Question

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

Answer 1

$\begin{array}{c}Given\\ x^y=e^{x-y}\\ applying\log bothsidesweget\\ y\log x=x-y\\ y\left(1+\log x\right)=x\\ y=\frac{x}{1+\log x}\\ Nowdifferenciatebothsidesweget\\ y\frac{dy}{dx}\log x+\log x\frac{dy}{dx}=\frac{d}{dx}x-\frac{d}{dx}y\\ \frac{y}{x}+\log x\frac{dy}{dx}=1-\frac{dy}{dx}\\ \frac{dy}{dx}\left(1+\log x\right)=1-\frac{y}{x}\\ nowsubstituteyvalueweget\\ \frac{dy}{dx}\left(1+\log x\right)=1-\frac{\frac{x}{1+\log x}}{x}\\ \frac{dy}{dx}\left(1+\log x\right)=1-\frac{1}{\log x}\\ \frac{dy}{dx}\left(1+\log x\right)=\frac{1+\log x-1}{1+\log x}\\ \frac{dy}{dx}\left(1+\log x\right)=\frac{\log x}{\left(1+\log x\right)}\\ \therefore \frac{dy}{dx}=\frac{logx}{{\left(1+\log x\right)}^2}\\ Hence,proved\end{array}$