Partial derivative techniques

Question

The question is as follows: $$ r = y^2, s = xy, t = -x^2, u = f(r,s,t)$$ , such that $f$ has continuous second order partial derivatives. Represent $u_{xy}$ in the following format.$$ u_{xy} = ...f_s(r,s,t)+...rf_{rs}(r,s,t)+ ...sf_{ss}(r,s,t)+...sf_{rt}(r,s,t)+ ...tf_{st}(r,s,t)$$ where $...$ are blanks that need to be filled as the answer. Instead of taking the actual values of $r, s, t$, I decided to do it this way- $$f_s = f(r,1,t), f_{ss} = f(r,0,t)$$ and so on. Is this the right way to do it? I tried the other way, using the actual values of $r, s, t$ but I didn't get the right answer.

Answer 1

I do not understand your choice of explicit values inside $f_s$ and $f_{ss}$. In general, as $r=r(y), s=s(x,y), t=t(x)$, then $u(x,y)=f(r(y),s(x,y),t(x))$. By the chain rule

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}:= \frac{\partial f}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\frac{dt}{dx}, $$

and

$$u_{xy}:=\frac{\partial^2 u}{\partial x\partial y}= \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}\right)+\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial t}\frac{dt}{dx}\right)=\\ \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial s}\right)\frac{\partial s}{\partial x}+\frac{\partial f}{\partial s}\frac{\partial}{\partial y}\left(\frac{\partial s}{\partial x}\right)+\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial t}\right)\frac{dt}{dx}=\\ \left(\frac{\partial g}{\partial r}\frac{dr}{dy}+\frac{\partial g}{\partial s}\frac{\partial s}{\partial y}\right)\frac{\partial s}{\partial x} +\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial x\partial y} +\left(\frac{\partial h}{\partial r}\frac{dr}{dy}+\frac{\partial h}{\partial s}\frac{\partial s}{\partial y}\right) \frac{d t}{dx}, $$

because $t$ is independent of $y$ and with $g:=\frac{\partial f}{\partial s}$ and $h:=\frac{\partial f}{\partial t}$.

In summary

$$u_{xy}=\left(\frac{\partial^2 f}{\partial s\partial r}\frac{d r}{dy}+\frac{\partial^2 f}{\partial s^2}\frac{\partial s}{\partial y}\right)\frac{\partial s}{\partial x} +\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial x\partial y} +\left(\frac{\partial^2 f}{\partial t\partial r}\frac{dr}{dy}+\frac{\partial^2 f}{\partial t\partial s}\frac{\partial s}{\partial y}\right) \frac{d t}{dx}. $$ All you need to do is to apply the definitions of $r$, $s$ and $t$, evaluating the corresponding (partial/ordinary) derivatives.