Question

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

Answer 1

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

Taking natural log on both the sides-

$y\log x=x-y$

$⟹x=y(\log x+1)$

$⟹\frac{y}{x}=\frac{1}{\log x+1}$

Differentiating both sides w.r.t x-

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

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

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

$⟹\frac{dy}{dx}=\frac{\log x}{(\log x+1{)}^2}$