Prove that $f$ is differentiable at $a$ Let $f$ be a function with domain of an open disk $A \subset \mathbb R^2$. Suppose that its partial derivatives exists at $a\in A$ and that $\frac{\partial f}{\partial x}$ is continuous in $A$. Prove that $f$ is differentiable at $a$
I know that if I can prove that $\frac{\partial f}{\partial y}$ is continuous in $A$, then I'm done. However, from what given I can't see anything that can lead to that conclusion 
 A: Write $a=(x_0,y_0)$, and set (for $u,v$ small enough)
$$\Delta(u,v)=f(x_0+u,y_0+v)-f(x_0,y_0)-\left(\frac{\partial f}{\partial x}(x_0,y_0)\, u+\frac{\partial f}{\partial x}(x_0,y_0)\, v\right)$$
One has to show that $\Delta(u,v)=o(\Vert (u,v)\Vert)$ as $(u,v)\to (0,0)$, where $\Vert\;\;\Vert$ is any norm on $\mathbb R^2$.
The idea is to write $$f(x_0+u,y_0+v)-f(x_0,y_0)=f(x_0+u,y_0+v)-f(x_0,y_0+v)+f(x_0,y_0+v)-f(x_0,y_0)\, .$$
Since $\frac{\partial f}{\partial x}$ exists in $A$, one can find $c_{u,v}$ between $x_0$ and $x_0+u$ such that 
$$f(x_0+u,y_0+v)-f(x_0,y_0+v)= \frac{\partial f}{\partial x}(c_{u,v}\, y_0+v) \, u\, . $$
This gives
\begin{eqnarray}\Delta(u,v)&=&\left(\frac{\partial f}{\partial x}(c_{u,v}, y_0+v)-\frac{\partial f}{\partial x}(x_0, y_0)\right)u+ \\
& &\left(f(x_0,y_0+v)-f(x_0,y_0)-\frac{\partial f}{\partial y}(x_0,y_0)\, v\right)
\end{eqnarray}
The first term in the right-hand side is $o(u)$ as $(u,v)\to (0,0)$ because $(c_{u,v},y_0+v)\to (x_0,y_0)$ and $\frac{\partial f}{\partial x}$ is continuous at $a=(x_0,y_0)$; and the second term is $o(v)$ by the definition of $\frac{\partial f}{\partial y}(x_0,y_0)$. Altogether, this gives $\Delta(u,v)=o(\Vert (u,v)\Vert)$ , as required.
