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(For those looking for a bit of an introduction, the component function writeup sort of segues into this one.)

While the differentiability of any function f at a point a is sufficient to guarantee the existence of the partial derivative of each component function of f at a, the converse does not necessarily hold. However, it is possible to obtain a condition under which the derivative of f at a certainly exists by making our starting condition slightly stronger. Specifically:

If f: Rn -> Rm, then Df(a) exists if all Djfi exist in an open set containing a and if each function Djfi is continuous at a (such a function f is called continuously differentiable at a).

Proof (taken from Michael Spivak's Calculus on Manifolds (1965) Cambridge, Mass.: Perseus Books): Consider the case where m = 1, so that f: Rn -> R. Then

f(a + h) - f(a) = f(a1 + h1, a2, ... , an) - f(a1, ... , an) + f(a1 + h1, a2 + h2, a3, ... , an) - f(a1 + h1, a2, ... , an) + ...
+ f(a1 + h1, ... , an + hn) - f(a1 + h1, ... , an - 1 + h n - 1, an)

Recall that D1f is the derivative of the function g defined by g(x) = f(x, a2, ... , an). Applying the mean value theorem to g we obtain

f(a1 + h1, a2, ... , an) - f(a1, ... , an) = h1*D1f(b1, a2, ... , an) for some b1 between a1 and a1 + h1. Similarly, the ith term in the sum equals

hi*Dif(a1 + h1, ... , ai - 1 + hi - 1, bi, ... , an) = hiDif(ci) for some ci. Then

lim      |f(a + h) - f(a) - Σ Dif(a)*hi|
h -> 0                 |h|

= lim |(Σ Dif(ci) - Dif(a))*hi| h -> 0 |h|

=< lim |Σ Dif(ci) - Dif(a)| * |hi| h -> 0 |h|

=< lim |Σ Dif(ci) - Dif(a)| h -> 0

= 0

Then the theorem follows for arbitrary m, since the ith row of f'(a) is (fi)'(a) (proven earlier in the book, and left here as an exercise for the reader) and each fi is differentiable.

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