96 CHAPTER 4. THE DERIVATIVE
4.5 Differentiability and C1 FunctionsThere is a way to get the differentiability of a function from the existence and continuityof the partial derivatives. This is very convenient because these partial derivatives are takenwith respect to a one dimensional variable. Of course, the determination of continuity isagain a multivariable consideration. The following theorem is the main result.
Definition 4.5.1 When f : U→Rp for U an open subset ofRn and the vector valuedfunctions, ∂ 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 thefunction is said to be Ck.
It turns out that for a C1 function, all you have to do is write the matrix described inTheorem 4.3.1 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 4.5.2 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. (4.6)
Here is why. Let h(t) = f(x+ tvkek) = f(x1, · · · ,xk + tvk,xk+1, · · · ,xn) . Then from thechain rule, h′ (t) = ∂ f
∂xk(x+ tvkek)vk. Therefore, since h′ is continuous, one can apply the
fundamental theorem of calculus to each component and write
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| suffi-ciently small. Thus v = ∑
nk=1 vkek. For the purposes of this argument, define ∑
nk=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
=n
∑i=1
∫ 1
0
(∂ f∂xi
(x+
n
∑k=i+1
vkek + tviei
)vi−
∂ f∂xi
(x)vi
)dt (4.7)
+n
∑i=1
∫ 1
0
∂ f∂xi
(x)vidt =n
∑i=1
∂ f∂xi
(x)vi +o(v)
and this shows f is differentiable at x. The reason for this is that each term in the sum in4.7 is o(v) . Indeed, letting |·| be the usual Euclidean norm,∣∣∣∣∣
∫ 1
0
(∂ f∂xi
(x+
n
∑k=i+1
vkek + tviei
)− ∂ f
∂xi(x)
)dtvi
∣∣∣∣∣