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The question is as follows:

Let $F:\mathbb{R}^2\rightarrow\mathbb{R}$ be a function given by $$F(x,y)=\bigg\{\begin{array}{cc} x^3\sin(1/x)+y^2 & x\neq 0\\ y^2&x=0. \end{array}$$ Show that $F$ is differentiable at the point $(0,0)$.

The piecewise nature of this function is throwing me off. I know that if all the partial derivatives are continuous in a neighborhood of the point, then the function is differentiable there. But $\frac{\partial}{\partial x}(x^3\sin(1/x)+y^2)=x(3x\sin(1/x)-\cos(1/x))$ is not continuous near $(0,0)$. So my inclination is to use the limit definition of derivative. However, $\lim_{(h,k)\rightarrow(0,0)}\frac{h^3\sin(1/h)+k^2+h-k}{\sqrt{h^2+k^2}}$ does not exist. Am I able to completely ignore the first piece since the function is not defined by it at $(0,0)$?

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You have $f_x(0,0)=f_y(0,0)=0$. Therefore, if $f$ is differentiable at $(0,0)$, then $f'(0,0)$ can only be the null function. On the other hand, asserting that $f'(0,0)$ is the null function is the same thing as asserting that$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.\tag1$$And this is indeed true, because if $\bigl\|(x,y)\bigr\|<1$ you have$$\bigl|f(x,y)\bigr|\leqslant|x|^3+y^2\leqslant x^2+y^2=\sqrt{x^2+y^2}^2$$and therefore$$\left|\frac{f(x,y)}{\sqrt{x^2+y^2}}\right|\leqslant\sqrt{x^2+y^2}.$$So, $(1)$ holds and $f'(0,0)$ is indeed the null function.

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  • $\begingroup$ There seems to be an error in your logic. You're claiming that differentiable implies null function, but then proving it is the null function. Does the converse of your claim hold? $\endgroup$ – Atsina Aug 15 '18 at 17:44
  • $\begingroup$ Oh, I see the word "only" hiding in there. So it is an if and only if statement? $\endgroup$ – Atsina Aug 15 '18 at 17:46
  • $\begingroup$ @Atsina You are right: I didn't express myself correctly. What I meant to say was that $f'(0,0)$ could only be the null function and then I proved that this is indeed the case. I've edited my answer. I hope that everything is clear now. $\endgroup$ – José Carlos Santos Aug 15 '18 at 17:46
  • $\begingroup$ Could you briefly elaborate on why $f_x(0,0)=f_y(0,0)=0$ gives that $f'(0,0)=$ null function $\Rightarrow f$ is differentiable at $(0,0)$? $\endgroup$ – Atsina Aug 15 '18 at 17:59
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    $\begingroup$ @Atsina Since we are deling with the limit at $(0,0)$, only vectors with a small norm matter. And I need to have $|x|<1$, so that the inequality $|x|^3\leqslant x^2$ holds. $\endgroup$ – José Carlos Santos Aug 15 '18 at 18:14

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