Question

If $xy=e^{x-y},\mathrm{find}\frac{dy}{dx}$.

Answer 1

$\mathrm{We}\mathrm{have},xy=e^{\left(x-y\right)}$
Taking log on both sides,
$$\begin{gathered} \log \left(xy\right)=\log \left(e^{x-y}\right) \\ \Rightarrow \log x+\log y=\left(x-y\right)\log e \\ \Rightarrow \log x+\log y=\left(x-y\right)\times 1 \\ \Rightarrow \log x+\log y=x-y \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(\log x\right)+\frac{d}{dx}\left(\log y\right)=\frac{d}{dx}\left(x\right)-\frac{dy}{dx} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx} \\ \Rightarrow \left(1+\frac{1}{y}\right)\frac{dy}{dx}=1-\frac{1}{x} \\ \Rightarrow \left(\frac{y+1}{y}\right)\frac{dy}{dx}=\frac{x-1}{x} \\ \Rightarrow \frac{dy}{dx}=\frac{y\left(x-1\right)}{x\left(y+1\right)} \end{gathered}$$