Calculus 2nd Edition Rogawski

SECTION 15.5 The Gradient and Directional Derivatives 821
CAUTION Do not confuse the Chain Rule for
Paths with the more elementary Chain Rule
for Gradients stated in Theorem 1 above.
THEOREM 2 Chain Rule for Paths
If f and c(t) are differentiable, then
d
dt f(c(t)) = ∇f c(t) · c ′ (t)
Explicitly, in the case of two variables, if c(t) = (x(t), y(t)), then
〈
d
∂f
dt f(c(t)) = ∂x , ∂f 〉
· 〈x ′ (t), y ′ (t) 〉 = ∂f dx
∂y
∂x dt + ∂f dy
∂y dt
Proof By definition,
d
f (x(t + h), y(t + h)) − f (x(t), y(t))
f(c(t)) = lim
dt h→0
h
To calculate this derivative, set
f = f (x(t + h), y(t + h)) − f (x(t), y(t))
x = x(t + h) − x(t),
y = y(t + h) − y(t)
The proof is based on the local linearity of f . As in Section 15.4, we write
f = f x (x(t), y(t))x + f y (x(t), y(t))y + e(x(t + h), y(t + h))
Now set h = t and divide by t:
f
t
= f x (x(t), y(t)) x
t + f y(x(t), y(t)) y e(x(t + t), y(t + t))
+
t t
Suppose for a moment that the last term tends to zero as t → 0. Then we obtain the
desired result:
d
f
f(c(t)) = lim
dt t→0 t
x
= f x (x(t), y(t)) lim
t→0 t + f y
y(x(t), y(t)) lim
t→0 t
= f x (x(t), y(t)) dx
dt + f y(x(t), y(t)) dy
dt
= ∇f c(t) · c ′ (t)
We verify that the last term tends to zero as follows:
(√ )
e(x(t + t), y(t + t)) e(x(t + t), y(t + t)) (x)
lim
= lim √ 2 + (y) 2
t→0 t
t→0 (x) 2 + (y) 2 t
(
) ⎛√ ⎞
(x )
e(x(t + t), y(t + t))
2 ( )
= lim √ lim ⎝
y 2
+ ⎠ = 0
t→0 (x) 2 + (y) 2 t→0 t t
} {{ }
Zero
The first limit is zero because a differentiable function is locally linear (Section 15.4). The
second limit is equal to √ x ′ (t) 2 + y ′ (t) 2 , so the product is zero.