Let $f:\mathbb{R}^n \to \mathbb{R}^n$ and $\{v_1, v_2, ..., v_n\}$ base of $\mathbb{R}^n$. Prove that if $\frac{\partial f}{\partial v_j}$ is continuous at $x_0$, $\forall 1 \leq j \leq n$ and for some $x_0 \in \mathbb{R}^n$, then $f$ is differentiable at $x_0$.
I tried what i thought is the natural candidate: if $x=\alpha_1v_1+...+\alpha_nv_n$, $Df(x) = \alpha_1\frac{\partial f}{\partial v_1}+...+\alpha_n\frac{\partial f}{\partial v_n}$, but i wasn't able to prove $\frac{R(x)}{\Vert x - x_0\Vert} \rightarrow 0$. Any ideas?
The key to this question is manipulating the differential quotient to get expressions involving the partial derivatives. You would do well to remember this.
To show that $Df(x)$ exists, we want to show that for every $\alpha \in \mathbb R^n$, the differential quotient $\lim_{h\to 0} \frac{f(x + h\alpha) - f(x)}{h}$ exists. We do something smart here: first, since $\alpha = \sum \alpha_iv_i$, where $\alpha_i$ are scalars and $v_i$ are the basis vectors, we have: \begin{split} \frac{f(x+h\alpha) - f(x)}{h} & = \frac{f(x + h(\sum_{i=1}^{\mathbf n} \alpha_iv_i)) - f(x+ h(\sum_{i=1}^{\mathbf{n-1}} \alpha_iv_i))}{h} \\& + \frac{f(x + h(\sum_{i=1}^{\mathbf{n-1}} \alpha_iv_i)) - f(x + h(\sum_{i=1}^{\mathbf {n-2}} \alpha_iv_i))}{h} + \frac{f(x + h(\sum_{i=1}^{\mathbf{n-2}} \alpha_iv_i)) - f(x + h(\sum_{i=1}^{\mathbf {n-3}} \alpha_iv_i))}{h} \\& + \ldots \\ & + \frac{f(x + h\alpha_1v_1) - f(x)}{h} \end{split}
Right, so what have we done?We broke the sum above, into various parts, where you can see that in each fraction, the difference between the two inputs of $f$ is just some scalar multiple of a basis vector.
This is the insight. Allow me to be sketchy from here on.
At this stage, you should remember the mean value theorem in one variable. Something similar holds for partial derivatives as well : if $\frac{\partial g}{\partial x_i}$ exists in a neighbourhood of $a$, then $g(a+\alpha x_i) - g(x_i) = \alpha \times \frac{\partial g}{\partial x_i} (a + \beta x_i)$ , for some $0 < \beta < \alpha$.
Use that theorem, on each of the above terms. Then, let $h \to 0$, and note that the partial derivatives are continuous, to conclude that the partial derivative is indeed what you expect it to be i.e. that given in your question statement. $()$