Differentiablity of $x \sin (y) at $(1,0) The question requires to determine the differentiability of
$x \sin(y)$  at   $(1,0)$.
My approach is to use the method of Linearization, which 
$L(1+h,0+k) = 0 + h\sin(y) - k\cos(y)$. 
As such, the error term limit can be written as ;
$$(1+h)\sin k - h\sin(0) - \frac{k\cos(0)}{\sqrt{h^2 + k^2}}$$ whether $h$ and $k$ tends to zero.
Yet I find the limit hard to evaluate, and as such I need your aid !!
I apologize for the messy question, i tried my best.
Thank You !
 A: Let us define $f(x,y)=x\sin(y)$ for $(x,y)\in\mathbb{R}^2$.
Saying that $f$ is differentiable at the point $(1,0)\in\mathbb{R}^2$ means by definition that we can write for all $(h,k)\in\mathbb{R}^2$
$$f(1+h,0+k)=f(1,0)+\ell((h,k))+o(\|h,k\|)$$
where $\ell\in\mathcal{L}_c\left(\mathbb{R}^2;\mathbb{R}\right)$ is a (bounded) linear map and with
$$\lim_{\|(h,k)\|\to0}\frac{o(\|h,k\|)}{|h,k\|}=0.$$
So for $(h,k)$ very small, we write
$$f(1+h,0+k)=(1+h)\sin(k)=\sin(k)+h\sin(k)=0+\sum_{n=0}^{+\infty}(-1)^n\frac{k^{2n+1}}{(2n+1)!}+h\sin(k)$$
$$=f(1,0)+k+\sum_{n=1}^{+\infty}(-1)^n\frac{k^{2n+1}}{(2n+1)!}+h\sin(k).$$
Now, the map $\ell:(h,k)\mapsto k$ is linear because for any $\lambda\in\mathbb{R}$ we have $\ell(\lambda(h,k))=\lambda k=\lambda\ell((h,k))$. So we only need to see if the remainder is very small when $(h,k)$ is : we have
$$\left|\sum_{n=1}^{+\infty}(-1)^n\frac{k^{2n+1}}{(2n+1)!}+h\sin(k)\right|\leq \left|\sum_{n=1}^{+\infty}(-1)^n\frac{k^{2n+1}}{(2n+1)!}\right|+\left|h\sin(k)\right|$$
$$\leq|k|^2\sum_{n=0}^{+\infty}\left|\frac{k^{2n+1}}{(2n+1)!}\right|+|hk|\leq|k|^2\mathrm{e}^k+|hk|$$
and
$$\frac{|k|^2\mathrm{e}^k+|hk|}{\|(h,k)\|}=\frac{|k|^2\mathrm{e}^k+|hk|}{\sqrt{h^2+k^2}}=\frac{|k|^2\mathrm{e}^k}{\sqrt{h^2+k^2}}+\frac{|hk|}{\sqrt{h^2+k^2}}$$
$$\leq\frac{|k|^2\mathrm{e}^k}{|k|}+\frac{1}{2}\frac{h^2+k^2}{\sqrt{h^2+k^2}}$$
where we used for the last line the inequalities $b\leq\sqrt{a^2+b^2}$ and $ab\leq\frac{1}{2}\left(a^2+b^2\right)$ for all $a,b\in\mathbb{R}_+$. Hence the remainder is $o\left(\|(h,k)\|\right)$ and the conclusion is that $f$ is differentiable at the point $(1,0)$, the differential being the (bounded) linear map
$$\mathrm{d}f((1,0)):(h,k)\longmapsto\ell((h,k))=k.$$
A very important last thing : there is NO silly question !
