If xy=e^x-y, then prove that dy/dx=y(x-1) …

given: xy = {e}^{x - y}xy=ex−y \begin{gathered}\frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)} \\\end{gathered}dxdy​=x(y+1)y(x−1)​​ solution: taking log on both side, log(xy) = log( {e}^{x - y} )log(xy)=log(ex−y) differentiate w.r.t x. \frac{d(logxy)}{dx} = \frac{d}{dx} \times log( {e}^{x - y} )dxd(logxy)​=dxd​×log(ex−y) \frac{d}{dx} \times ( log(x) + log(y) ) = \frac{d}{dx} \times log( log( {e}^{x - y} ) )dxd​×(log(x)+log(y))=dxd​×log(log(ex−y)) \frac{1}{x} + \frac{1}{y} \times \frac{dy}{dx} = \frac{1}{ {e}^{x - y} } - \frac{dy}{dx}x1​+y1​×dxdy​=ex−y1​−dxdy​ \frac{1}{y} \times \frac{dy}{dx} + \frac{dy}{dx} = \frac{1}{ {e}^{x - y} } - \frac{1}{x}y1​×dxdy​+dxdy​=ex−y1​−x1​ \frac{dy}{dx} \times ( \frac{1}{y} + 1) = \frac{1}{ {e}^{x - y} } - \frac{1}{x}dxdy​×(y1​+1)=ex−y1​−x1​ \frac{dy}{dx} = \frac{1}{ {e}^{x - y} } - \frac{1}{x} \div ( \frac{1}{y} + 1)dxdy​=ex−y1​−x1​÷(y1​+1)