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
\(\frac{d}{dx}(logx)+\frac{d}{dx}(logy)=\frac{d}{dx}(x)-\frac{dy}{dx}\)
\(⇒\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}\)
\(⇒(1+\frac{1}{y})\frac{dy}{dx}=1-\frac{1}{x}\)
\(⇒(\frac{y+1}{y})\frac{dy}{dx}=\frac{x-1}{x}\)
\(∴\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}\)