Question

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

Answer 1

$x^y=e^{x-y}$
On taking log both sides
$\log x^y=\log e^{x-y}$
$y\log x=(x-y)\log e=x-y$
$y+y\log x=x$
$y=\frac{x}{1+\log x}$
On differentiating both sides with respect to $x$
$\frac{dy}{dx}=\frac{(1+\log x)1-x\left(\frac{1}{x}\right)}{{(1+\log x)}^2}$
$=\frac{1+\log x-1}{{(1+\log x)}^2}=\frac{\log x}{{(1+\log x)}^2}$