Question: If ${\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}{\mathbf{e}}^{\mathbf{-}{\mathbf{t}}^{\mathbf{2}}}\mathbf{dt}\mathbf{=}\mathbf{x}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{u}}^{\mathbf{v}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{y}\mathbf{,}\mathbf{}\mathbf{find}{\left(\frac{\partial u}{\partial x}\right)}_{\mathbf{y}}\mathbf{,}{\left(\frac{\partial u}{\partial y}\right)}_{\mathbf{x}}\mathbf{,}\mathbf{and}\mathbf{}{\left(\frac{\partial y}{\partial x}\right)}_{\mathbf{u}}\mathbf{at}\mathbf{}\mathbf{u}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{v}\mathbf{=}\mathbf{0}$

Given that${\int }_{\mathrm{u}}^{\mathrm{v}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}=\mathrm{x}\mathrm{and}{\mathrm{u}}^{\mathrm{v}}=\mathrm{y}$

We know that $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\left(x\right)}^{\mathbf{v}\left(x\right)}\mathit{f}\left(x,t\right)\mathit{d}\mathit{t}\mathbf{=}\mathit{f}\left(x,v\right)\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\left(x,u\right)\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathit{d}\mathit{t}$

$$\begin{gathered} \frac{\partial x}{\partial u}=e^{-v^2}\frac{d}{du}\left(v\right)-e^{-u^2}\frac{d}{du}\left(u\right)+{\int }_u^{v}0dt \\ =0-e^{-u^2}\left(1\right)+0 \\ =-e^{-u^2} \\ =-e^{-u^2}du \end{gathered}$$

Solving further

$$\begin{gathered} \frac{\partial x}{\partial v}=e^{-v^2}\frac{d}{dv}\left(v\right)-e^{-u^2}\frac{d}{dv}\left(u\right)+{\int }_u^{v}0dt \\ =e^{-v^2}\left(1\right)-0+0 \\ =e^{-v^2} \\ =e^{-v^2}dv \\ {u}^v=y \end{gathered}$$

Take logarithm on both sides

$v\mathrm{In}\left(\mathrm{u}\right)=In\left(y\right)$

Take differential of the equation $v\frac{1}{u}du+\mathrm{In}\left(u\right)dv=\frac{dy}{y}$

$\frac{v}{u}du+\mathrm{In}\left(u\right)dv=\frac{dy}{y}$

Substitute the values$u=2,v=0\mathrm{in}dx=-e^{-u^2}du\mathrm{and}dx=e^{-v^2}dv$

We get,

$$\begin{gathered} dx=e^{-4}du \\ dx=e^{-v^2}dv \\ dx=dv \\ \frac{v}{u}du+In\left(u\right)dv=\frac{dy}{y} \\ u=2,v=0 \\ y=u^v=2°=1 \end{gathered}$$

In(2) dv = dy

From equation (1)

$$\begin{gathered} -e^{-4}du=dx \\ {\left(\frac{\partial u}{\partial x}\right)}_y=-e^4 \end{gathered}$$

To Find ${\left(\frac{\partial u}{\partial y}\right)}_x,\mathrm{put}dx=0$in equation (1) and (2)

$$\begin{gathered} -e^{-4}du=0\mathrm{and}dv=0 \\ -e^{-4}du+dv=0 \end{gathered}$$

From equation (3) In(2) dv = dy

Solve these equations for du

$$\begin{gathered} d{u}_x=\frac{\left|01 \\ \mathrm{dy}\mathrm{In}\left(2\right)\right|}{\left|-e^{-4}1 \\ 0\mathrm{In}\left(2\right)\right|} \\ =\frac{-dy}{-e^{-4}\mathrm{In}\left(2\right)} \\ =\left(\frac{{\mathrm{e}}^4}{\mathrm{In}\left(2\right)}\right)\mathrm{dy} \\ {\left(\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\right)}_{\mathrm{x}}=\frac{{\mathrm{e}}^4}{\mathrm{In}\left(2\right)} \end{gathered}$$

From equation (2) and (3)

$$\begin{gathered} dy=\mathrm{In}\left(2\right)dx \\ {\left(\frac{\partial y}{\partial x}\right)}_u=\mathrm{In}\left(2\right) \end{gathered}$$

Hence${\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}\right)}_{\mathrm{y}}=-{\mathrm{e}}^4,{\left(\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\right)}_x=\frac{{\mathrm{e}}^4}{\mathrm{In}\left(2\right)}\mathrm{and}{\left(\frac{\partial \mathrm{y}}{\partial \mathrm{x}}\right)}_{\mathrm{u}}=\mathrm{In}\left(2\right)$