Question

If $f\left(x\right)=co{t}^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$, then $f^{\prime }\left(1\right)$ equals

• A. -1
• B. 1
• C. log 2
• D. -log 2

Answer 1

The correct option is A -1

Putting $u=x^x$ and using logarithmic differentiation, we get log u = x log x.
$\Rightarrow \frac{1}{u}\frac{du}{dx}=x.\frac{1}{x}+logx\Rightarrow \frac{du}{dx}=x^x(1+logx)$
Similarly, if $v=x^{-x}$, we get
$\frac{dv}{dx}=-x^{-x}(1+logx)$
$\therefore f^{\prime }\left(x\right)=-\frac{1}{1+{\left(\frac{x^x-x^{-x}}{2}\right)}^2}\frac{d}{dx}\left(\frac{x^x-x^{-x}}{2}\right)$
$=-\frac{4}{x^{2x}+x^{-2x}+2}\left[\frac{1}{2}\left[\right(x^x+x^{-x}\left)\right(1+logx\left)\right]\right]$
$\Rightarrow f^{\prime }\left(1\right)=-\frac{2}{1+1+2}\left[2\right(1+log1\left)\right]=-1$