Question

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

Answer 1

Gicen, $x^y=e^{x-y}$
Taking $\log$ on both sides,
$y\log x=x-y$ ... (i)
$y(1+\log x)=x$
$y=\frac{x}{(1+\log x)}$ .... (ii)
Differentiating both sides of equation $\left(i\right)$ using quotient rule, we get
$\frac{dy}{dx}=\frac{(1+\log x).1-x.\left(\frac{1}{x}\right)}{(1+\log x{)}^2}$

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

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