Question

If $f\left( x \right) = {\cot^{ - 1}}\left( {\dfrac{{{x^x} - {x^{ - x}}}}{2}} \right)$ ,then ${f'}\left( 1 \right)$ is equal to
A. $ - 1$
B. $1$
C. $\log 2$
D. $ - \log 2$

Answer 1

Hint: In this question, we divide the given function into two separate functions, differentiate them both with respect to x, and then divide both outputs again to get differentiation of one function with respect to another.

Formula Used:
1. $\tan \left( {2\theta } \right) = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
2. ${\tan^{ - 1}}(x) \cdot \tan \left( x \right) = 1$
3. ${\cot^{ - 1}}\left( x \right) = - \dfrac{1}{{1 + {x^2}}}$
4. $\log {x^x} = x\log x$
5. $\log 1 = 0$

Complete step by step solution:
We are given that $f\left( x \right) = {\cot^{ - 1}}\left( {\dfrac{{{x^x} - {x^{ - x}}}}{2}} \right)$
Now we solve the given equation in simplified form, and we get
$f\left( x \right) = {\cot^{ - 1}}\left( {\dfrac{{\left( {{x^x} - \dfrac{1}{{{x^x}}}} \right)}}{2}} \right)$
$f\left( x \right) = {\cot^{ - 1}}\left( {\dfrac{{\left( {\dfrac{{{{\left( {{x^x}} \right)}^2} - 1}}{{{x^x}}}} \right)}}{2}} \right)$
$f\left( x \right) = {\cot^{ - 1}}\left( {\dfrac{{{{\left( {{x^x}} \right)}^2} - 1}}{{2{x^x}}}} \right)$
Now we know the condition that if ${x^x}$ is greater than zero then ${\cot^{ - 1}}$ changes to ${\tan^{ - 1}}$
So,
$f\left( x \right) = {\tan^{ - 1}}\left( {\dfrac{{2{x^x}}}{{1 - {{\left( {{x^x}} \right)}^2}}}} \right)$
Now we assume that ${x^x} = \tan \theta $ and $f\left( x \right) = y$
After substituting the assumed values, we get
$y = {\tan^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$
Now we know that $\tan \left( {2\theta } \right) = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
Thus, the value of $y$ becomes
$y = {\tan^{ - 1}}.\tan \left( {2\theta } \right)$
Now, by using the formula ${\tan^{ - 1}}(x) \cdot \tan \left( x \right) = 1$, we get
$y = 2\theta $
Or
$y = 2{\tan^{ - 1}}{x^x}$
Now when we differentiate with respect to x, we get
$\dfrac{{dy}}{{dx}} = \dfrac{2}{{1 + {{\left( {{x^x}} \right)}^2}}} \cdot \dfrac{d}{{dx}}\left( {{x^x}} \right)...\left( i \right)$
Now, again assuming $z = {x^x}$ and taking its differentiation, we get
$z = {x^x}$
Taking the log on both sides, we get
$\log z = \log \left( {{x^x}} \right)$
$\log z = x\log x$
Differentiate with respect to x we get
$\dfrac{1}{z}\dfrac{{dz}}{{dx}} = \left( {\log x + x \cdot \dfrac{1}{x}} \right)$
$\dfrac{1}{z}\dfrac{{dz}}{{dx}} = \left( {\log x + 1} \right)$
$\dfrac{{dz}}{{dx}} = z\left( {\log x + 1} \right)$
Now substitute the value of $z$, we get
$\dfrac{d}{{dx}}\left( {{x^x}} \right) = {x^x}\left( {\log x + 1} \right)$
Now substitute this value in equation $\left( i \right)$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{1 + {{\left( {{x^x}} \right)}^2}}} \cdot {x^x}\left( {\log x + 1} \right)$
$f\left( x \right) = \dfrac{{ - 2}}{{1 + {{\left( {{x^x}} \right)}^2}}} \cdot {x^x}\left( {\log x + 1} \right)$
Now, by substituting $x = 1$ in the above equation, we get
$f\left( 1 \right) = \dfrac{{ - 2}}{{1 + {{\left( {{1^1}} \right)}^2}}} \cdot {1^1}\left( {\log 1 + 1} \right)$
$f\left( 1 \right) = \dfrac{{ - 2}}{{1 + 1}}.1\left( {0 + 1} \right)\left( {\because \log 1 = 0} \right)$
After simplifying, we get
Therefore, the value of $f\left( 1 \right) = - 1$

Option ‘A’ is correct

Note: Whenever we come across such problems always remember the concept of derivative which is the rate of change of function u with regard to function v. Here, when substituting $z = x^x$ remember that using logarithmic differentiation, we get $log z = x log x$ which is the turning point of the problem.