If \(xy = e^{(x-y)}\), then value of \(\displaystyle\frac{dy}{dx}\) at (1,1) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
Taking logarithms on both sides (logarithmic differentiation):
\(\log x + \log y = x - y\)
Differentiating implicitly with respect to x:
\( \displaystyle \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx} \)
The point (1, 1) lies on the curve, since \(1 \cdot 1 = e^{0}\). Putting x = 1, y = 1:
\( \displaystyle 1 + \frac{dy}{dx} = 1 - \frac{dy}{dx} \)
\( \displaystyle 2\frac{dy}{dx} = 0,\quad \frac{dy}{dx} = 0 \)