468 CHAPTER 22. DERIVATIVE OF A FUNCTIONS OF MANY VARIABLES

22.4 C1 FunctionsMost of the time, there is an easier way to conclude that a derivative exists and to find it. Itinvolves the notion of a C1 function.

Definition 22.4.1 When f : U → Rp for U an open subset of Rn and the vectorvalued functions ∂f

∂xiare all continuous, (equivalently each ∂ fi

∂x jis continuous), the function

is said to be C1 (U). If all the partial derivatives up to order k exist and are continuous,then the function is said to be Ck.

It turns out that for a C1 function, all you have to do is write the matrix described inTheorem 22.2.4 and this will be the derivative. There is no question of existence for thederivative for such functions. This is the importance of the next theorem.

Theorem 22.4.2 Suppose f : U → Rp where U is an open set in Rn. Suppose alsothat all partial derivatives of f exist on U and are continuous. Then f is differentiable atevery point of U.

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

f (x+vkek)−f (x) =∫ 1

0

∂f

∂xk(x+ tvkek)vkdt. (22.7)

Here is why. Let h(t) = f (x+ tvkek). Then

h(t +h)−h(t)h

=f (x+ tvkek +hvkek)−f (x+ tvkek)

hvkvk

and so, taking the limit as h → 0 yields

h′ (t) =∂f

∂xk(x+ tvkek)vk

Therefore,

f (x+vkek)−f (x) = h(1)−h(0) =∫ 1

0h′ (t)dt =

∫ 1

0

∂f

∂xk(x+ tvkek)vkdt.

Now I will use this observation to prove the theorem. Let v = (v1, · · · ,vn) with |v|sufficiently small. Thus v = ∑

nk=1 vkek. For the purposes of this argument, define

n

∑k=n+1

vkek ≡ 0.

Then with this convention,

f (x+v)−f (x) =n

∑i=1

(f

(x+

n

∑k=i

vkek

)−f

(x+

n

∑k=i+1

vkek

))

=n

∑i=1

∫ 1

0

∂f

∂xi

(x+

n

∑k=i+1

vkek + tviei

)vidt