Let $z=f(x,y)=e^{x}\sin(xy)$, $x=g(s,t)$ and $y=h(s,t)$. If $k(s,t)=f(g(s,t),h(s,t))$, find $\displaystyle\frac{\partial k}{\partial s}$.
Until now, I have found that:
$\displaystyle\frac{\partial k}{\partial s} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial s}+\frac{\partial f}{\partial h}\frac{\partial h}{\partial s}$
I'm not so sure on how to proceed. The actual computation of the partial derivatives shouldn't pose a problem, but I'm stuck on this small issue. I would say that $\displaystyle\frac{\partial f}{\partial g} = \frac{\partial f}{\partial x}$, but I'm not sure if my claim is correct. If this is the case, then it would as well follow that $\displaystyle\frac{\partial f}{\partial h} = \frac{\partial f}{\partial y}$. If this is the case, then why would it be true?
Thank you for your time.
Your $\frac {\partial f}{\partial g}$ can be considered as a shortcut for a more correct $$ \frac {\partial f}{\partial x}\Big|_{x=g(s,t)} $$ But I don't think it may cause any problem if you understand your own notaitons.