Page 56 - Vector Analysis

52 CHAPTER 2. Differentiation of Functions of Several Variables

Suppose that f is differentiable. Then F is differentiable, and the chain rule implies that
]Bx Bx [
[BF BF ] = [Bf Bf   Bf B f ] [ θ ]
By Bx cos ´r sin θ
Br Bθ Bx B r B θ =
B y B y .
B y sin θ r cos θ

Br Bθ

Therefore, we arrive at the following form of chain rule

B = B x B + B y B and B = B x B + B y B
Br Br Bx Br By Bθ Bθ Bx Bθ By

which is commonly seen in Calculus textbook.

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Example 2.51. Let f : R Ñ( R and) F : R2 Ñ R be differentiable, and F x, f (x) = 0 and
BF ´ Fx(x, f (x)) , BF BF .
By ‰ 0. Then f 1(x) = where Fx = Bx and Fy =
Fy x, f (x) By
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Example 2.52. Let γ :)(0, 1) Ñ Rn and f : Rn Ñ R be differentiable. Let F (t) = f γ(t) .
(
Then F 1(t) = (Df ) γ(t) γ 1(t).

Example 2.53. Let f (u, v, w) = u2v + wv2 and g(x, y) = (xy, sin x, ex). Let h = f ˝ g :

R2 Ñ R. Find Bh .

Bx

Way I: Compute Bh directly: Since

Bx

h(x, y) = f (g(x, y)) = f (xy, sin x, ex) = x2y2 sin x + ex sin2 x ,

we have Bh = 2xy2 sin x + x2y2 cos x + ex sin2 x + 2ex sin x cos x .
Bx

Way II: Use the chain rule:

Bh = Bf B g1 + Bf B g2 + Bf B g3 = 2uv ¨y + (u2 + 2wv) ¨ cos x + v2 ¨ ex
Bx Bu Bx Bv Bx Bw Bx

= 2xy2 sin x + (x2y2 + 2ex sin x) cos x + ex sin2 x.

Example 2.54. Let F (x, y) = f (x2 + y2), f : R Ñ R, F : R2 Ñ R. Show that x BF = y BF .

By Bx
Proof: Let g(x, y) = x2 + y2, g : R2 Ñ R, then F (x, y) = (f ˝ g)(x, y). By the chain rule,
[ ][ ]
BF Bg Bg [ ]
BF By = f 1(g(x, y)) ¨ Bx = f 1(g(x, y)) 2x 2y
Bx By

which implies that

BF = 2xf 1(g(x, y)), BF = 2yf 1(g(x, y)) .

Bx By

So y BF = f 1(g(x, y))2xy = x BF .

Bx By