Question

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

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

Answer 1

Answer: (A)

$-1$

Substitute $u=x^x$ and using logarithmic differentiation. we get $\log u=x\log x$,

$\Rightarrow$ $\frac{1}{u}\frac{du}{dx}=x\frac{1}{x}+\log x$ $\Rightarrow$ $\frac{du}{dx}=x^x(1+\log x)$

Similarly, if $v=x^{-x}$, we get

$\frac{dv}{dx}=-x^{-x}(1+\log x)$

$\therefore$ $f^{\prime }\left(x\right)=-\frac{1}{1+{\left(\frac{x^x-x^{-x}}{2}\right)}^2}\frac{d}{dx}\left(\frac{x^x-x^{-x}}{2}\right)$

$=-\frac{4}{x^{2x}+x^{-2x}+2}\left[\frac{1}{2}\left[(x^x+x^{-x})(1+\log x)\right]\right]$

$\Rightarrow$ $f^{\prime }\left(1\right)=-\frac{2}{1+1+2}\left[2(1+\log 1)\right]=-1$