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AP EAMCET · Maths · Differentiation

For \(-1 < x < 1\), if \(f(x)=\cos ^2\left(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\), then \(f^{\prime}(x)=\)

  1. A \(\frac{1}{2}\)
  2. B 1
  3. C -1
  4. D \(\frac{-1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

Given function \(f(x)=\cos ^2\left(\tan ^1 \sqrt{\frac{1-x}{1+x}}\right)\), for \(-1 < x < 1\). Let \(x=\cos 2 \theta\) Then, \(\quad \frac{1-x}{1+x}=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\tan ^2 \theta\) So, \(f(x)=\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}=\frac{1+x}{2}\) So,…
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