(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: R^{n} -> R^{m}, then Df(a) exists if all D_{j}f^{i} exist in an open set containing a and if each function D_{j}f^{i} 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: R^{n} -> R. Then

f(a + h) - f(a) = f(a_{1} + h_{1}, a_{2}, ... , a_{n}) - f(a_{1}, ... , a_{n}) + f(a_{1} + h_{1}, a_{2} + h_{2}, a_{3}, ... , a_{n}) - f(a_{1} + h_{1}, a_{2}, ... , a_{n}) + ...

+ f(a_{1} + h_{1}, ... , a_{n} + h_{n}) - f(a_{1} + h_{1}, ... , a_{n - 1} + h _{n - 1}, a_{n})

Recall that D_{1}f is the derivative of the function g defined by g(x) = f(x, a_{2}, ... , a_{n}). Applying the mean value theorem to g we obtain

f(a_{1} + h_{1}, a_{2}, ... , a_{n}) - f(a_{1}, ... , a_{n}) = h_{1}*D_{1}f(b_{1}, a_{2}, ... , a_{n}) for some b_{1} between a_{1} and a_{1} + h_{1}. Similarly, the ith term in the sum equals

h_{i}*D_{i}f(a_{1} + h_{1}, ... , a_{i - 1} + h_{i - 1}, b_{i}, ... , a_{n}) = h_{i}D_{i}f(c_{i}) for some c_{i}. Then

lim __|f(a + h) - f(a) - Σ D___{i}f(a)*h_{i}|
h -> 0 |h|
= lim __|(Σ D___{i}f(c_{i}) - D_{i}f(a))*h_{i}|
h -> 0 |h|

=< lim |Σ D_{i}f(c_{i}) - D_{i}f(a)| * __|h___{i}|
h -> 0 |h|

=< lim |Σ D_{i}f(c_{i}) - D_{i}f(a)|
h -> 0

= 0

Then the theorem follows for arbitrary m, since the ith row of f'(a) is (f

_{i})'(a) (proven earlier in the book, and left here as an exercise for the reader) and each f

_{i} is differentiable.