5. If and $\mathit{u}\mathbf{=}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{3}}\mathit{z}$,$\mathit{x}\mathbf{=}\mathbf{\text{sin}}\left(s+t\right)\mathbf{,}\mathit{y}\mathbf{=}\mathbf{\text{cos}}\left(s+t\right)$,$\mathit{z}\mathbf{=}{\mathit{e}}^{\mathbf{s}\mathbf{t}}$find$\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{s}}$and$\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{t}}$.

Begin by using the differentials of the equation $u=x^2{y}^{3}z$then substitute from one equation into the other to determine $\frac{\partial u}{\partial s}$as follows:

$$\begin{gathered} du=2x{y}^{3}zdx+3x^2{y}^{2}zdy+x^2{y}^{3}dx \\ \frac{\partial u}{\partial s}=2x{y}^{3}z\frac{\partial x}{\partial s}+3x^2{y}^{2}z\frac{\partial y}{\partial s}+x^2{y}^3\frac{\partial z}{\partial s}...\left(1\right) \end{gathered}$$

Differentiate the functions x,y and z partially as follows:

$$\begin{gathered} \frac{\partial x}{\partial s}=\frac{\partial }{\partial s}\left(\sin \left(s+t\right)\right),\frac{\partial y}{\partial s}=\frac{\partial }{\partial s}\left(\cos \left(s+t\right)\right),\frac{\partial z}{\partial s}=\frac{\partial }{\partial s}\left(e^{st}\right) \\ \frac{\partial x}{\partial s}=\cos \left(s+t\right), \frac{\partial y}{\partial s}=-\sin \left(s+t\right), \frac{\partial z}{\partial s}=t{e}^{st}...\left(2\right) \end{gathered}$$

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

$$\begin{gathered} \frac{\partial u}{\partial s}=2x{y}^{3}z\cos \left(s+t\right)-3x^2{y}^{2}z\sin \left(s+t\right)+x^2{y}^{3}t{e}^{st} \\ \frac{\partial u}{\partial s}=x^2{y}^{3}z\left[\frac{2\cos \left(s+t\right)}{x}-\frac{3\sin \left(s+t\right)}{y}+\frac{t{e}^{st}}{z}\right] \end{gathered}$$

Utilize the relation $u=x^2{y}^{3}z$with the equation in (2) as follows:

$$\begin{gathered} \frac{\partial u}{\partial s}=u\left[\frac{2y}{x}-\frac{3x}{y}+\frac{tz}{z}\right] \\ \frac{\partial u}{\partial s}=u\left[\frac{2y}{x}-\frac{3x}{y}+t\right] \end{gathered}$$

Multiply the top and bottom of the above equation by xy as follows:

$\frac{\partial u}{\partial s}=\frac{u}{xy}\left(2y^2-3x^2+xyt\right)$

Thus, the required answer is $\frac{\partial u}{\partial s}=\frac{u}{xy}\left(2y^2-3x^2+xyt\right)$.

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

$$\begin{gathered} du=2x{y}^{3}zdx+3x^2{y}^{2}zdy+x^2{y}^{3}dz \\ \frac{\partial u}{\partial t}=2x{y}^{3}z\frac{\partial x}{\partial t}+3x^2{y}^{2}z\frac{\partial y}{\partial t}+x^2{y}^3\frac{\partial z}{\partial t}...\left(1\right) \end{gathered}$$

Differentiate the functions x,y, and z partially as follows:

$$\begin{gathered} \frac{\partial x}{\partial t}=\frac{\partial }{\partial t}\left(\sin \left(s+t\right)\right),\frac{\partial y}{\partial t}=\frac{\partial }{\partial t}\left(\cos \left(s+t\right)\right),\frac{\partial z}{\partial t}=\frac{\partial }{\partial t}\left(e^{st}\right) \\ \frac{\partial x}{\partial t}=\cos \left(s+t\right), \frac{\partial y}{\partial t}=-\sin \left(s+t\right), \frac{\partial z}{\partial t}=s{e}^{st}...\left(2\right) \end{gathered}$$

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

$$\begin{gathered} \frac{\partial u}{\partial t}=2x{y}^{3}z\cos \left(s+t\right)-3x^2{y}^{2}z\sin \left(s+t\right)+x^2{y}^{3}s{e}^{st} \\ \frac{\partial u}{\partial t}=x^2{y}^{3}z\left[\frac{2\cos \left(s+t\right)}{x}-\frac{3\sin \left(s+t\right)}{y}+\frac{s{e}^{st}}{z}\right] \end{gathered}$$

Utilize the relation $u=x^2{y}^{3}z$with the equation in (2) as follows:

$$\begin{gathered} \frac{\partial u}{\partial t}=u\left[\frac{2y}{x}-\frac{2x}{y}+\frac{sz}{z}\right] \\ \frac{\partial u}{\partial t}=u\left[\frac{2y}{x}-\frac{2x}{y}+s\right] \end{gathered}$$

Multiply the top and bottom of the above equation by xy as follows:

$\frac{\partial u}{\partial t}=\frac{u}{xy}\left(2y^2-3x^2+xys\right)$

Thus, the required answer is $\frac{\partial u}{\partial s}=\frac{u}{xy}\left(2y^2-3x^2+xyt\right),\frac{\partial u}{\partial t}=\frac{u}{xy}\left(2y^2-3x^2+xys\right)$.