I have the following function:
$$F: \mathbb{R}^2 \rightarrow \mathbb{R}, ~~ (x,y) \rightarrow xy\frac{x^2-y^2}{x^2+y^2}$$
for $(x,y) \ne 0$ and $F(0,0) = 0$. I want to show that $F$ is partially differentiable twice.
So far I've figured out that for the function to be differentiable I have to show the differential quotient exists for both – $x$ and $y$. So starting with $x$ I'd treat $y$ as a constant and check for the limit:
$$\lim_{h \rightarrow 0} \frac{F(x_0+h) -F(x_0)}{h} = y\frac{(x_0+h)}{h}\frac{(x_0+h)^2-y^2}{(x_0+h)^2+y^2} – \frac{x_0y}{h}\frac{x_0^2-y^2}{x_0^2+y^2}$$
And unfortunately that is where I am stuck at the moment – I don't know how to get rid of the $h$ after expanding – maybe I've done something wrong?
Thanks for helping out.
FunkyPeanut
Best Answer
Yes. But in doing so, you should not forget single-variable calculus. Your strategy should be: