If I have $f: \mathbb R^2 \to \mathbb R$, a function whose partial derivatives are well defined in a ball centered in $(a, b)$, such that one of the partials is continuous at that point $(a, b)$. How can I prove that $f$ is differentiable in $(a, b)$?
I understand that if both partials were continuous it would be enough reasoning but in this case only one is continuous at $(a,b)$.
We denote $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ by $f_{x}$ and $f_{y}$ respectively. Choose $\delta>0$ such that $f_{x}$ and $f_{y}$ exist on $(a-\delta,a+\delta)\times(b-\delta,b+\delta)$. We assume that $f_{y}$ is continuous at $(a,b)$.
Define $R:(-\delta,\delta)\times(-\delta,\delta)\rightarrow\mathbb{R}$ by $R(h,k)=f(a+h,b+k)-f(a,b)-f_{x}(a,b)h-f_{y}(a,b)k$. We go to prove that $\lim_{(h,k)\rightarrow(0,0)}\frac{R(h,k)}{\sqrt{h^{2}+k^{2}}}=0$.
Let $(h,k)\in(-\delta,\delta)\times(-\delta,\delta)$ with $(h,k)\neq(0,0)$ be given.
Note that the function $f(a+h,\cdot)$ is continuous on $[b,b+k]$ and differentiable on $(b,b+k)$ because $f_{y}(a+h,\cdot)$ exists on $(b-\delta,b+\delta)$. By mean-value theorem, there exists $\xi_{h,k}\in(b,b+k)$ such that $f(a+h,b+k)-f(a+h,b)=f_{y}(a+h,\xi_{h,k})k$. (In the above, we have assumed that $k>0$. If $k<0$, we write the intervals as $[b+k,b]$ and $(b+k,b)$ accordingly. If $k=0$, we choose $\xi_{h,k}=b$, then we still have $f(a+h,b+k)-f(a+h,b)=f_{y}(a+h,\xi_{h,k})k.$)
Therefore, we have that \begin{eqnarray*} \frac{R(h,k)}{\sqrt{h^{2}+k^{2}}} & = & \frac{1}{\sqrt{h^{2}+k^{2}}}\left\{ \left[f(a+h,b+k)-f(a+h,b)\right]+\left[f(a+h,b)-f(a,b)\right]-\left[f_{x}(a,b)h+f_{y}(a,b)k\right]\right\} \\ & = & \left[f_{y}(a+h,\xi_{h,k})-f_{y}(a,b)\right]\cdot\frac{k}{\sqrt{h^{2}+k^{2}}}+\left[f(a+h,b)-f(a,b)-f_{x}(a,b)h\right]\cdot\frac{1}{\sqrt{h^{2}+k^{2}}}. \end{eqnarray*} Observe that $(a+h,\xi_{h,k})\rightarrow(a,b)$ as $(h,k)\rightarrow(0,0)$ and $f_{y}$ is continuous at $(a,b)$. Therefore, as $(h,k)\rightarrow(0,0)$, \begin{eqnarray*} \left|\left[f_{y}(a+h,\xi_{h,k})-f_{y}(a,b)\right]\cdot\frac{k}{\sqrt{h^{2}+k^{2}}}\right| & \leq & \left|f_{y}(a+h,\xi_{h,k})-f_{y}(a,b)\right|\\ & \rightarrow & 0. \end{eqnarray*} On the other hand, by the definition of partial derivative, $\lim_{h\rightarrow0}\frac{f(a+h,b)-f(a,b)}{h}-f_{x}(a,b)=0$. Therefore, for $h\neq0$, \begin{eqnarray*} \left|\left[f(a+h,b)-f(a,b)-f_{x}(a,b)h\right]\cdot\frac{1}{\sqrt{h^{2}+k^{2}}}\right| & = & \left|\left[\frac{f(a+h,b)-f(a,b)}{h}-f_{x}(a,b)\right]\cdot\frac{h}{\sqrt{h^{2}+k^{2}}}\right|\\ & \leq & \left|\frac{f(a+h,b)-f(a,b)}{h}-f_{x}(a,b)\right|\\ & \rightarrow & 0 \end{eqnarray*} as $h\rightarrow0$. If $h=0$, we clearly have $\left[f(a+h,b)-f(a,b)-f_{x}(a,b)h\right]\cdot\frac{1}{\sqrt{h^{2}+k^{2}}}=0$. It follows that $\left[f(a+h,b)-f(a,b)-f_{x}(a,b)h\right]\cdot\frac{1}{\sqrt{h^{2}+k^{2}}}\rightarrow0$ as $(h,k)\rightarrow(0,0)$ (because $(h,k)\rightarrow(0,0)$$\Rightarrow h\rightarrow0$).
This shows that $\frac{R(h,k)}{\sqrt{h^{2}+k^{2}}}\rightarrow0$ as $(h,k)\rightarrow(0,0)$ and hence $f$ is differentiable at $(a,b)$.
