Question

please answer this question...
correct answer would be marked as brainlist ​

Answer 1

Step-by-step explanation:

Given,

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

Taking log both side,

$ln( {x}^{y} ) = ln( {e}^{x - y} )$

$= > y .ln(x) = (x - y)....(i)$

Differentiating both side, we get,

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

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

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

$= > \frac{dy}{dx} = \frac{y. ln(x) }{x(1 + ln(x)) }$