Question

If $e^{x+y}=xy$, show that $\frac{dy}{dx}=\frac{y(1-x)}{x(y-1)}$.

Answer 1

$e^{x+y}=xy$ .........(i)

Differentiating both sides

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

$[1+\frac{dy}{dx}]\times \left(xy\right)=y+x.\frac{dy}{dx}$ [from eqn(i)]

$xy+xy.\frac{dy}{dx}=y+x.\frac{dy}{dx}$

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

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