Question

if xy=e^x-y then show that dy/dx=y(x-1)/x(y+1)​

Answer 1

given:

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

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

solution:

taking log on both side,

$log(xy) = log( {e}^{x - y} )$

differentiate w.r.t x.

$\frac{d(logxy)}{dx} = \frac{d}{dx} \times log( {e}^{x - y} )$

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

$\frac{1}{x} + \frac{1}{y} \times \frac{dy}{dx} = \frac{1}{ {e}^{x - y} } - \frac{dy}{dx}$

$\frac{1}{y} \times \frac{dy}{dx} + \frac{dy}{dx} = \frac{1}{ {e}^{x - y} } - \frac{1}{x}$

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

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

the value of dy/ dx.