$f: \Bbb R^2 \to \Bbb R$ whose partials exist. Show: $\partial _xf \:\:\mathrm{continuous} \Rightarrow f \:\:\mathrm {differentiable}$ Let $f: \Bbb R^2 \to \Bbb R$ be a function whose partial derivatives exist.
Now i have to show: $$\partial _xf \:\:\mathrm{continuous} \Rightarrow f  \:\:\mathrm {differentiable}$$
Any tipps on how to approach this? (I really don't even have an intuitive view of why this should work) Thanks in advance!
 A: Hint:
Write
$$
f(x + h_1, y + h_2) - f(x, y) = f(x + h_1, y + h_2) - f(x, y + h_2) + f(x, y + h_2) - f(x, y) = \dfrac{\partial f}{\partial x}(\xi, y + h_2) h_1 + \dfrac{\partial f}{\partial y}(x, y) h_2 + o(h),
$$
where $\xi \in (x, x + h_1)$. What happens as $h \rightarrow 0$? Note that we have not assumed that $\dfrac{\partial f}{\partial y}$ is continuous so $f$ might not be $C^1$.
A: This is a classical theorem in analysis, assuming both (all) partial derivatives are continuous
By definition, $f$ is differentiable at $x_0$ if there exists a linear map $J$ such that
$$\lim_{\mathbf{h}\to \mathbf{0}} \frac{\|\mathbf{f}(\mathbf{x_0}+\mathbf{h}) - \mathbf{f}(\mathbf{x_0}) - \mathbf{J}\mathbf{(h)}\|_{\mathbf{R}^{n}}}{\| \mathbf{h} \|_{\mathbf{R}^{m}}}=0.$$
Recall that the natural differential $J$ for $f$ is given by the Jacobi matrix, whose entries are the partial derivatives of $f$.
Now write out the numerator in the definition of differentiable at $x_0$ so that it reads like $\sum_{i=1}^n a_i(x^i-x_0^i)$ using the https://en.wikipedia.org/wiki/Mean_value_theorem .
Then you can bound the above term using continuity of the partial derivatives. 
The solution can be found at page 201 of the following skript (in German, but yeah, math is math) https://people.math.ethz.ch/~struwe/Skripten/Analysis-I-II-final-6-9-2012.pdf
