If \(xy = e^{x - y}\), where x > 0 and y > 0, then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{y(x - 1)}{x(y + 1)}\)
Taking logarithms of both sides:
\(\log_e x + \log_e y = x - y\)
Differentiating implicitly with respect to x:
\( \displaystyle \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx} \)
\( \displaystyle \frac{dy}{dx}\left(\frac{1}{y} + 1\right) = 1 - \frac{1}{x} \)
\( \displaystyle \frac{dy}{dx} \cdot \frac{1 + y}{y} = \frac{x - 1}{x} \)
\( \displaystyle \frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)} \)