By J. H. Curtiss
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Fm ). This “variable” notation is convenient in computations, but may cause ambiguities (see the next example). Example. Let u = g(x, y, z), where z depends on x and y: z = f3 (x, y). Set f1 (x, y) = x, f2 (x, y) = y, f = (f1 , f2 , f3 ), and h = g ◦ f ; so h(x, y) = g(x, y, z). By (8′ ), ∂u ∂x ∂u ∂y ∂u ∂z ∂u = + + . ∂x ∂x ∂x ∂y ∂x ∂z ∂x Here ∂x ∂f1 ∂y = = 1 and = 0, ∂x ∂x ∂x for f2 does not depend on x. Thus we obtain (9) ∂u ∂u ∂z ∂u = + . ) The trouble with (9) is that the variable u “poses” as both g and h.
Differentiable Functions 12. Do Problem 8(i) in case g is only continuous at p, and f (p) = 0. Find dh. 13. Do Problem 8(i) for dot products h = f · g of functions f, g : E ′ → E m (C m ). 14. Prove the following. (i) If φ ∈ L(E n , E 1 ) or φ ∈ L(C n , C), then φ = |v|, with v as in §2, Theorem 2(ii). (ii) If f : E n → E 1 (f : C n → C 1 ) is differentiable at p, then df (p, · ) = |∇f (p)|. Moreover, in case f : E n → E 1 , |∇f (p)| ≥ Du f (p) if |u| = 1 and |∇f (p)| = Du f (p) when u = thus ∇f (p) ; |∇f (p)| |∇f (p)| = max Du f (p).
As f is differentiable at p, so are its components fi (Problem 9 in §3), and si = dfi (p; ek ) = Dk fi (p) by Theorem 2(ii) in §3. Using also Corollary 1 in §3, we get m dg(q; s) = m si Di g(q) = i=1 Dk fi (p)Di g(q). i=1 But as s = df (p; ek ), formula (7) yields dg(q; s) = dh(p; ek ) = Dk h(p) by Theorem 2(ii) in §3. Thus the result follows. Note 3. Theorem 2 is often called the chain rule for functions of several variables. It yields Theorem 3 in Chapter 5, §1, if m = n = 1. 31 §4. The Chain Rule.