91Ó°ÊÓ

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

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

Differentiate the functions r and s partially as follows:

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

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

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

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

$\frac{∂w}{∂u}=-2w(rv+s)$

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

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

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

Differentiate the function r and s partially as follows:

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

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

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

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

$\frac{∂w}{∂v}=-2w\left(ru+2s\right)$

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