Question

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

Answer 1

$\mathrm{We}\mathrm{have},y^x=e^{y-x}$
Taking log on both sides,
$$\begin{gathered} \log y^x=\log e^{\left(y-x\right)} \\ \Rightarrow x\log y=\left(y-x\right)\log e \\ \Rightarrow x\log y=y-x...\left(i\right) \end{gathered}$$

Differentiating with respect to x,
$$\begin{gathered} \frac{d}{dx}\left(x\log y\right)=\frac{d}{dx}\left(y-x\right) \\ \Rightarrow \left[x\frac{d}{dx}\left(\log y\right)+\log y\frac{d}{dx}\left(x\right)\right]=\frac{dy}{dx}-1 \\ \Rightarrow x\left(\frac{1}{y}\right)\frac{dy}{dx}+\log y\left(1\right)=\frac{dy}{dx}-1 \\ \Rightarrow \frac{dy}{dx}\left(\frac{x}{y}-1\right)=-1-\log y \\ \Rightarrow \frac{dy}{dx}\left(\frac{y}{\left(1+\log y\right)y}-1\right)=-\left(1+\log y\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{dy}{dx}\left[\frac{1-1-\log y}{\left(1+\log y\right)}\right]=-\left(1+\log y\right) \\ \Rightarrow \frac{dy}{dx}=-\frac{{\left(1+\log y\right)}^2}{-\log y} \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(1+\log y\right)}^2}{\log y} \end{gathered}$$