Question

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

Answer 1

Given $xy=e^{x-y}$

Applying logarithm on both sides

$⟹\ln x+\ln y=x-y$

Differentiating on both sides wrt $x$

$⟹\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$

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

$⟹\frac{dy}{dx}\times \frac{y+1}{y}=\frac{x-1}{x}$

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