Question: Find $\frac{\partial z}{\partial s}$ by the chain rule at $(r,s)=(1,1)$ if $z=\sin(2u)\cos(3t)$, where $ u=r+s−2$, $ t=r−s$.
My attempt goes as such:
You could define $z=f(u,t)$
$\frac{\partial z}{\partial s}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial s}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial s}$
$\frac{\partial f}{\partial u}=2\cos(2u)\cos(3t)$, $\frac{\partial u}{\partial s}=1$, $\frac{\partial f}{\partial t}=-3\sin(2u)\sin(3t)$, $\frac{\partial t}{\partial s}=-1$
Hence,
$\frac{\partial z}{\partial s}=2\cos(2u)\cos(3t)+3\sin(2u)\sin(3v)$
Since we want it at $(r,s)=(1,1)$ that means that $u=t=0$
$\frac{\partial z}{\partial s}=2$.
Would like some confirmation as to whether I followed the chain rule correctly, I have not done Calculus III in a LONG time! Thank you.