Question

$\mathrm{If}x=e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.},\mathrm{prove}\mathrm{that}\frac{dy}{dx}=\frac{x-y}{x\log x}$

Answer 1

$$\begin{gathered} x=e^{\frac{x}{y}} \\ \mathrm{Taking}\mathrm{logarithm}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log x=\frac{x}{y} \\ \Rightarrow y\log x=x \\ \Rightarrow \log x\frac{dy}{dx}+\frac{y}{x}=1 \\ \Rightarrow \log x\frac{dy}{dx}=1-\frac{y}{x} \\ \Rightarrow \log x\frac{dy}{dx}=\frac{x-y}{x} \\ \Rightarrow x\log x\frac{dy}{dx}=x-y \\ \Rightarrow \frac{dy}{dx}=\frac{x-y}{x\log x} \end{gathered}$$