How do I show that partial derivatives exist everywhere?

Question

I'm having trouble with a certain multi-variable calculus question. $$ f(x,y) = \begin{cases} \large\frac{2xy^2}{x^2 + y^4}, & \text{$(x,y)\neq 0$} \\[2ex] 0, & \text{$(x,y) = 0$} \end{cases}$$

I need to show that both $\large\frac{∂f}{∂x}$ and $\large\frac{∂f}{∂y}$ exist everywhere.

I can easily manage to find both partial derivatives, but I'm not really sure what the question means when it asks to show that they "exist everywhere".

Any help would be appreciated, thanks.

Answer 1

The potential problem is at the origin. But note that

$$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{2h(0^2)}{h^2+0^4}-0}{h}=0$$

and

$$f_y(0,0)=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{2(0)(h^2)}{0^2+h^4}-0}{h}=0$$

Therefore, $f_x(0,0)=f_y(0,0)$. For $x^2+y^2>0$, we can simply note that $f(x,y)$ is composition of differentiable functions with

$$\begin{align} \frac{\partial f(x,y)}{\partial x}&=\frac{2y^2(y^4-x^2)}{(x^2+y^4)^2}\\\\ \frac{\partial f(x,y)}{\partial y}&=\frac{4xy(x^2-y^4)}{(x^2+y^4)^2} \end{align}$$

Hence, we see that

$$\begin{align} \frac{\partial f(x,y)}{\partial x}=\begin{cases}\frac{2y^2(y^4-x^2)}{(x^2+y^4)^2}&,x^2+y^2>0\\\\ 0&,x=y=0 \end{cases} \end{align}$$

$$\begin{align} \frac{\partial f(x,y)}{\partial y}=\begin{cases}\frac{4xy(x^2-y^4)}{(x^2+y^4)^2}&,x^2+y^2>0\\\\ 0&,x=y=0 \end{cases} \end{align}$$

NOTE: While the first partial derivatives, $f_x$ and $f_y$, exist everywhere, neither is continuous at the origin.