Multivariate Calculus - Bruce E. Shapiro

84 LECTURE 12. THE CHAIN RULE
Solution.
dw
dt
= ∂w dx
∂x dt + ∂w dy
∂y dt + ∂w dy
∂z dt
∂(xy + yz + xz) d(t 2 ) ∂(xy + yz + xz)
= +
∂x dt
∂y
∂(xy + yz + xz) d(1 − t)
+
∂z dt
= (y + z)(2t) + (x + z)(−2t) + (y + x)(−1)
= 2t(y + z − x − z) − y − x
= 2t(y − x) − y − x
= 2t(1 − t 2 − t 2 ) − (1 − t 2 ) − t 2
= 2t(1 − 2t 2 ) − 1
= 2t − 4t 3 − 1
d(1 − t 2 )
dt
Example 12.4 Find ∂w/∂t and ∂w/∂s using the chain rule and express the result
as a function of s and t, for w = 5x 2 − y ln x, x = s + t, and y = s 3 t.
Solution. Here w only depends on two variables x and y, so the chain rules become
∂w
∂t = ∂w ∂x
∂x ∂t + ∂w ∂y
∂y ∂t
∂w
∂s = ∂w ∂x
∂x ∂s + ∂w ∂y
∂y ∂s
The second factor in each term is a partial derivative because x and y are functions
of two variables, s and t, and not just functions of a single variable. From the first
equation,
Similarly,
∂w
∂t
∂w
∂s
= ∂w ∂x
∂x ∂t + ∂w ∂y
∂y ∂t
= ∂(5x2 − y ln x)
∂x
∂(s + t)
∂t
= (10x − y/x)(1) + (− ln x)(s 3 )
= 10(s + t) − s3 t
− [ln(s + t)]s3
s + t
= ∂w ∂x
∂x ∂s + ∂w ∂y
∂y ∂s
= ∂(5x2 − y ln x)
∂x
∂(s + t)
∂s
= (10x − y/x)(1) + (− ln x)(3s 2 t)
= 10(s + t) − s3 t
s + t − 3s2 t ln(s + t)
+ ∂(5x2 − y ln x) ∂(s 3 t)
∂y ∂t
+ ∂(5x2 − y ln x) ∂(s 3 t)
∂y ∂s
Revised December 6, 2006. Math 250, Fall 2006