Find the partial derivatives of second order of $f(x,y)=\varphi(xy,\frac{x}{y})$

Ok guys, I'm given this smooth function $\varphi(u,v)$ defined in $R^2$. So that $f(x,y)=\varphi(xy,\frac{x}{y})$. I have to find all partial derivatives of second order of $f$ using the partial derivatives of $\varphi$. I know how to find the partial derivative of "normal" functions like $\frac{xy}{x+y}$ or something like that, but this kind of problem I have no idea how to do. Any ideas and solutions are welcomed $\ddot \smile$

Use the chain rule: $$\frac{\partial}{\partial x} \phi(u(x,y),v(x,y)) = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v} \frac{\partial v}{\partial x},$$
$$\frac{\partial}{\partial y} \phi(u(x,y),v(x,y)) = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial \phi}{\partial v} \frac{\partial v}{\partial y}.$$
• So if I figured it out correctly I'll have $$f^{'}_x=\frac{\partial\varphi}{\partial u}y + \frac{\partial\varphi}{\partial v}\frac{1}{y}$$ , where $u=xy$ and $v=\frac{x}{y}$. So for the second derivative I'll have $$\frac{\partial}{\partial x}(\frac{\partial\varphi}{\partial u}y + \frac{\partial\varphi}{\partial v}\frac{1}{y})=\frac{\partial^2 \varphi}{\partial u \partial u}y^2 + \frac{\partial^2 \varphi}{\partial v \partial v}\frac{1}{y^2}+2\frac{\partial^2 \varphi}{\partial u \partial v}$$ Is this right or am I wrong again? – randomname Jun 20 '14 at 16:28
• That seems right for the second derivative with respect to $x$. The other second derivatives $\frac{\partial^2 f}{\partial y^2}$ and $\frac{\partial^2 f}{\partial x\partial y}$ are a little more complicated because you have to use the product rule as well as the chain rule. – p.s. Jun 20 '14 at 22:02