$\mathbf{If}\mathbf{}{\mathbf{xe}}^{\mathbf{y}}\mathbf{=}{\mathbf{ye}}^{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{find}\mathbf{}\mathbf{dy}\mathbf{/}\mathbf{dx}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{d}}^{\mathbf{2}}\mathbf{y}\mathbf{/}{\mathbf{dx}}^{\mathbf{2}}\mathbf{}\mathbf{for}\mathbf{}\mathbf{y}\mathbf{\ne }\mathbf{1}\mathbf{.}$

The provided equation is$x{e}^y=y{e}^x.$

In implicit differentiation, we differentiate each side of an equation with two variables (Usually xandy) by treating one of the variables as a function of the other.

Consider the equation below:

$x{e}^y=y{e}^x$

Take precautions on all sides (In) .

$$\begin{gathered} \log \left(x{e}^y\right)=\log \left(y{e}^x\right) \\ \log x+\log e^y=\log y+\log e^x \end{gathered}$$

Therefore, differentiate the provided equation with respect to x .

$$\begin{gathered} \frac{1}{x}+\frac{d}{dx}\left({\log }_e{e}^y\right)=\frac{1}{y}.y\text{'}+\frac{d}{dx}\left({\log }_e{e}^x\right) \\ \frac{1}{x}+\frac{d}{dx}\left(\log e^y\right)=\frac{1}{y}.y\text{'}+\frac{d}{dx}\left(x.{\log }_{e}e\right) \\ \frac{1}{x}+\frac{d}{dx}\left(y.1\right)=\frac{1}{y}.y\text{'}+\frac{d}{dx}\left(x.1\right) \end{gathered}$$

Now,

$$\begin{gathered} \frac{1}{x}+y^{\text{'}}=\frac{1}{y}.y\text{'}+1 \\ y\text{'}-\frac{y\text{'}}{y}=1-\frac{1}{x} \\ y\text{'}=\left(\frac{x-1}{x}\right)\times \left(\frac{y}{y-1}\right) \end{gathered}$$

Hence,

$$\begin{gathered} \frac{dy}{dx}=\frac{y}{x}\left(\frac{x-1}{y-1}\right) \\ Ify=1,then \\ \frac{dy}{dx}=\left(\frac{1}{x}\right)\times \left(\frac{x-1}{0}\right) \\ =\infty \end{gathered}$$

(Does not exist)

For$$\begin{gathered} y\text{'}=\frac{y}{x}\left(\frac{x-1}{y-1}\right) \\ \log y\text{'}=\log \frac{y}{x}\left(\frac{x-1}{y-1}\right) \\ =\log \left[y\left(x-1\right)\right]-\log \left[x\left(y-1\right)\right] \\ =\log y+\log \left(x-1\right)-\left[\log x+\log \left(y-1\right)\right] \end{gathered}$$

Again, differentiate the provided equation with respect to x .

$\frac{1}{y}.y\text{'}=\frac{1}{y}.y\text{'}+\frac{1}{x-1}.\left(1-0\right)-\left[\frac{1}{x}+\frac{1}{y-1}\left(y\text{'}-0\right)\right]$

Implies that,

$$\begin{gathered} \frac{y"}{y\text{'}}=\frac{y\text{'}}{y}+\frac{1}{x-1}-\frac{1}{x}-\frac{y\text{'}}{y-1} \\ =y\text{'}\left[\frac{1}{y}-\frac{1}{y-1}\right]+\left[\frac{1}{x-1}-\frac{1}{x}\right] \\ =\frac{y{\text{'}}^2}{y\left(y-1\right)}+y\text{'}\frac{1}{x\left(x-1\right)} \end{gathered}$$

Then,

$$\begin{gathered} \frac{y\text{'}}{y\text{'}}=-\frac{1}{y\left(y-1\right)}\times {\left[\frac{y\left(x-1\right)}{x\left(y-1\right)}\right]}^2+y\text{'}\frac{1}{x\left(x-1\right)} \\ =-\frac{y{\left(y-1\right)}^2}{x^2{\left(y-1\right)}^3}+y\frac{1}{x^2\left(y-1\right)} \end{gathered}$$

Hence,

$\frac{d^{2}y}{d{x}^2}=\frac{y\left(y^2-x^2+2x-2y\right)}{x^2{\left(y-1\right)}^3}aty\ne 1$