# Check this function is continuously differentiable on $\mathbb{R}^2$

by http://mathworld.wolfram.com/GreensTheorem.html (Green's theorem)

I choose $Q=0$ and defind P(x,y) : $$P(x,y) = \left\{ \begin{array}{ll} \frac{e^{-xy}\sin x}{x} & \text{if} ~~x\ne 0 \\ 1 & \textrm{if}~~ x=0 \\ \end{array} \right.$$

How to check this function is continuously differentiable on $\mathbb{R}^2$

For $x\neq0$ we have $$P(x,y)=\exp(-xy)\sum_{k=0}^\infty(-1)^k\frac{1}{(2k+1)!}x^{2k+1}/x\\=\exp(-xy)\sum_{k=0}^\infty(-1)^k\frac{1}{(2k+1)!}x^{2k}.$$ This power series can be extentended to $x=0$ by $P(0,y)=1$ and it converges on $\mathbb{R}^2$. The exponential term does not play any role here.