How do I show that the following function is differentiable at $(0,0)$? $$ \begin{cases} \dfrac{\sin(xy)}{y}, & \text{if }y \neq 0 \\ \\ 0, & \text{if }y = 0 \end{cases} $$ I calculated the partial derivatives and

  • $f(x) = \cos(xy)$ exists near $(0,0)$ and is continuous
  • $f(y) = \dfrac{xy \cos(xy) - \sin(xy)}{y^2}$ exists, but how do I show that it is continuous?
  • $\begingroup$ If you know power series, $g(u)=\frac{\sin(u)}{u}=\sum_{k\geq 0}(-1)^k\frac{u^{2k}}{(2k+1)!}$ (with $g(0)=1$) is $C^{\infty}$ on $\mathbb{R}$, now your function is $f(x,y)=xg(xy)$, and you are done. $\endgroup$ – Kelenner Jul 2 '17 at 8:05
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    $\begingroup$ Use the definition of the partial derivative to find $\partial _x f(0,0)$ and $\partial _y f(0,0)$. Then show they are continuous in $(0,0)$. $\endgroup$ – windircurse Jul 2 '17 at 8:10
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    $\begingroup$ Are you sure that the condition is $(x,y)\neq (0,0)$? Is it maybe $y\neq 0$? $\endgroup$ – Nikolaos Skout Jul 2 '17 at 8:12

As you defined it your function is not even continuous at $(0,0)$. Note that $$\lim_{y\to0}{\sin(xy)\over y}=x\ .$$ I'm assuming this will be corrected. Then you can argue as follows: When $y\ne0$ you have $${\sin(xy)\over y}=\int_0^x \cos(y t)\>dt\ ,$$ whereby the RHS is obviously $C^1$ in a neighborhood of $(0,0)$.

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