derivativespartial derivative
Show that $f(x,y)$ defined by:
$$f(x,y) = \begin{cases}\dfrac{x^2y^2}{\sqrt{x^2+y^2}}&\text{ if }(x,y)\not =(0,0)\\0 &\text{ if }(x,y)=(0,0)\end{cases}$$
is differentiable at $(x,y) = (0,0)$
I tried to solve this problem by applying the theorem that if partial derivatives are continuous then the function is differentiable. Therefore, I calculated partial derivated but not I am stuck in showing they are indeed continuous. Help me!
There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
First we determine whether or not $f$ is continuous at $(0,0)$. If we approach $(0,0)$ on the direction $x=y$, we have $$\lim_{x\to 0,y\to 0, x=y}f(x,y)\lim_{x\to 0}\frac{x^3}{x^4+x^2}=\lim_{x\to 0}\frac{1}{x+x^{-1}}=0$$ If we approach on the direction $y=x^2$, then $$\lim_{x\to 0,y\to 0, y=x^2}f(x,y)=\lim_{x\to 0}\frac{x^2\cdot x^2}{x^4+(x^2)^2}=\frac{1}{2}$$ Thus we conclude that $f$ is not continuous at $(0,0)$.
For the partial derivative. To determine whether $f_x(0,0)$ exists, simply apply the definition of partial derivative:
$$\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}=0$$
Thus $f_x(0,0)=0$
Similarly
$$\lim_{y\to 0}\frac{f(0,y)-f(0,0)}{y}=0$$ thus $f_y(0,0)=0$
Best Answer
You have $$x^2+y^2-2 \vert xy \vert=(\vert x \vert - \vert y \vert)^2 \ge 0$$ Hence $$\vert xy \vert \le \frac{x^2+y^2}{2}$$ and $$0 \le \frac{\vert f(x,y) \vert}{\sqrt{x^2+y^2}} = \frac{x^2y^2}{x^2+y^2} \le \frac{1}{4}(x^2+y^2)$$ As $\lim_{(x,y) \to (0,0)} x^2+y^2 = 0$, this proves that $f$ is differentiable at $(0,0)$ and that its Fréchet derivative is equal to $0$. Which means that $f_x(0,0)=f_y(0,0)=0$.