Question

If $x^y=e^{x-y},$ then show that $\frac{dy}{dx}=\frac{\log x}{(1+\log x{)}^2}=\left(\log x\right)(\log ex{)}^{-2}$

Answer 1

$x^y=e^{x-y}$.
Taking log, we get

$y\log x=(x-y)$ ($\because {\log }_{e}e=1$)
$\therefore y(1+\log x)=x$

or $y=\frac{x}{(1+\log x)}$

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

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

$=\frac{\log x}{(\log e+\log x{)}^2}$

$\Rightarrow \frac{dy}{dx}=\left(\log x\right)(\log ex{)}^{-2}$