3. If $\mathit{z}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{y}}$andlocalid="1664273019902" $\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{s}$,find $\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{s}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{t}}$.

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

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

Differentiate the functions x and y partially as follows:

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

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

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

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

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

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

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

Differentiate the function x and y partially as follows:

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

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

$\frac{\partial z}{\partial t}=e^{-y}\sin ht$

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