Multivariate Calculus - Bruce E. Shapiro

LECTURE 12. THE CHAIN RULE 85
Example 12.5 Find ∂w/∂t and ∂w/∂s using the chain rule and express the result
as a function of s and t, where
w = x 3 + y + z 2 + xy
x = st
y = s + t
z = s + 4t
Solution. Now w is a function of 3 variables: x, y and z, so there are three terms
in each of the chain rule formulas. All of the derivatives are partial because we are
asked to find the partial derivative:
To find ∂w/∂t we calculate
∂w
∂t
∂w
∂t = ∂w ∂x
∂x ∂t + ∂w ∂y
∂y ∂t + ∂w ∂z
∂z ∂t
∂w
∂s = ∂w ∂x
∂x ∂s + ∂w ∂y
∂y ∂s + ∂w ∂z
∂z ∂s
= ∂w ∂x
∂x ∂t + ∂w ∂y
∂y ∂t + ∂w ∂z
∂z ∂t
= ∂(x3 + y + z 2 + xy) ∂(st)
∂x ∂t
+ ∂(x3 + y + z 2 + xy) ∂(s + 4t)
∂z
∂t
= (3x 2 + y)(s) + (1 + x)(1) + (2z)(4)
= s(3x 2 + y) + 1 + x + 8z
Substituting for x, y, and z gives
∂w
∂t
= s(3x 2 + y) + 1 + x + 8z
To find ∂w/∂s, we similarly calculate
∂w
∂s
+ ∂(x3 + y + z 2 + xy)
∂y
= s ( 3(st) 2 + (s + t) ) + 1 + st + 8(s + 4t)
= 3s 3 t 2 + 3s + 3t + 1 + st + 8s + 32t
= 3s 3 t 2 + 3s + 35t + 1 + st + 8s
= ∂w ∂x
∂x ∂s + ∂w ∂y
∂y ∂s + ∂w ∂z
∂z ∂s
= ∂(x3 + y + z 2 + xy) ∂(st)
∂x ∂s
+ ∂(x3 + y + z 2 + xy) ∂(s + 4t)
∂z
∂s
= = (3x 2 + y)(t) + (1 + x)(1) + 2z(1)
= = (3x 2 + y)t + 1 + x + 2z
+ ∂(x3 + y + z 2 + xy)
∂y
∂(s + t)
∂t
∂(s + t)
∂s
Math 250, Fall 2006 Revised December 6, 2006.