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Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that the partial derivatives with respect to $x$ and $y$ exist and one of them is continuous. Prove that $f$ is differentiable.

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Why are there so many downvotes without any comment? – 23rd Apr 25 '13 at 17:42
Dear André, what have you done so far? – Philippe Malot Apr 25 '13 at 17:55
Partly due to a rapid increase in the volume of questions asked here, members of the community are concerned about an influx of questions that lack information except the problem statement itself (the downvotes here are probably a consequence). I encourage you to edit your question and include the source of the problem, and any partial progress (like in your comment above). – Douglas S. Stones Apr 25 '13 at 18:09
Hint: show that this reduces to the case when the function depends on one variable only. – 1015 Apr 25 '13 at 19:43
This is an interesting question. But you should have made more efforts in asking: motivation, personal thoughts, and so on... It is currently a very hot topic on meta. A lot of people think that the questions which are plain copy/paste from homework or from a book should be banned. That's the reason of the downvotes, and of the close votes. – 1015 Apr 25 '13 at 22:42

In short: the problem reduces to the easy case when $f$ depends solely on one variable. See the greyish box below for the formula that does the reduction.

It suffices to show that $f$ is differentiable at $(0,0)$ with the additional assumption that $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$. First pass from $(x_0,y_0)$ to $(0,0)$ by considering the function $g(x,y)=f(x+x_0,y+y_0)$. Then work on $h(x,y)=g(x,y)-x\frac{\partial f}{\partial x}(0,0)-y\frac{\partial f}{\partial y}(0,0)$.

So let us assume assume that $\frac{\partial f}{\partial x}$ exists and is continuous on $\mathbb{R}^2$ (only continuity in an open neighborhood of $(0,0)$ is really needed for the local argument), that $\frac{\partial f}{\partial y}$ exists at $(0,0)$, and that $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$. We need to show that $f$ is differentiable at $(0,0)$. Note that the derivative must be $0$ given our assumptions.

Now observe that for every $x,y$, we have, by the fundamental theorem of calculus:

$$ f(x,y)=f(0,y)+\int_0^x \frac{\partial f}{\partial x}(s,y)ds. $$

I let you check properly that $(x,y)\longmapsto f(0,y)$ is differentiable at $(0,0)$ with zero derivative, using only $\frac{\partial f}{\partial y}(0,0)=0$. For the other term, just note that it is $0$ at $(0,0)$ and that for every $0<\sqrt{x^2+y^2}\leq r$ $$ \frac{1}{\sqrt{x^2+y^2}}\Big|\int_0^x \frac{\partial f}{\partial x}(s,y)ds\Big|\leq \frac{|x|}{\sqrt{x^2+y^2}}\sup_{0\leq \sqrt{x^2+y^2}\leq r}\Big| \frac{\partial f}{\partial x}(s,t)\Big|\leq \sup_{0\leq \sqrt{x^2+y^2}\leq r}\Big| \frac{\partial f}{\partial x}(s,t)\Big|. $$ By continuity of $\frac{\partial f}{\partial x}$ at $(0,0)$, the rhs tends to $0$ when $(x,y)$ tends to $(0,0)$. This proves that the function $(x,y)\longmapsto \int_0^x \frac{\partial f}{\partial x}(s,y)ds$ is differentiable at $(0,0)$ with zero derivative. And this concludes the proof.

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Well this is one of the reason I think we shouldn't close questions like these ! – Kasper Apr 25 '13 at 22:52
@Kasper Yes, I'm glad the question was asked. I only knew that if all partial derivatives are continuous, then the function is differentiable. And I first looked for a counterexample...So I've learned something today again. – 1015 Apr 25 '13 at 22:57
Can this be generalized to $f: \mathbb {R} ^n \rightarrow \mathbb {R} $ with all but one continuous partial derivative? – Marco Flores Jan 26 '15 at 1:32

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