Question

If $y=\cot ^{-1}(\cos 2 x)^{\dfrac{1}{2}}$ then the value of $\dfrac{d y}{d x}$ at $x=\dfrac{\pi}{6}$ will be
A. $\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}$
B. $\left(\dfrac{1}{3}\right)^{\dfrac{1}{2}}$
C. $(3)^{\dfrac{1}{2}}$
D. $(6)^{\dfrac{1}{2}}$

Answer 1

Hint: Given, $y = {\cos ^{ - 1}}{(\cos 2x)^{\dfrac{1}{2}}}$. We have to find the value of $\dfrac{{dy}}{{dx}}$ at $x = \dfrac{\pi }{6}$. Firstly, we will differentiate $y = {\cos ^{ - 1}}{(\cos 2x)^{\dfrac{1}{2}}}$ with respect to $x$ while differentiating we use $\dfrac{{d({{\cot }^{ - 1}}x)}}{{dx}} = \dfrac{{ - 1}}{{1 + {x^2}}}\dfrac{{dx}}{{dx}}$ . After get simplified value of $\dfrac{{dy}}{{dx}}$ we put $x = \dfrac{\pi }{6}$ to get required answer and use $\cos \dfrac{\pi }{3} = \dfrac{1}{2},\,\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ for simplification.

Complete step by step solution:
Given, $y = {\cot ^{ - 1}}{(\cos 2x)^{\dfrac{1}{2}}}$
$y = {\cot ^{ - 1}}\sqrt {(\cos 2x)} $
Differentiating with respect to $x$
We know that $\dfrac{{d({{\cot }^{ - 1}}x)}}{{dx}} = \dfrac{{ - 1}}{{1 + {x^2}}}\dfrac{{dx}}{{dx}}$
$\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{1 + \cos 2x}}\dfrac{{d(\sqrt {(\cos 2x)} )}}{{dx}}$
After solving
$\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{1 + \cos 2x}} \times \dfrac{1}{{2\sqrt {\cos 2x} }} \times 2\sin 2x$
After simplifying above expression
$\dfrac{{dy}}{{dx}} = \dfrac{{\sin 2x}}{{(1 + \cos 2x)\sqrt {\cos 2x} }}$
We have to find the value of $\dfrac{{dy}}{{dx}}$ at $x = \dfrac{\pi }{6}$
${\dfrac{{dy}}{{dx}}_{x = \dfrac{\pi }{6}}} = \dfrac{{\sin 2 \times \dfrac{\pi }{6}}}{{(1 + \cos 2 \times \dfrac{\pi }{6})\sqrt {\cos 2 \times \dfrac{\pi }{6}} }}$
After simplifying the angle above expression becomes
${\dfrac{{dy}}{{dx}}_{x = \dfrac{\pi }{6}}} = \dfrac{{\sin \dfrac{\pi }{3}}}{{(1 + \cos \dfrac{\pi }{3})\sqrt {\cos \dfrac{\pi }{3}} }}$ (1)
We know $\cos \dfrac{\pi }{3} = \dfrac{1}{2},\,\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Putting $\cos \dfrac{\pi }{3} = \dfrac{1}{2},\,\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ in the equation (1)
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\left( {1 + \dfrac{1}{2}} \right)\sqrt {\dfrac{1}{2}} }}$
After solving above equation
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{3}{2} \times \sqrt {\dfrac{1}{2}} }}$
$\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt 3 \times \sqrt 2 }}{3}$
Multiplying numerator and denominator by $\sqrt 3 $
$\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}$
We know that $\sqrt a = {\left( a \right)^{\dfrac{1}{2}}}$
Using $\sqrt a = {\left( a \right)^{\dfrac{1}{2}}}$
$\dfrac{{dy}}{{dx}} = {\left( {\dfrac{2}{3}} \right)^{\dfrac{1}{2}}}$

Option ‘A’ is correct

Note: We need to know the basic differentiation of the trigonometric function in order to solve questions. We have used differentiation by parts to find the value of dx. Differentiation is a method by which we can measure per unit of a function in the given independent variable.