There is a way to get the differentiability of a function from the existence and continuity of the partial
derivatives. This is very convenient because these partial derivatives are taken with respect to a one
dimensional variable. The following theorem is the main result.

Definition 3.4.1When f : U → ℝ^{p}for U an open subset of ℝ^{n}and the vector valued functions,

∂∂xfi

are all continuous, (equivalently each

∂∂fxij

is continuous), the function issaid to be C^{1}

(U)

. Ifall the partial derivatives up to order k exist and are continuous, then the function is said to be C^{k}.

It turns out that for a C^{1} function, all you have to do is write the matrix described in Theorem 3.3.1
and this will be the derivative. There is no question of existence for the derivative for such functions. This
is the importance of the next theorem.

Theorem 3.4.2Suppose f : U → ℝ^{p}where U is an open set in ℝ^{n}. Suppose also that all partialderivatives of f exist on U and are continuous. Then f is differentiable at every point of U.

Proof: If you fix all the variables but one, you can apply the fundamental theorem of calculus as
follows.

∫
1 ∂f--
f (x+vkek)− f (x) = 0 ∂xk (x + tvkek)vkdt. (3.5)

(3.5)

Here is why. Let h

(t)

= f

(x+ tvkek)

. Then

h(t+ h)− h (t) f (x+ tv e + hv e )− f (x +tv e )
--------------= -------k-k----k-k----------k-k-vk
h hvk

and so, taking the limit as h → 0 yields

′ ∂f--
h (t) = ∂xk (x + tvkek)vk

Therefore,

∫ 1 ∫ 1 ∂f
f (x+vkek )− f (x) = h (1) − h(0) = h′(t)dt = ----(x+ tvkek)vkdt.
0 0 ∂xk

Now I will use this observation to prove the theorem. Let v =

(v1,⋅⋅⋅,vn)

with

|v|

sufficiently small.
Thus v = ∑_{k=1}^{n}v_{k}e_{k}. For the purposes of this argument, define