Question

If \[ f(x)=\cot^{-1}\!\left(\sqrt{\cos 2x}\right), \] then \[ f'\!\left(\frac{\pi}{6}\right)= \]

• A. \(\dfrac{1}{\sqrt3}\)
• B. \(\dfrac{2}{\sqrt3}\)
• C. \(\sqrt{\dfrac23}\)
• D. \(-\dfrac{2}{\sqrt3}\)

Hint:

For composite inverse trigonometric functions, first differentiate the outer inverse function and then apply the chain rule carefully to the inner square-root expression.

Answer 1

The Correct Option is C

Concept: Use the chain rule and the derivative \[ \frac{d}{dx}\left(\cot^{-1}u\right) = -\frac{u'}{1+u^2}. \]

Step 1: Differentiate the function. Let \[ u=\sqrt{\cos 2x}. \] Then \[ f(x)=\cot^{-1}(u). \] Hence, \[ f'(x) = -\frac{u'}{1+u^2}. \] Now, \[ u=(\cos 2x)^{1/2}. \] Therefore, \[ u' = \frac12(\cos 2x)^{-1/2}(-2\sin 2x) = -\frac{\sin 2x}{\sqrt{\cos 2x}}. \] Thus, \[ f'(x) = \frac{\sin 2x} {\sqrt{\cos 2x}\,(1+\cos 2x)}. \]

Step 2: Substitute \(x=\dfrac{\pi}{6}\). \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt3}{2}, \] \[ \cos\left(\frac{\pi}{3}\right) = \frac12. \] Hence, \[ f'\left(\frac{\pi}{6}\right) = \frac{\frac{\sqrt3}{2}} {\sqrt{\frac12}\left(1+\frac12\right)}. \] \[ = \frac{\frac{\sqrt3}{2}} {\frac{1}{\sqrt2}\cdot\frac32}. \] \[ = \frac{\sqrt3}{2}\cdot\frac{2\sqrt2}{3}. \] \[ = \frac{\sqrt6}{3}. \] \[ = \sqrt{\frac23}. \] Therefore, \[ \boxed{f'\left(\frac{\pi}{6}\right)=\sqrt{\frac23}} \] \[ \boxed{\text{Answer = (C)}} \]