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Question: The derivative of the function \({{\cot }^{-1}}\\{{{(\cos 2x)}^{1/2}}\\}\)at \(x=\pi /6\) is A. \(...

The derivative of the function cot⁡−1(cos⁡2x)1/2{{\cot }^{-1}}\\{{{(\cos 2x)}^{1/2}}\\}at x=π/6x=\pi /6 is
A. (2/3)1/2{{(2/3)}^{1/2}}
B. (1/3)1/2{{(1/3)}^{1/2}}
C. 31/2{{3}^{1/2}}
D. 612{{6}^{\dfrac{1}{2}}}

Explanation

Solution

Hint: Here we are using these formula ddx(cot⁡−1x)=−11+x2,ddxcos⁡x=−sin⁡x\dfrac{d}{dx}({{\cot}^{-1}}x)=\dfrac{-1}{1+{{x}^{2}}},\dfrac{d}{dx}\cos x=-\sin x and ddxxn=n.xn−1\dfrac{d}{dx}{{x}^{n}}=n.{{x}^{n-1}}. We will apply chain rule here as It is given as a composite function.
Complete step-by-step answer:
Function is cot⁡−1(cos⁡2x)1/2{{\cot }^{-1}}\\{{{(\cos 2x)}^{1/2}}\\}, differently \dfrac{d}{dx}\left[ {{\cot }^{-1}}\left\\{ {{(\cos 2x)}^{\dfrac{1}{2}}} \right\\} \right]
=\dfrac{-1}{1+\left\\{ {{\left( \cos 2x \right)}^{\dfrac{1}{2}}} \right\\}}.\dfrac{d}{dx}{{(\cos 2x)}^{\dfrac{1}{2}}}
=−11+cos⁡2x.12.(cos⁡2x)−12.ddx(cos⁡2x)=\dfrac{-1}{1+\cos 2x}.\dfrac{1}{2}.{{(\cos 2x)}^{\dfrac{-1}{2}}}.\dfrac{d}{dx}(\cos 2x)
=−11+cos⁡2x.12.(cos⁡2x)−12.sin2x.ddx(2x)=\dfrac{-1}{1+\cos 2x}.\dfrac{1}{2}.{{(\cos 2x)}^{\dfrac{-1}{2}}}.sin2x.\dfrac{d}{dx}(2x)
Consider that in all these steps, we are applying chain rule along with application of formulas. Chain rule helps to differentiate functions that are composite in nature. In short, it states that
f(g(x))=f′(g(x)).g′(x)f(g(x))={{f}^{'}}(g(x)).{{g}^{'}}(x)
Continuing our question, we get that
=−11+cos⁡2x.12.(cos⁡2x)−12.(−sin2x).ddx(2x)=\dfrac{-1}{1+\cos 2x}.\dfrac{1}{2}.{{(\cos 2x)}^{\dfrac{-1}{2}}}.(-sin2x).\dfrac{d}{dx}(2x)
=11+cos⁡2x.12.1(cos⁡2x)12.sin2x.2=\dfrac{1}{1+\cos 2x}.\dfrac{1}{2}.\dfrac{1}{{{(\cos 2x)}^{\dfrac{1}{2}}}}.sin2x.2
We are given in the question the value of x to be π6\dfrac{\pi }{6} , put in the equation
=11+cos⁡2(π6).12.1(cos⁡2(π6))12.sin2(π6).2=\dfrac{1}{1+\cos 2\left( \dfrac{\pi }{6} \right)}.\dfrac{1}{2}.\dfrac{1}{{{\left( \cos 2\left( \dfrac{\pi }{6} \right) \right)}^{\dfrac{1}{2}}}}.sin2\left( \dfrac{\pi }{6} \right).2
=11+cos⁡π3.12.1(cos⁡π3)12.sinπ3.2=\dfrac{1}{1+\cos \dfrac{\pi }{3}}.\dfrac{1}{2}.\dfrac{1}{{{\left( \cos \dfrac{\pi }{3} \right)}^{\dfrac{1}{2}}}}.sin\dfrac{\pi }{3}.2
=11+12.12.1(12)12.32.2=\dfrac{1}{1+\dfrac{1}{2}}.\dfrac{1}{2}.\dfrac{1}{{{\left( \dfrac{1}{2} \right)}^{\dfrac{1}{2}}}}.\dfrac{\sqrt{3}}{2}.2
=23.(2)121.32=\dfrac{2}{3}.\dfrac{{{(2)}^{\dfrac{1}{2}}}}{1}.\dfrac{\sqrt{3}}{2}
=(2)12(3)12=(23)12=\dfrac{{{(2)}^{\dfrac{1}{2}}}}{{{(3)}^{\dfrac{1}{2}}}}={{\left( \dfrac{2}{3} \right)}^{\dfrac{1}{2}}}
Thus the value of derivatives at x=π/6x=\pi /6 is (23)12{{\left( \dfrac{2}{3} \right)}^{\dfrac{1}{2}}} and this is option A.
Note: The step by step application of chain rule can often lead to mistakes. It is advised to carefully open each step and proceed forward. The value of x, if stated in the question should be put after the last step. Remember all the formulas of derivatives for quick solutions and no mistake.