Second order partial derivatives

Question

Suppose that $z=g(x,y),\, x=s+t,$ and $y=st$, where all first and second order partial derivatives of $g$ exist and are continuous.

Show that $$\frac{\partial ^2 z}{\partial s\partial t}=\frac{\partial ^2 g}{\partial x^2}+x\frac{\partial^2g}{\partial x\partial y}+y\frac{\partial^2 g}{\partial y^2}+\frac{\partial g}{\partial y}.$$

Someone told me I have to use the chain rule twice, but I still don't quite understand what I'm meant to do.

Answer 1

So, firstly using the chain rule once; \begin{eqnarray} \frac{\partial z}{\partial s} &=& \frac{\partial g}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial s} \\ &=& \frac{\partial g}{\partial x} +t \frac{\partial g}{\partial y} \\ \end{eqnarray} So, now we use the chain rule a second time by applying it to the above, namely \begin{eqnarray} \frac {\partial^{2} g}{\partial s \partial t} &=& \frac{\partial^{2} g}{\partial x^{2}}\frac{\partial x}{\partial t}+\frac{\partial^{2} g}{\partial x \partial y} \frac{\partial y}{\partial t} + \frac{\partial g}{\partial y} + t \left( \frac{\partial^{2} g}{\partial x \partial y} \frac{\partial x}{\partial t} + \frac{\partial g}{\partial y^{2}}\frac{\partial y}{\partial t} \right) \\ &=& \frac{\partial^{2} g}{\partial x^{2}} + \frac{\partial^{2} g}{\partial x \partial y}s + \frac{\partial g}{\partial y} + t \left( \frac{\partial^{2} g}{\partial x \partial y} + \frac{\partial g}{\partial y^{2}}s \right) \\ &=& \frac{\partial^{2} g}{\partial x^{2}} + \frac{\partial^{2} g}{\partial x \partial y}(s+t) + st \frac{\partial^{2} g}{\partial y^{2}} \\ &=& \frac{\partial^{2} g}{\partial x^{2}} + x\frac{\partial^{2} g}{\partial x \partial y} + y \frac{\partial^{2} g}{\partial y^{2}} \end{eqnarray}

Job done.

EDIT: @Kevin is correct, I needed to replace full derivaties w.r.t $s, t$ with partials. The remaining parts of the answer is correct. I would reply to Kevin as a comment about this but the "add comment" function on my PC seems to be disabled.