Finding the second partial derivatives of $w=\sqrt{u^2+v^2}$

Question

I'm trying to find all the second partial derivatives of $w=\sqrt{u^2+v^2}$, only I've gotten stuck on the first one. I found $w_u= \frac{2u}{2\sqrt{u^+v^2}}$ with the chain rule and v as a constant. Then for $$w_{uu} =-\frac{1}{4}(u^2+v^2)^{-1.5}*2u*2u+(u^2+v^2)^{-0.5}$$=$$\frac{u^2}{(u^+v^2)^{1.5}}+\frac{1}{(u^2+v^2)^{0.5}}$$

Quite a bit off from the answer of $\frac{v^2}{(u^2+v^2)^{1.5}}$

Can someone please tell me where I'm messing up here?

Answer 1

Your second derivative is wrong:

$$ w_u=\frac{u}{\sqrt{u^2+v^2}} \Rightarrow w_{uu}=\frac{\sqrt{u^2+v^2}-u\frac{u}{\sqrt{u^2+v^2}}}{u^2+v^2}=\frac{u^2+v^2-u^2}{(u^2+v^2)\sqrt{u^2+v^2}} $$