, the Chain Rule has several versions. Assume
that all functions below are dierentiable.
Case 1.
z = f (x, y), where x = g(t) and y = h(t). Then
dz
∂z dx ∂z dy
=
+
dt
∂x dt ∂y dt
√
Example 1. If z = x3 y 2 + x2 y 3 , where x = 1 + t and y = et . Find z 0 (t).
Example 2. The radius of a right circular cone is increasing at a rate of 1.8 cm/s while
its height is decreasing at a rate 2.5 cm/s. At what rate is the volume of the cone
changing when the radius is 120 cm and the height is 140 cm.
Case 2.
z = f (x, y), where x = g(s, t) and y = h(s, t). Then
∂z ∂z ∂x ∂z ∂y
=
+
∂s ∂x ∂s ∂y ∂s
∂z ∂z ∂x ∂z ∂y
=
+
∂t ∂x ∂t ∂y ∂t
Example 3. If z = x2 sin y , where x = s2 + t2 and y = 2st. Find
∂z
∂z
and .
∂s
∂t
The Chain Rule (general version) u = f (x1, . . . , xn), where each xj is a function
of the m variables t1 , t2 , . . . , tm . Then
∂u
∂u ∂x1
∂u ∂x2
∂u ∂xn
=
+
+ ··· +
∂ti
∂x1 ∂ti
∂x2 ∂ti
∂xn ∂ti
for each i = 1, 2, . . . , m.
Example 4. If u = x2 y + y 3 z 2 where x = rset , y = r + s2 e−t and z = rs sin t. Find us
when (r, s, t) = (1, 2, 0).
Implicit Dierentiation
Suppose that an equation F (x, y) = 0 denes y implicitly as a dierentiable function
of x, i.e., y = f (x). In order to nd dy/dx, we dierentiate both sides of the equation
F (x, y) = 0 with respect to x. If
∂F
6= 0 then
∂y
∂F
dy
Fx
= − ∂x = −
∂F
dt
Fy
∂y
Example 5. Find y 0 if x2 − xy + y 2 = 8.
Suppose that z is given implicitly as a function z = f (x, y) by an equation of the form
F (x, y, z) = 0. We can use the Chain Rule to dierentiate the equation F (x, y, z) = 0
and nd
∂F
Fx
∂z
= − ∂x = −
∂F
∂x
Fz
∂z
∂F
∂z
Fy
∂y
=−
=−
∂F
∂y
Fz
∂z
Example 6. Find
∂z
∂z
and
if x3 + y 3 + z 3 + 6xyz = 1.
∂x
∂y
Example 7. Find
∂x
∂x
and
if ln(xy + yz + 1) = yz 2 .
∂y
∂z