91Ó°ÊÓ

Begin by using the differentials of the equation $z=x{e}^{-y}$then substitute from one equation into the other to determine $\frac{∂z}{∂s}$as follows:

$$\begin{gathered} ∂z=e^{-y}dx-x{e}^{-y}dy \\ \frac{∂z}{∂s}=e^{-y}\frac{∂x}{∂s}-x{e}^{-y}\frac{∂y}{∂s}...\left(1\right) \end{gathered}$$

Differentiate the functions x and y partially as follows:

$$\begin{gathered} \frac{∂x}{∂s}=\frac{∂}{∂t}\left(\cos ht\right),\frac{∂y}{∂s}=\frac{∂}{∂s}\left(\cos s\right) \\ \frac{∂x}{∂s}=0,\frac{∂y}{∂s}=-\sin s...\left(2\right) \end{gathered}$$

Substitute the equation (2) into equation (1) as follows:

$\frac{∂z}{∂s}=-x{e}^{-y}\left(-\sin s\right)$

Utilize the relation $z=x{e}^{-y}$as follows:

Thus, the required answer is $\frac{∂z}{∂s}=z\sin s$.

Using the exactly same method as in the preceding part, determine $\frac{∂z}{∂t}$as follows:

$\begin{array}{l}dz=e^{-y}dx-x{e}^{-y}dy\\ \frac{∂z}{∂t}=e^{-y}\frac{∂z}{∂t}-x{e}^{-y}\frac{∂y}{∂t}...\left(1\right)\end{array}$

Differentiate the function x and y partially as follows:

localid="1664274706089" $$\begin{gathered} \frac{∂x}{∂t}=\frac{∂}{∂t}\left(\cos ht\right),\frac{∂y}{∂t}=\frac{∂}{∂t}\left(\cos s\right) \\ \frac{∂x}{∂t}=\cos ht,\frac{∂y}{∂t}=0...\left(2\right) \end{gathered}$$

Substitute the equation (2) into the equation (1) as follows:

$\frac{∂z}{∂t}=e^{-y}\sin ht$

Thus, the required answer is $\frac{∂z}{∂s}=z\sin s,\frac{∂z}{∂t}=e^{-y}\sin ht$.