Showing the derivative of f: $\mathbb R^n \rightarrow \mathbb R$ is equal to the function g. $f: B(0,1) \rightarrow \mathbb R$ where $B(0,1) \subset \mathbb R^n$ where $B(0,1)$ is the open ball of radius $1$ centered at $0$. Assume there exists a continuous map $g: B(0,1) \rightarrow L(\mathbb R^n, \mathbb R)$ such that for any $x,y \in B(0,1)$ we have that 
$f(y) - f(x) = \int_0^1 (g(ty + (1-t)x)(y-x))dt$. We have to prove that $f$ has a derivative equal to $g$. i.e that $Df = g$
 A: It is enough to prove it for $x=0$.
$$\frac{|f(y)-f(0)-g(0)(y)|}{\|y\|}=\frac{|\int_{0}^{1}g(ty)(y)dt-g(0)(y)|}{\|y\|}$$
Let the matrix of $g(x)$ in the standard basis be $(g_1(x),g_2(x),...,g_n(x))$. We are given that each $g_i$ is continuous real function.
Therefore $$\begin{align}\frac{\left|\int_{0}^{1}g(ty)(y)dt-g(0)(y)\right|}{\|y\|}&=\frac{\left|\int_{0}^{1}\sum_{k=1}^{n}g_k(ty)y_kdt-\sum_{k=1}^{n}g_k(0)y_k\right|}{\|y\|}\\&=\frac{\left|\sum_{k=1}^{n}y_k(\int_{0}^{1}g_k(ty)dt-g_k(0))\right|}{\|y\|}\\&\leq\left\|\left(\int_{0}^{1}(g_k(ty)-g_k(0))dt\right)\right\|\end{align}$$
where the last inequality is due to Cauchy's inequality.
Since each $g_k$ is continuous this tends to zero as $y\to 0$. In fact, given $\epsilon>0$ there is $\delta_k>0$ such that if $\|y\|<\delta$ then $|g_k(y)-g_k(0)|<\epsilon/\sqrt{n}$. Since $\|y\|<\delta\implies \|ty\|<\delta$, for $t\in[0,1]$, then we also have $|g_k(ty)-g(0)|<\epsilon/\sqrt{n}$. Taking $\delta=\min_{1\leq k\leq n}\delta_k$, we get that for $\|y\|<\delta$, then the right-hand side is $<\epsilon$
