Consider the function $f(x,y) = (x^2y)^\frac{1}{3}$ continuous is $\mathbb{R}^2$ and its partial derivatives:
$$\begin{cases}
f_x(x,y) = \frac{2}{3}\left(\frac{y}{x}\right)^\frac{1}{3} \\
f_y(x,y) = \frac{1}{3}\left(\frac{x}{y}\right)^\frac{2}{3} \\
\end{cases}$$
Clearly, $f_x$ is defined for $x \neq 0$. Evaluating limits, then:
$$\lim_{x \to 0} f_x(x,y) = \begin{cases}
\infty & \text{if}~y\neq 0 \\
\frac{0}{0} & \text{if}~y = 0
\end{cases}$$
By the way:
$$\lim_{x \to 0} f_x(x,0) = \lim_{h \to 0} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0} \frac{((0+h)^20)^\frac{1}{3}-0}{h} = 0.$$
Should I conclude that $f_x(x,y)$ is continuous in $(0,0)$ and then that $f(x,y)$ has the partial derivative with respect to $x$?
A very similar question arises when dealing with $f_y$, concluding that $$\lim_{y \to 0} f_y(x,y) = \begin{cases}
\infty & \text{if}~x\neq 0 \\
0 & \text{if}~x = 0
\end{cases}$$
Since (I presume!) $f$ is derivable in $(0,0)$e, then I can conclude that the function is differentiable in $(0,0)$.
But when I evaluate the limit to establish differentiability, then …
$$\lim_{(h,k) \to 0} \frac{(h^2k)^\frac{1}{3}}{(h^2+k^2)^\frac{1}{2}} = \lim_{\rho \to 0} (\cos(\theta)^2\sin(\theta))^\frac{1}{3}$$
where at the end I transformed $(x,y)$ to polar coordinates $(\rho, \theta)$ as usual. Of course, the limit does not exist.
Summary: I have a function which I presume is derivable everywhere in$\mathbb{R}^2$. But I also prove that the function is not differentiable.
What am I doing wrongly?
Errata: I have a function which I presume is derivable in $(0,0)$. But I also prove that the function is not differentiable in $(0,0)$.
What am I doing wrongly?
Best Answer
There is an issue in the
By the way: $$\lim_{x \to 0} f_x(x,0) = \lim_{h \to 0} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0} \frac{((0+h)^20)^\frac{1}{3}-0}{h} = 0.$$
You have indeed $f^\prime_x(0,0)=0$,which is what is computed in the line above. But that doesn't allow to speak of $\lim_{x \to 0} f_x(x,0)$.
So the partial derivatives are not continuous at the origin. And as you mentionned the function is not differentiable at the origin. Therefore, no contradiction.