Question

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

Answer 1

The correct answer is $∴\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$
The given function is $xy=e^{(x-y)}$
Taking logarithm on both the sides,we obtain
$log(xy)=log(e^{x-y})$
$⇒log(x)+log(y)=(x-y)loge$
$⇒logx+logy=(x-y)\times 1$
$⇒logx+logy=x-y$
Differentiating both sides with respect to $x$,we obtain
$\frac{d}{dx}(logx)+\frac{d}{dx}(logy)=\frac{d}{dx}(x)-\frac{dy}{dx}$
$⇒\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$
$⇒(1+\frac{1}{y})\frac{dy}{dx}=1-\frac{1}{x}$
$⇒(\frac{y+1}{y})\frac{dy}{dx}=\frac{x-1}{x}$
$∴\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$