If \(ye^{x} + xe^{y} = 2\), then the value of \(\displaystyle\frac{dy}{dx}\) at the point \((0, 2)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-(2 + e^{2})\)
The point lies on the curve, since \(2e^{0} + 0 = 2\). Writing y′ for \(\frac{dy}{dx}\) and using the product rule on each term (implicit differentiation):
\( \left(y'e^{x} + ye^{x}\right) + \left(e^{y} + xe^{y}y'\right) = 0 \)
At x = 0, y = 2, using \(e^{0} = 1\):
\( y' + 2 + e^{2} + 0 = 0 \)
\( y' = -(2 + e^{2}) \)