Question

If $e^y=y^x,$ prove that $\frac{dy}{dx}=\frac{{\left(\log y\right)}^2}{\log y-1}$

Answer 1

$\mathrm{We}\mathrm{have},e^y=y^x$
Taking log on both sides,
$$\begin{gathered} \log e^y=\log y^x \\ \Rightarrow y\log e=x\log y \\ \Rightarrow y=x\log y...\left(i\right) \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(x\log y\right) \\ \Rightarrow \frac{dy}{dx}=x\frac{dy}{dx}\left(\log y\right)+\log y\frac{d}{dx}\left(x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{x}{y}\frac{dy}{dx}+\log y \\ \Rightarrow \frac{dy}{dx}\left(1-\frac{x}{y}\right)=\log y \\ \Rightarrow \frac{dy}{dx}\left(\frac{y-x}{y}\right)=\log y \\ \Rightarrow \frac{dy}{dx}=\frac{y\log y}{y-x} \\ \Rightarrow \frac{dy}{dx}=\frac{y\log y}{\left(y-{\displaystyle \frac{y}{\log y}}\right)}\left[\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{dy}{dx}=\frac{y\log y\left(\log y\right)}{y\log y-y} \\ \Rightarrow \frac{dy}{dx}=\frac{y{\left(\log y\right)}^2}{y\left(\log y-1\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(\log y\right)}^2}{\left(\log y-1\right)} \end{gathered}$$