Section 12.5: The Chain Rule
Theorem: (Chain Rule for One Parameter)
Suppose that z = f (x, y) is a differentiable function of x and y, where x = x(t) and y = y(t)
are differentiable functions of t. Then z is a differentiable function of t and
∂z dx ∂z dy
dz
=
+
.
dt
∂x dt
∂y dt
Example: Let z = x2 y 3 + 2xy 2 , where x = sin t and y = e2t . Find
dz
.
dt
By the Chain Rule,
dz
∂z dx ∂z dy
=
+
dt
∂x dt
∂y dt
3
= (2xy + 2y 2 ) cos t + (3x2 y 2 + 4xy)2e2t
= 2 cos t(e6t sin t + e4t ) + 2e2t (3e4t sin2 t + 4e2t sin t).
Example: The radius of a circular cylinder is decreasing by 1.2 cm/sec and its height is
increasing at a rate of 3 cm/sec. At what rate is the volume changing when the radius is 80
cm and the height is 150 cm?
The volume of the cylinder is V = πr2 h. By the Chain Rule,
dV
dt
∂V dr ∂V dh
+
∂r dt
∂h dt
= 2πrh(−1.2) + πr2 (3)
= 2π(80)(150)(−1.2) + π(802 )(3)
= −9600π.
=
The volume is decreasing at a rate of 9600π cm3 /sec.
Theorem: (Chain Rule for Two Parameters)
Suppose that z = f (x, y) is a differentiable function of x and y, where x = x(s, t) and
y = y(s, t) are differentiable functions of s and t. Then z is a differentiable function of s and
t and
∂z
∂z ∂x ∂z ∂y
=
+
,
∂s
∂x ∂s ∂y ∂s
∂z ∂x ∂z ∂y
∂z
=
+
.
∂t
∂x ∂t
∂y ∂t
Example: Let z = x2 − 3x2 y 3 , where x = set and y = se−t . Find
∂z
∂z
and
.
∂s
∂t
By the Chain Rule,
∂z ∂x ∂z ∂y
∂z
=
+
∂s
∂x ∂s ∂y ∂s
= (2x − 6xy 3 )et − 9x2 y 2 e−t
= 2se2t (1 − 3s3 e−3t ) − 9s4 e3t ,
∂z
∂z ∂x ∂z ∂y
=
+
∂t
∂x ∂t
∂y ∂t
= (2x − 6xy 3 )set + 9x2 y 2 se−t
= 2s2 e2t (1 − 3s3 e−3t ) + 9s5 e3t .
Theorem: (Chain Rule, General Version)
Suppose that u is a differentiable function of x1 , . . . , xn and each xj is a differentiable function
of t1 , . . . , tm . Then
∂u ∂x1
∂u ∂x2
∂u ∂xn
∂u
=
+
+ ··· +
,
∂ti
∂x1 ∂ti
∂x2 ∂ti
∂xn ∂ti
i = 1, . . . , m.
∂u ∂u
Example: Let u = x4 y + y 2 z 3 , where x = rset , y = rs2 e−t , and z = r2 s sin t. Find
,
,
∂r ∂s
∂u
and
for r = 2, s = 1, and t = 0.
∂t
By the Chain Rule,
∂u
∂u ∂x ∂u ∂y ∂u ∂z
=
+
+
∂r
∂x ∂r ∂y ∂r ∂z ∂r
= (4x3 y)(set ) + (x4 + 2yz 3 )(s2 e−t ) + (3y 2 z 2 )(2rs sin t)
∂u
∂u ∂x ∂u ∂y ∂u ∂z
=
+
+
∂s
∂x ∂s ∂y ∂s ∂z ∂s
= (4x3 y)(ret ) + (x4 + 2yz 3 )(2rse−t ) + (3y 2 z 2 )(r2 sin t)
∂u
∂u ∂x ∂u ∂y ∂u ∂z
=
+
+
∂t
∂x ∂t
∂y ∂t
∂z ∂t
3
t
4
= (4x y)(rse ) + (x + 2yz 3 )(−rs2 e−t ) + (3y 2 z 2 )(r2 s cos t).
For r = 2, s = 1, and t = 0, we have x = 2, y = 2, and z = 0. Thus,
∂u
= (64)(1) + (16)(1) + (0)(0) = 80
∂r
∂u
= (64)(2) + (16)(4) + (0)(0) = 192
∂s
∂u
= (64)(2) + (16)(−2) + (0)(4) = 96.
∂t