If $ y=\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}} $, then $ \frac{d y}{d x} $ is equal to

$ \frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\left[\frac{3}{2} \cdot \frac{1-\cos x}{1-\sin x}-\frac{1}{2 x}\right] $

$ \frac{2(x-\sin x)^{1 / 2}}{\sqrt{x}}\left[\frac{3}{2} \cdot \frac{1-\cos x}{x-\sin x}-\frac{1}{2 x}\right] $

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y=\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\left[\frac{3}{2} \cdot \frac{1-\cos x}{x-\sin x}-\frac{1}{2 x}\right]

Option B is correct because it follows from logarithmic differentiation and simplification of the derivative of the given function.