Question

If $f(x) = {\cos ^{ - 1}}\left[ {\dfrac{{1 - {{\left( {\log x} \right)}^2}}}{{1 + {{\left( {\log x} \right)}^2}}}} \right]$, then $f'(e)$ is equal to
1. $\dfrac{2}{e}$
2. $\dfrac{1}{e}$
3. $\dfrac{3}{e}$
4. None of these

Answer 1

Hint:Start making given term similar to inverse trigonometric formula i.e., ${\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{\tan ^{ - 1}}x$. For this, let $\log x$ be equal to $\theta $. Apply the formula then differentiate with respect to $x$ and in the last put $e$ at the place of $x$ to know the value of $f'(e)$.

Formula Used:
Inverse Trigonometric formula –
${\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{\tan ^{ - 1}}x$
Derivative of inverse trigonometric functions –
$\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = \dfrac{1}{{1 + {x^2}}}$
Chain Rule-
$\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)) \times g'(x)$
Logarithm Formula –
$\log e = 1$

Complete step by step Solution:
Given that,
$f(x) = {\cos ^{ - 1}}\left[ {\dfrac{{1 - {{\left( {\log x} \right)}^2}}}{{1 + {{\left( {\log x} \right)}^2}}}} \right] - - - - - - (1)$
Let, $\log x = \theta $
$ \Rightarrow x = {\log ^{ - 1}}\theta $
Put the value of $\log x$in equation (1)
$f(x) = {\cos ^{ - 1}}\left[ {\dfrac{{1 - {{\left( \theta \right)}^2}}}{{1 + {{\left( \theta \right)}^2}}}} \right]$
Using trigonometric inverse formula $\left[ {{{\cos }^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 2{{\tan }^{ - 1}}x} \right]$
$f(x) = 2{\tan ^{ - 1}}\theta $
$ \Rightarrow f(x) = 2{\tan ^{ - 1}}(\log x)$
Differentiate the above equation with respect to $x$,
$\dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}\left[ {2{{\tan }^{ - 1}}(\log x)} \right]$
$f'(x) = 2\left[ {\dfrac{1}{{1 + {{(\log x)}^2}}}} \right] \times \dfrac{1}{x} - - - - - - (2)$
Now put $x = e$ in equation (2)
$f'(e) = 2\left[ {\dfrac{1}{{1 + {{(\log e)}^2}}}} \right] \times \dfrac{1}{e}$
$f'(e) = 2\left[ {\dfrac{1}{{1 + {{(1)}^2}}}} \right] \times \dfrac{1}{e}$
$f'(e) = \dfrac{1}{e}$

Hence, the correct option is 2.

Note:When two or more differentiable functions will be there for derivative apply the Chain rule. Chain rule is basically a rule in which we start finding the derivative of given function by beginning from the very first function and keep derivates like a chain till last $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)) \times g'(x)$. Also $\dfrac{d}{{dx}}\left[ {{{(f(x))}^n}} \right] = n{(f(x))^{n - 1}} \times f'(x)$.