15.5 The Chain Rule
Theorem (The Chain Rule - case 1) Suppose z = f (x, y) is a differentiable function and x = x(t), y = y(t) are both
differentiable functions of t. Then z is differentiable with respect to t and
dz
∂f dx ∂f dy
=
+
dt
∂x dt
∂y dt
(Proof)
∆z
=
∂f
∂f
∆x +
∆y + 1 ∂x + 2 ∂y
∂x
∂y
∆z
∆t
=
∂f ∆x
∂f ∆y
∆x
∆y
+
+ 1
+ 2
∂x ∆t
∂y ∆t
∆t
∆t
=
∂f dx
∂f dy
dx
dy
∂f dx
∂f dy
+
+
lim 1 +
lim 2 =
+
∂x dt
∂y dt
dt ∆t→0
dt ∆t→0
∂x dt
∂y dt
lim
∆t→0
∆z
∆t
because lim 1 = lim 2 = 0
∆t→0
∆t→0
Example Use the Chain Rule to find dz/dt or dw/dt :
x = 5t4 , y = 1/t
(a) z = cos(x + 4y) ;
(b) w = tan−1
(Answers)
p
x2 + y 2 + z 2 ;
x = sin t, y = cos t, z = tan t
dx
dy
dz
3
2
= (cos(x + 4y))x
+ (cos(x + 4y))y
= − sin(x + 4y) · 20t − 4 sin(x + 4y) · (−1/t )
dt
dt
dt
dx
dy
dz
p
p
p
dw
−1
−1
−1
(b)
x2 + y 2 + z 2
x2 + y 2 + z 2
x2 + y 2 + z 2
= tan
+ tan
+ tan
x dt
y dt
z dt
dt
(a)
=
x
(1 + x2 + y 2 + z 2 )
+
+
=
p
x2 + y 2 + z 2
· cos t
y
(1 + x2 + y 2 + z 2 )
p
x2 + y 2 + z 2
p
x2 + y 2 + z 2
z
(1 + x2 + y 2 + z 2 )
· (− sin t)
2
· sec t
x cos t − y sin t + z sec2 t
p
x2 + y 2 + z 2
(1 + x2 + y 2 + z 2 )
Theorem (The Chain Rule - case 2) Let z = f (x, y) be a differentiable function of x and y where x = x(s, t), y = y(s, t)
are differentiable functions of s and t. Then z = f (x(s, t), y(s, t)) is a differentiable function of s and t and
∂z
∂f ∂x ∂f ∂y
=
+
,
∂s
∂x ∂s
∂y ∂s
∂z
∂f ∂x ∂f ∂y
=
+
∂t
∂x ∂t
∂y ∂t
Example Use the Chain Rule to find ∂z/∂s and ∂z/∂t
(a) z = arcsin(xy) ;
x = s2 + t2 , y = 1 − 2st
(b) z = er cos θ ; r = st, θ =
(Answers)
(a)
(b)
√
s2 + t2
∂z
y
x
2ys − 2xt
= (arcsin(xy))x · xs + (arcsin(xy))y · ys = p
· 2s + p
· (−2t) = p
∂s
1 − x2 y 2
1 − x2 y 2
1 − x2 y 2
∂z
2ty − 2sx
Similarly,
= p
∂t
1 − x2 y 2
∂z
s
r
r
r
r
= (e cos θ)r · rs + (e cos θ)θ · θs = e cos θ · t − e sin θ · √
∂s
s2 + t2
∂z
t sin θ
r
Similarly,
=e
s cos θ − √
∂t
s2 + t2
1
Theorem (The Chain Rule - generalized) Suppose z = f (x1 , · · · , xn ) be a differentiable function where

x1 = x1 (t1 , · · · , tm )



 x2 = x2 (t1 , · · · , tm )
are differentiable functions of t1 , · · · , tm . Then z = f (x1 (t1 , · · · , tm ), · · · , xn (t1 , · · · , tm ))
.

 ..


xn = xn (t1 , · · · , tm )
is a differentiable function of t1 , · · · , tm and
n
X ∂f ∂xk
∂f ∂x1
∂f ∂x2
∂f ∂xn
∂z
=
+
+ ··· +
=
∂ti
∂x1 ∂xi
∂x2 ∂ti
∂xn ∂ti
∂xk ∂ti
k=1
Example Use the Chain Rule to find
∂u ∂u ∂u
,
,
when
∂p ∂r ∂θ
u = x2 + yz ;
(Answers)
x = pr cos θ, y = pr sin θ, z = p + r
• up = (x2 + yz)x · xp + (x2 + yz)y · yp + (x2 + yz)z · zp = 2x · r cos θ + z · r sin θ + y · 1
• ur = (x2 + yz)x · xr + (x2 + yz)y · yr + (x2 + yz)z · zr = 2x · p cos θ + z · p sin θ + y · 1
• uθ = (x2 + yz)x · xθ + (x2 + yz)y · yθ + (x2 + yz)z · zθ = 2x · (−pr sin θ) + z · pr cos θ + y · 0
Theorem (Implicit Differentiation)
(a) Given f (x, y) = C where y is an implicitly defined function of x, we have
f (x, y) = C
⇓
fx dx + fy dy = 0
provided fx and fy exist.
(b) Given f (x, y, z) = C where z is an implicitly defined function of x and y, we have
f (x, y, z) = C
⇓
fx + fz
∂z
∂z
= 0 and fy + fz
=0
∂x
∂y
provided fx , fy , and fz exist.
(Explanation)
(a) Since y = y(x), we can view f (x, y) and C as functions of x. Since f (x, y) and C are equal at all x, their derivatives with respect to x
are equal as well. Differentiate both sides of f (x, y(x)) = C with respect to x :
d
f (x, y)
dx
fx
dx
dy
+ fy
dx
dx
fx + fy
dy
dx
fx dx + fy dy
=
d
C
dx
=
0
=
0
=
0
(Chain Rule)
2
(b) Since z = z(x, y) and x,y are independent variables, we can view f (x, y, z(x, y)) and C as functions of two variables x and y. The
two functions f (x, y, z(x, y)) and C are equal at all x and y. Hence their partial derivatives with respect to x(and y) are equal at all x
and y. Differentiate both sides of f (x, y, z(x, y)) = C with respect to x :
∂
f (x, y, z)
∂x
fx
=
∂
C
∂x
∂y
∂z
∂x
+ fy
+ fz
∂x
∂x
∂x
=
0
∂z
∂x
=
0
fx + 0 + fz
(Chain Rule)
Similarly when we differentiate functions in both sides of f (x, y, z) = C with respect to y, we get the result fy + fz
Example
2
(a) Use partial derivatives to find dy/dx when y 5 + x2 y 3 = 1 + yex .
(b) Use partial derivatives to find ∂z/∂x and ∂z/∂y when xyz = cos(x + y + z).
(Answers)
2
(a) Let f (x, y) = y 5 + x2 y 3 − 1 − yex . Then we have f (x, y) = 0.
2
2
(y 5 + x2 y 3 − 1 − yex )dx + (y 5 + x2 y 3 − 1 − yex )dy
x2
(2xy 3 − 2xye
)dx + (5y 4 + 3x2 y 2 − e
x2
)dy
=
0
=
0
2
2xy 3 − 2xyex
dy
=−
.
Hence
dx
5y 4 + 3x2 y 2 − ex2
(b) Let f (x, y, z) = xyz − cos(x + y + z). We have f (x, y, z) = 0. Differentiate both sides with respect to x :
(xyz − cos(x + y + z))x + (xyz − cos(x + y + z))z
(yz + sin(x + y + z)) + (xy + sin(x + y + z))
∂z
∂x
∂z
∂x
∂z
yz + sin(x + y + z)
Hence
=−
.
∂x
xy + sin(x + y + z)
xz + sin(x + y + z)
∂z
=−
.
By the same way we achieve
∂y
xy + sin(x + y + z)
3
=
0
=
0
∂z
= 0.
∂y