If f x = cot - 1 x x - x - x 2 , then f ' 1 equals

We have, $f\left(x\right)={\cot }^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$

Differentiating both sides w.r.t. $x$, we get
$f^{\text{'}}\left(x\right)=-\frac{1}{1+{\left(\frac{x^x-x^{-x}}{2}\right)}^2}·\frac{d}{dx}\left(\frac{x^x-x^{-x}}{2}\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-2}{4+{\left(x^x-x^{-x}\right)}^2}·\frac{d}{dx}\left(x^x-x^{-x}\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-2}{{\left(x^x+x^{-x}\right)}^2}·\frac{d}{dx}\left(e^{x\log x}-e^{-x\log x}\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-2}{{\left(x^x+x^{-x}\right)}^2}\times \left\{e^{x\log x}·\frac{d}{dx}\left(x\log x\right)-e^{-x\log x}·\frac{d}{dx}\left(-x\log x\right)\right\}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-2}{{\left(x^x+x^{-x}\right)}^2}\left\{x^x\left(1+\log x\right)+x^{-x}\left(1+\log x\right)\right\}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-2\left(1+\log x\right)}{x^x+x^{-x}}$

Put $x=1$, then we get

$f^{\text{'}}\left(1\right)=\frac{-2}{\left(1+1\right)}=-1$