MHT CET · Maths · Differentiation
The derivative of the function \(\cot ^{-1}[\cos 2 x]^{1 / 2}\) at \(x=\pi / 6\) is
- A \(\left(\frac{1}{3}\right)^{1 / 2}\)
- B \(\left(\frac{2}{3}\right)^{1 / 2}\)
- C \(\left(\frac{3}{2}\right)^{1 / 2}\)
- D \((3)^{1 / 2}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{2}{3}\right)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\cot ^{-1}\left[(\cot 2 x)^{\frac{1}{2}}\right]=\cot ^{-1}(\sqrt{\cos 2 x}) \)
\( \therefore \quad f^{\prime}(x)=\frac{-1}{1+(\sqrt{\cos 2 x})^2} \times \frac{d}{d x}(\sqrt{\cos 2 x}) \)
\( =\frac{-1}{1+\cos 2 x} \times \frac{1}{2 \sqrt{\cos 2 x}} \times(-2 \sin 2 x)=\) \(\frac{2 \sin 2 x}{2(1+\cos 2 x) \sqrt{\cos 2 x}} \)
\( \therefore \left[f^{\prime}(x)\right]_{x=\frac{\pi}{6}}=\frac{\sin \left(\frac{\pi}{3}\right)}{\left(1+\cos \frac{\pi}{3}\right) \sqrt{\cos \frac{\pi}{3}}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}\)
\( \therefore \quad f^{\prime}(x)=\frac{-1}{1+(\sqrt{\cos 2 x})^2} \times \frac{d}{d x}(\sqrt{\cos 2 x}) \)
\( =\frac{-1}{1+\cos 2 x} \times \frac{1}{2 \sqrt{\cos 2 x}} \times(-2 \sin 2 x)=\) \(\frac{2 \sin 2 x}{2(1+\cos 2 x) \sqrt{\cos 2 x}} \)
\( \therefore \left[f^{\prime}(x)\right]_{x=\frac{\pi}{6}}=\frac{\sin \left(\frac{\pi}{3}\right)}{\left(1+\cos \frac{\pi}{3}\right) \sqrt{\cos \frac{\pi}{3}}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}\)
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