Question

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

Answer 1

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

$\Rightarrow y\log x=(x-y)$ by taking $\log$ on both sides

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

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

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

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