Question

If $x^y=e^{x-y}$ then $\frac{dy}{dx}$ is

Answer 1

$x^y=e^{x-y}$

$\Rightarrow \log x^y=\log e^{x-y}$

$\Rightarrow y\log x=x-y$

$\Rightarrow y\log x+y=x$

$\Rightarrow y(\log x+1)=x$

$\Rightarrow y(\frac{1}{x}+0)+(\log x+1)\frac{dy}{dx}=1$

$\Rightarrow (\log x+1)\frac{dy}{dx}=1-\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{x-y}{x(\log x+1)}$

$\therefore \frac{dy}{dx}=\frac{x-y}{x(\log x+1)}$