Multivariate Calculus - Bruce E. Shapiro

90 LECTURE 12. THE CHAIN RULE
Example 12.7 Repeat the previous example using theorem 1 instead of implicit
differentiation.
Solution. We had f(x, y) = x 3 + 2x 2 y − 10y 5 so that
(
∂
dy
dx = −∂f/∂x ∂f/∂y = − ∂x x 3 + 2x 2 y − 10y 5)
∂
∂y (x3 + 2x 2 y − 10y 5 ) = − 3x2 + 4xy
2x 2 − 50y 4
Example 12.8 Find dy/dx if x 2 cos y − y 2 sin x = 0.
Solution.Writing f(x, y) = x 2 cos y − y 2 sin x and usingtheorem 1,
(
∂
dy
dx = −∂f/∂x ∂f/∂y = − ∂x x 2 cos y − y 2 sin x )
∂
∂y (x2 cos y − y 2 sin x) = − 2x cos y − y2 cos x
−x 2 sin y − 2y sin x
Differentials
Definition 12.1 The differential of a function f(x, y, z) is the quantity
df = ∇f · dr
where
namely,
dr = (dx, dy, dz)
df = ∂f ∂f ∂f
dx + dy +
∂x ∂y ∂z dz
The differential gives an estimate of the change in the function f(x, y, z) when r is
perturbed by a small amount from (x, y, z) to (x + dx, y + dy, z + dz).
Example 12.9 Estimate the change in f(x, y) = x 2 + y 2 as you move from (1, 1)
to (1.01, 1.01) using differentials, and compare with the exact change.
Solution. We have
and
so that
Therefore
The exact change is
∆r = (0.01, 0.01)
∇f = (2x, 2y)
∇f(1, 1) = (2, 2)
∆f = ∇f · ∆r = (2, 2) · (0.01, 0.01) = .04
f(1.01, 1.01) − f(1, 1) = 1.01 2 + 1.01 2 − 1 − 1 = 0.0402
Revised December 6, 2006. Math 250, Fall 2006