Question

Find $\frac{dy}{dx}$ of $xy=e^{x-y}$

Answer 1

Given, $xy=e^{x-y}$
Taking logarithm on both the sides, we obtain
$\log \left(xy\right)=\log \left(e^{x-y}\right)$
$\Rightarrow \log x+\log y=(x-y)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
$\Rightarrow (1+\frac{1}{y})\frac{dy}{dx}=1-\frac{1}{x}$
$\Rightarrow \left(\frac{y+1}{y}\right)\frac{dy}{dx}=\frac{x-1}{x}$
$\therefore \frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$