If $z = f(x,y)$ has continuous second order partial derivatives and $x = r^2 + s^2$ and $y = 2rs$, find $ \frac{\partial^2 z}{\partial r^2} $ So, $ \frac{\partial^2 z}{\partial r^2} = \frac{\partial}{\partial r} \left(\frac{\partial z}{\partial r}\right) = \frac{\partial}{\partial r}\left(2r \frac{\partial z}{\partial x} + 2s \frac{\partial z}{\partial y}\right) = 2\frac{\partial z}{\partial x} + 2r\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial x}\right) + 2s\frac{\partial }{\partial r}\left(\frac{\partial z}{\partial y}\right) $
I understand that the product rule has been used to attain the last equality, however, this would imply that $\frac{\partial }{\partial r} (2s) = 0$ Why? Can I not write $s$ as a function of $r$ by rearranging the relations I have at the top of the page? Many thanks