Question

Find $\frac{dy}{dx}$ for $x^y=e^{x-y}$.

Answer 1

Given,

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

or, $e^{y\log x}=e^{x-y}$

Now differentiating both sides with respect to $x$ we get,

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

or, $(x^y\log x+e^{x-y})\frac{dy}{dx}=e^{x-y}-y{x}^{y-1}$

or, $\frac{dy}{dx}=\frac{e^{x-y}-y{x}^{y-1}}{x^y\log x+e^{x-y}}$