Chain rule notation when there is multivariables

Question

Use the chain rule to determinate $u_s$ and $u_t$.

$u = x \csc(yz)$

$x = rs$

$y = s^2 t$

z = $\frac{s}{t^2}$

Determinating $u_s$ means applying the chain rule using $t$ as a constant? How would be the proccess?

Answer 1

$$\begin{align} u_s&=x_s\csc(yz)+x\big(\csc(yz)\big)_s&\text{(product rule)}\\[1ex] &=x_s\csc(yz)-x\csc(yz)\cot(yz)\big(yz\big)_s&\text{(chain rule)}\\[1ex] &=x_s\csc(yz)-x\csc(yz)\cot(yz)(y_sz+yz_s)&\text{(product rule)} \end{align}$$

The remaining derivatives are more immediate:

$$\begin{cases}x=rs\\y=s^2t\\z=\frac s{t^2}\end{cases}\implies\begin{cases}x_s=r\\y_s=2st\\z_s=\frac1{t^2}\end{cases}$$

Then the ultimate result follows by substituting everything in $u$:

$$\begin{align} u_s&=r\csc\left(\frac{s^3}t\right)-rs\csc\left(\frac{s^3}t\right)\cot\left(\frac{s^3}t\right)\left(\frac{2s^2}t+\frac{s^2}t\right)\\[1ex] &=r\csc\left(\frac{s^3}t\right)-\frac{3rs^3}t\csc\left(\frac{s^3}t\right)\cot\left(\frac{s^3}t\right) \end{align}$$

which agrees with the derivative you added in the comments, but the final result should be free of the intermediate dependent variables $x,y,z$.