Let $f(x,y)=\frac{x^4y}{x^4+y^2}$ whenever $(x,y)\ne(0,0)$. and $f(x,y)=(0,0)$ otherwise.
Prove or disprove: $f$ is differentiable in $(0,0)$.
My attempt:
I am trying to prove it by showing that $\epsilon \to 0$ when $\Delta x,\Delta y \to 0$ in this definition:
$f(x+\Delta x, y+\Delta y)-f(x,y)=f_x(x,y)\Delta x+f_y \Delta y + \epsilon\sqrt{(\Delta x)^2+(\Delta y)^2}$.
I have checked by definition that $f_x(0,0)=f_y(0,0)=0$.
And so $f(\Delta x, \Delta y)=\epsilon\sqrt{\Delta x^2+ \Delta y^2}$ Which gives me:
$$\frac{(\Delta x)^4(\Delta y)}{[(\Delta x)^4 + (\Delta y)^2]\sqrt{(\Delta x)^2+ (\Delta y)^2}}=\epsilon$$But now I got stuck trying to show that $\epsilon \to 0$, I am not sure how to deal with this limit and would appreciate any help and explanations on your thoughts (What did you think when you saw this limit, like did you know it converges to $0$ just by looking at it? What I'm trying to say is I would appreciate explanations on your thoughts "behind the scenes" that led you to use your method).
Thanks in advance!
Best Answer
Clearly, $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$. So, if $f$ is differentiable at $(0,0)$, then $Df_{(0,0)}$ is the null function, which happens if and only if$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$And this is indeed true. For any two real numbers $a$ and $b$ you have$$ab\leqslant\frac{a^2+b^2}2,$$and therefore$$\left(\forall(x,y)\in\Bbb R^2\right):|x^2y|\leqslant\frac{x^4+y^2}2.\tag1$$But then, if $(x,y)\ne(0,0)$,\begin{align}0&\leqslant\left|\frac{f(x,y)}{\sqrt{x^2+y^2}}\right|\\&=\frac{x^4|y|}{(x^4+y^2)\sqrt{x^2+y^2}}\\&\leqslant\frac{x^2}{2\sqrt{x^2+y^2}}\\&\leqslant\frac{|x|}2,\end{align}and therefore, by the squeeze theorem, $(1)$ holds.