Page 52 - Vector Analysis

48 CHAPTER 2. Differentiation of Functions of Several Variables

Then fx(0, 0) = 1 and fy(0, 0) = 0. However,

ˇ[ ] [] ˇ |x|y2
ˇf (x, y) ´ f (0, 0) ´ 1 0 x ˇ
ˇ y ˇ

ax2 + y2 = (x2 + y 2 ) 3 Û 0 as (x, y) Ñ (0, 0).
2

Therefore, f is not differentiable at (0, 0). On the other hand, f is continuous at (0, 0) since

ˇˇf (x, y) ´ f (0, 0)ˇˇ = ˇˇf (x, y)ˇˇ ď |x| Ñ 0 as (x, y) Ñ (0, 0).

2.5.2 The Product Rules

Proposition 2.45. Let U Ď Rn be an open set, and f : U Ñ Rm and g : U Ñ R be
differentiable at x0 P A. Then gf : A Ñ Rm is differentiable at x0, and

D(gf )(x0)(v) = g(x0)(Df )(x0)(v) + (Dg)(x0)(v)f (x0) . (2.3)

Moreover, if g(x0) ‰ 0, then f :A Ñ Rm is also differentiable at x0, and D( f )(x0) : Rn Ñ
g g
Rm is given by

(f ) () (x0)
D g (x0)(v) g(x0) (Df )(x0)(v) ´ (Dg)(x0)(v)f
= g2(x0) . (2.4)

Proof. We only prove (2.3), and (2.4) is left as an exercise.
Let A be the Jacobian matrix of gf at x0; that is, the (i, j)-th entry of A is

B (gfi) (x0) = g(x0) B fi (x0) + Bg (x0)fi(x0) .
B xj B xj B xj

Then Av = g(x0)(Df )(x0)(v) + (Dg)(x0)(v)f (x0); thus
()

(gf )(x) ´ (gf )(x0) ´ A(x ´ x0) = g(x0) f (x) ´ f (x0) ´ (Df )(x0)(x ´ x0)
()

+ g(x) ´ g(x0) ´ (Dg)(x0)(x ´ x0) f (x)
( )( )

+ (Dg)(x0)(x ´ x0) f (x) ´ f (x0) .

Since (Dg)(x0) P B(Rn, R), }(Dg)(x0)}B(Rn,R) ă 8; thus using the inequality

ˇˇ(Dg)(x0)(x ´ x0)ˇˇ ď ››(Dg)(x0)››B(Rn,R)}x ´ x0}Rn

and the continuity of f at x0 (due to Theorem 2.40), we find that

ˇ ˇˇ(Dg)(x0)(x ´ x0)ˇˇ ˇ
ˇ
lim ››f (x) ´ f (x0)››Rm ˇ ď lim ››(Dg)(x0)››B(Rn,R)››f (x) ´ f (x0)››Rm = 0.
ˇ }x ´ x0}Rn ˇ
xÑx0 xÑx0