Question

If $f\left(x\right)={\cot }^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$, then $f\text{'}\left(1\right)$ is equal to

• A. $-1$
• B. $1$
• C. $\log 2$
• D. $-\log 2$

Answer 1

The correct option is A

$-1$

Explanation for the correct option.

Step 1. Simplify the given function.

The given function $f\left(x\right)={\cot }^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$ can be simplified using properties of algebra and exponent as:

$\begin{array}{rcl}f\left(x\right)& =& {\cot }^{-1}\left(\frac{x^x-{\displaystyle \frac{1}{x^x}}}{2}\right)\\ & =& {\cot }^{-1}\left(\frac{{\left(x^x\right)}^2-1}{2x^x}\right)\end{array}$

Now, substitute $x^x=\cot \theta$.

$\begin{array}{rcl}f\left(x\right)& =& {\cot }^{-1}\left(\frac{{\cot }^2\theta -1}{2\cot \theta }\right)\\ & =& {\cot }^{-1}\left(\cot 2\theta \right)\left[\cot 2\theta =\frac{{\cot }^2\theta -1}{2\cot \theta }\right]\\ & =& 2\theta \\ & =& 2{\cot }^{-1}\left(x^x\right)\left[x^x=\cot \theta \right]\end{array}$

Step 2. Find the value of $f\text{'}\left(1\right)$.

Differentiate the function $f\left(x\right)=2\mathrm{co}{t}^{-1}\left(x^x\right)$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(2{\cot }^{-1}\left(x^x\right)\right)\\ & =& 2\times \frac{-1}{1+{\left(x^x\right)}^2}{x}^x\left(1+\log x\right)\end{array}$

Now, substitute $x=1$.

$\begin{array}{rcl}f\text{'}\left(1\right)& =& 2\times \frac{-1}{1+{\left(1^1\right)}^2}{1}^1\left(1+\log 1\right)\\ & =& 2\times \frac{-1}{1+1}\times 1\left(1+0\right)\\ & =& 2\times \frac{-1}{2}\\ & =& -1\end{array}$

Hence, the correct option is A.