Question

If $xy=e^{(x-y)}$, then find $\frac{dy}{dx}$.

• A. $\frac{y(x-1)}{x(y+1)}$
• B. $\frac{y(y+1)}{x(y+1)}$
• C. $\frac{x(y-1)}{y(x+1)}$
• D. None of these

Answer 1

The correct option is A $\frac{y(x-1)}{x(y+1)}$

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