Question

If $f\left( x \right) = \left( {{{\log }_{\cot x}}\tan x} \right){\left( {{{\log }_{\tan x}}\cot x} \right)^{ - 1}} + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$ then $f'\left( 1 \right)$ is
1. $0$
2. $ - 2$
3. $\dfrac{1}{{\sqrt 3 }}$
4. $\sqrt 3 $

Answer 1

Hint: Here, start solving the given function by using logarithm formulas to open the first term of function use ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ and $\log \left( {\dfrac{1}{a}} \right) = - \log a$ solve further and to solve the ${\tan ^{ - 1}}$ term put $x = 2\sin \theta $. Lastly, differentiate the required term with respect to $x$ and find $f'\left( 1 \right)$ at $x = 1$.

Formula used:
Logarithm formula –
${\log _a}b = \dfrac{{\log b}}{{\log a}}$
$\log \left( {\dfrac{1}{a}} \right) = - \log a$
Derivative of inverse –
$\dfrac{d}{{dx}}{\sin ^{ - 1}}\theta = \dfrac{1}{{\sqrt {1 - {x^2}} }}$

Complete step by step solution: 
Given that,
$f\left( x \right) = \left( {{{\log }_{\cot x}}\tan x} \right){\left( {{{\log }_{\tan x}}\cot x} \right)^{ - 1}} + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$
$ = \dfrac{{\log \tan x}}{{\log \cot x}}{\left( {\dfrac{{\log \cot x}}{{\log \tan x}}} \right)^{ - 1}} + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$
$ = \dfrac{{\log \tan x}}{{\log \cot x}} \times \dfrac{{\log \tan x}}{{\log \cot x}} + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$
$ = \dfrac{{\log \tan x}}{{\log \left( {\dfrac{1}{{\tan x}}} \right)}} \times \dfrac{{\log \tan x}}{{\log \left( {\dfrac{1}{{\tan x}}} \right)}} + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$
$ = \dfrac{{\log \tan x}}{{\left( { - \log \tan x} \right)}} \times \dfrac{{\log \tan x}}{{\left( { - \log \tan x} \right)}} + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$
$ = 1 + {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {4 - {x^2}} }}} \right)$
Put $x = 2\sin \theta \Rightarrow \theta = {\sin ^{ - 1}}\dfrac{x}{2}$
$ = 1 + {\tan ^{ - 1}}\left( {\dfrac{{2\sin \theta }}{{\sqrt {4 - {{\left( {2\sin \theta } \right)}^2}} }}} \right)$
$ = 1 + {\tan ^{ - 1}}\left( {\dfrac{{2\sin \theta }}{{2\sqrt {1 - {{\sin }^2}\theta } }}} \right)$
$ = 1 + {\tan ^{ - 1}}\left( {\dfrac{{2\sin \theta }}{{2\sqrt {{{\cos }^2}\theta } }}} \right)$
$ = 1 + {\tan ^{ - 1}}\left( {\tan \theta } \right)$
$ = 1 + \theta $
$ = 1 + {\sin ^{ - 1}}\dfrac{x}{2} - - - - - (1)$
Differentiate equation (1) with respect to $x$
$f'(x) = \dfrac{1}{{\sqrt {1 - \dfrac{{{x^2}}}{4}} }} \times \dfrac{1}{2}$
$ = \dfrac{1}{{\sqrt {4 - {x^2}} }}$
At $x = 1$
$f'\left( 1 \right) = \dfrac{1}{{\sqrt {4 - {{\left( 1 \right)}^2}} }} = \dfrac{1}{{\sqrt 3 }}$
$ \Rightarrow $ Option (3) is the correct answer.

Note: In such questions, to solve the inverse term let $x$ be the trigonometric function so that the inverse will cancel, and it will be easier to find the derivative If you want to solve with the inverse only go ahead it will just make the calculation lengthy only. Also, while finding the derivative of the expression like $f(g(x))$always apply chain rule. First derive the whole function and then derive the function which is inside the original one.