Question

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

• A. $-1$
• B. $1$
• C. $log2$
• D. $-log2$

Answer 1

Answer: (A)

Let $f\left(x\right)=co{t}^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$
Take out$x^{-x}$ common
$f\left(x\right)=co{t}^{-1}\left(\frac{x^{2x}-1}{2x^x}\right)$
Put$x^x=tan\theta$
$\therefore f\left(x\right)=co{t}^{-1}\left\{\frac{ta{n}^2\theta -1}{2tan\theta }\right\}=co{t}^{-1}\left(-cot2\theta \right)$
$=\pi -co{t}^{-1}\left(cot2\theta \right)\left(\because tan2\theta =\frac{2tan\theta }{1-ta{n}^2\theta }\right)$
$\Rightarrow f\left(x\right)=\pi -2\theta =\pi -2ta{n}^{-1}\left(x^x\right)$
Differentiate w.r.t. x, we get
$f^{\prime }x=-\frac{2}{1+x^{2x}}.x^x\left(1+logx\right)$
$\therefore$ At $x=1$
$f^{\prime }\left(1\right)=\frac{-2}{1+1}\left(1+0\right)=-1.$