4. If$\mathit{w}\mathbf{=}{\mathit{e}}^{\mathbf{-}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{s}}^{\mathbf{2}}}$,$\mathit{r}\mathbf{=}\mathit{u}\mathit{v}$,$\mathit{s}\mathbf{=}\mathit{y}\mathbf{+}\mathbf{2}\mathit{v}$, find $\frac{\mathbf{\partial }\mathbf{w}}{\mathbf{\partial }\mathbf{u}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\frac{\mathbf{\partial }\mathbf{w}}{\mathbf{\partial }\mathbf{v}}$.

Begin by using the differentials of the equation $w=e^{-r^2-s^2}$then substitute from one equation into the other to determine $\frac{\partial w}{\partial u}$as follows:

localid="1664277491703" $$\begin{gathered} dw=-2e^{-r^2-s^2}(rdr+sds) \\ \frac{\partial w}{\partial u}=-2e^{-r^2-s^2}(r\frac{\partial r}{\partial u}+s\frac{\partial s}{\partial u})...\left(1\right) \end{gathered}$$

Differentiate the functions r and s partially as follows:

localid="1664277719595" $$\begin{gathered} \frac{\partial r}{\partial u}=\frac{\partial }{\partial u}\left(uv\right),\frac{\partial s}{\partial u}=\frac{\partial }{\partial u}(u+2v) \\ \frac{\partial r}{\partial u}=v,\frac{\partial r}{\partial u}=1...\left(2\right) \end{gathered}$$

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

$\frac{\partial w}{\partial u}=-2e^{-r^2-s^2}\left(rv+s\right)$

Utilize the relation $w=e^{-r^2-s^2}$as follows:

$\frac{\partial w}{\partial u}=-2w(rv+s)$

Thus, the required answer is $\frac{\partial w}{\partial u}=-2w(rv+s)$.

Using the exact same method as in the preceding part, determine $\frac{\partial w}{\partial u}$as follows:

$$\begin{gathered} dw=-2e^{-r^2-s^2}(rdr+sdr) \\ \frac{\partial w}{\partial v}=-2e^{-r^2-s^2}(r\frac{\partial r}{\partial s}+s\frac{\partial s}{\partial v})...\left(1\right) \end{gathered}$$

Differentiate the function r and s partially as follows:

$$\begin{gathered} \frac{\partial r}{\partial v}=\frac{\partial }{\partial v}\left(uv\right),\frac{\partial s}{\partial v}=\frac{\partial }{\partial v}(u+2v) \\ \frac{\partial r}{\partial v}=u,\frac{\partial s}{\partial v}=2...\left(2\right) \end{gathered}$$

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

$\frac{\partial w}{\partial v}=-2e^{-r^2-s^2}\left(ru+2s\right)$

Utilize the relation $w=e^{-r^2-s^2}$as follows:

$\frac{\partial w}{\partial v}=-2w\left(ru+2s\right)$

Thus, the required answer is $\frac{\partial w}{\partial u}=-2w\left(rv+s\right),\frac{\partial w}{\partial v}=-2w\left(ru+2s\right)$.