Is the function $f(x,y)= \sqrt{|x^3y|}$ differentiable at $y>0, x=0$? I would like to calculate if the following function is differentiable, at $y>0, x=0$ points, without polar coordinates.
I've tried using the sandwich rule, but with no real success.
$ f(x,y)= \sqrt{|x^3y|}$
Thank you in advance! Much appreciated.
 A: We have that for any $y$
$$\lim_{(x,y)\to (0,y)} \sqrt{|x^3y|}=0$$
and
$$f_x(0,y)=\lim_{h\to 0} \frac{\sqrt{|h^3y|}}{h}=0$$
$$f_y(0,y)=\lim_{h\to 0} \frac{\sqrt{|0\,(y+h)|}}{h}=0$$
then by definition
$$\lim_{(h,k)\to(0,0)} \frac{\sqrt{|h^3(y+k)|}}{\sqrt{h^2+k^2}}=0$$
indeed eventually
$$\sqrt{\frac{|h^3(y+k)|}{h^2+k^2}}= \sqrt{|h|}\sqrt{\frac{|h^2(y+k)|}{h^2+k^2}}\le\sqrt{|h|}\sqrt{|y+k|} \to 0$$
therefore the given function is differentiable.
A: As usual, you compute the partial derivatives by the definition. For $y\ne 0$, we have
\begin{align*}
\frac{\partial f}{\partial x}(0,y) &= \lim_{h\to 0}\frac{\sqrt{|y|}(|h|^{3/2}-0)}h = 0,\quad \text{and}\\
\frac{\partial f}{\partial y}(0,y) &= \lim_{h\to 0} \frac 0h = 0.
\end{align*}
Now we ask whether
$$\lim_{(h,k)\to (0,0)} \frac{f(h,y+k)-f(0,y)-0h-0k}{\sqrt{h^2+k^2}} = 0.$$
(Here I inserted the partial derivatives at the point $(0,y)$ in the linear approximation.)
Well, this quantity is
$$\frac{\sqrt{|h^3(y+k)|}}{\sqrt{h^2+k^2}} = \frac{|h|}{\sqrt{h^2+k^2}}\cdot\sqrt{|h||y+k|}\le 1\cdot \sqrt{|h||y+k|},$$
and the last expression goes to $0$ as $(h,k)\to 0$ (indeed, as $h\to 0$), so, by squeeze or sandwich, the original expression goes to $0$.
A: hint
With polar cordinates
$$x=r\cos(t)$$
$$y=r\sin(t)$$
the function becomes
$$f(x,y)=r\sqrt{|\cos^3(t)\sin(t)|}$$
thus
$$|f(x,y)|\le r$$
A: Fix $a>0$, then define $l(t)=f(u_1t,a+u_2t)$ where $u_1,u_2\in \mathbb{R}$ such that $u_1^2+u_2^2=1$. Then $l'(0)$ is the directional derivative of $f$ at $(0,a)$ is the direction determined by the unit vector $\vec{u}=\big<u_1,u_2\big>$, namely $f_{\vec{u}}(0,a)$. Now notice $$l'(0)=\lim_{t\rightarrow 0}\Bigg(\frac{l(t)-l(0)}{t-0}\Bigg)=\sqrt{|au_1^3|} \cdot \lim_{t \rightarrow 0}\Bigg(\frac{\sqrt{|t^3|}}{t}\Bigg)=0$$ Since this limits exists for any $a>0$ and for any unit vector $\vec{u}=\big<u_1,u_2\big>$, we have that $f$ is differentiable at $(0,a)$.
