Question

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

Answer 1

Apply log on both the sides

$y\log x=x-y$

$y=\frac{x}{1+\log x}$

Differentiating both wrt x we get :

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