Find if two variables function is differentiable: $f(x,y)=\frac{x(1-\cos x)}{\sin(x^2+y^2)}$ I really stamped on this question.
$$f(x,y)=\left\{\begin{matrix} \frac{x(1-\cos x)}{\sin(x^2+y^2)},  (x,y)\neq (0,0)
 & \\ 0, (x,y)=(0,0)
 & 
\end{matrix}\right.$$
the function continuous at $(0,0)$.
the derivative by x at $(0,0)$ is $\frac{1}{2}$.
the derivative by y at $(0,0)$ is $0$.
I dont know how to solve the limit of the differentiable, I tried some ways but didn't get any answer.
$$\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\frac{\Delta x(1-\cos(\Delta x))}{\sin((\Delta x)^2+(\Delta y)^2)}-\frac{\Delta x}{2}}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}$$
 A: I shall concentrate on the origin. Note that there are also problems on the circles $x^2+y^2=n\pi$.
Write $f$ in the form
$$f(x,y)={{\rm sinc}^2(x/2)\over2\,{\rm sinc}(x^2+y^2)}\>{x^3\over x^2+y^2}\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$
It then becomes obvious that we have to analyze the function
$$g(x,y):={x^3\over x^2+y^2}$$
at the origin. One computes $g_x(0,0)=1$, $\>g_y(0,0)=0$. If $g$ wants to be differentiable at $(0,0)$ the derivative $dg(0,0)$ would therefore have to be given by
$$dg(0,0).(X,Y)=X\ .$$
This means that we have to check whether
$$g(x,y)-x={-xy^2\over x^2+y^2}={-r^3\cos\phi\sin^2\phi\over r^2}=o(r)\qquad\bigl(r:=\sqrt{x^2+y^2}\to 0+\bigr)\ .$$
Since $${g(x,y)-x\over r}=-\cos\phi\sin^2\phi$$
does not converge to $0$ independently of $\phi$ when $r\to0+$ it follows that $g$, hence $f$, is not differentiable at $(0,0)$.
A: HINT: In order to study the continuity of the function, you shall compute
$$
\lim\limits_{(x,y) \to (0,0)} f(x,y)
$$
Try to find a function tending to $0$ that is an upper bound for $f(x,y)$.
In order to compute the derivative with respect to $x$ (or to $y$), you need to study the limit of the difference quotient $\frac{f(\Delta x,0)-f(0,0)}{\Delta x}$ for $(\Delta x,0)$ (or $(0,\Delta y)$, that is when starting from $(0,0)$, you have an increment only along the variable $x$ (or $y$).
