Question

\(\frac{d}{dx}[sin^2\{cot^{-1}\sqrt{\frac{1-x}{1+x}}\}] =\) ...

• A. \(-1\)
• B. \(\frac{1}{2}\)
• C. \(\frac{-1}{2}\)
• D. \(1\)

Hint:

Put \(x=\cos2\phi\); the expression simplifies to \(\frac{1+x}{2}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Simplify the inner expression first, then differentiate.

Step 2: Substitute:
Let \(x = \cos2\phi\). Then \(\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{2\sin^2\phi}{2\cos^2\phi}} = \tan\phi\).
So \(\cot^{-1}(\tan\phi) = \frac{\pi}{2} - \phi\), and \(\sin^2\left(\frac{\pi}{2}-\phi\right) = \cos^2\phi = \frac{1+\cos2\phi}{2} = \frac{1+x}{2}\).

Step 3: Differentiate:
\[ \frac{d}{dx}\left(\frac{1+x}{2}\right) = \frac12 \]

Final Answer:
The derivative is \(\frac12\), option (B). \[ \boxed{\frac12} \]