# Anyone have an idea for a function, that its derivative equals 0 along y=x, y=-x (|y|=|x|)

I am looking for a differentiable function, thats its partial derivatives equals 0 along $$|y|=|x|$$ I have thought about $$u(x,y)=\frac{(x-y)^2(x+y)^2}{2}$$ but I am looking for another one

if anyone has a method/tips of "building" functions like that I'd be happy to learn...

Thank you!

-
What's your criterion for "another" function? If you have one, you can multiply it by any real number and you'll get another one. –  Alan Guo Nov 17 '12 at 15:12
I know that! :) I am looking for something different if possible –  YNWA Nov 17 '12 at 15:15

A source of examples is $u(x,y)=(x^2-y^2)^n$ where $n \ge 2$. Then the $x$ partial is $n(x^2-y^2)^{n-1} \cdot 2x$, which vanishes on $|x|=|y|$, and similarly for the $y$ partial. (Note this was inspired by did's answer.)
EDIT: Actually as did basically suggested, if $W(t)$ satisfies $W'(0)=0$ then the function $u(x,y)=W(x^2-y^2)$ is an example, since on $x^2=y^2$ its partial wrt $x$ say is $W'(x^2-y^2)\cdot 2x$ which is zero provided $x^2=y^2$ since $W'(0)=0$.
For any function $U$ such that $U'(0)=0$, $u:(x,y)\mapsto U(x^2-y^2)$.
hmm you sure? I cant see why :\ we will need to demand that $$U_(x^2-y^2)=0$$ ? (derivative of U at ... ) –  YNWA Nov 17 '12 at 16:02