If \(y=x\sin y\), then \(\displaystyle\frac{dy}{dx}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\sin^2 y}{\sin y-y \cos y}\)
By implicit differentiation with respect to x, using the product rule on \(x\sin y\):
\(\displaystyle \frac{dy}{dx} = \sin y + x\cos y\,\frac{dy}{dx}\)
\(\displaystyle \frac{dy}{dx}\,(1 - x\cos y) = \sin y\)
\(\displaystyle \frac{dy}{dx} = \frac{\sin y}{1 - x\cos y}\)
This form is not among the options; x is eliminated using the given relation \(x = \frac{y}{\sin y}\):
\(\displaystyle 1 - x\cos y = 1 - \frac{y\cos y}{\sin y} = \frac{\sin y - y\cos y}{\sin y}\)
\(\displaystyle \frac{dy}{dx} = \sin y\cdot\frac{\sin y}{\sin y - y\cos y}\)
\(\displaystyle = \frac{\sin^2 y}{\sin y - y\cos y}\)