Show that $f(x,y) = \sin( x )|y|$ is differentible at $(0,0)$. Show that $$f(x,y) = \sin( x )|y|$$ is differentible at $(0,0)$.

Trying to find the partial derivatives, I found out that $$f'_y = \sin(x)\frac{|y|}{y}$$ is not defined at $(0,0)$. However, if the partial derivative does not exist then we cannot say $f$ is differentible since the function is differentible at some point iff all partial derivatives exist and are continuous.
Any help is appreciated.
 A: You have
$$\left\vert \frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}}\right\vert \le  \frac{\vert x y \vert}{\sqrt{x^2+y^2}} \le \frac{1}{2} \sqrt{x^2+y^2}$$
The fact that $$\lim\limits_{(x,y) \to (0,0)}\sqrt{x^2+y^2} = 0$$ implies from the definition that $f$ is differentiable at $(0,0)$ and that its derivative is equal to zero (i.e. the always vanishing linear map).
You don't need to look at partial derivatives in this case as you're just asked to prove differentiability at $(0,0)$.
A: Hint: $f_y'=\sin x \frac {|y|} y$ for $y \neq 0$ and this tends to $0$ as $(x,y) \to (0,0)$ since $|\sin x \frac {|y|} y| =|\sin x|$. Now verify directly fron=m the  definition that the partial derivatives at $(0,0)$ are both $0$. Show also that $f_x' \to 0$ as $(x,y) \to (0,0)$. Conclude that $f$ has continuous partial derivatives.
[$ \frac {\partial } {\partial x} f(0,0) =\lim \frac {f(0+h,0)-f(0,0)} h =0$. Similarly, $ \frac {\partial } {\partial y} f(0,0)=0$].
A: For $x=0$, the derivative of the function $f'_y$ is continuous at $y=0$.
A: $$|f(x,y)-f(0,0)|=|f(x,y)-0| = |\sin x||y| \le |x||y| \le \frac12 (|x|^2 + |y|^2) = \frac12 \|(x,y) -(0,0)\|^2 $$
By the definition of (Frechet) differentiability, $f$ is differentiable at $(0,0)$ with gradient equal to the $0$ vector.
