Question

Find $\frac{dy}{dx},if{x}^y=e^{x-y}$

Answer 1

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

$ylogx=x-y$

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

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

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

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