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Q.
If $f \left(x\right)=cot^{-1}\left(\frac{x^{x}-x^{-x}}{2}\right),$ then $f '(1)$ is equal to
Continuity and Differentiability
Solution:
Let $f \left(x\right)=cot^{-1}\left(\frac{x^{x}-x^{-x}}{2}\right)$
Take out$ x^{-x}$ common
$f \left(x\right)=cot^{-1}\left(\frac{x^{2x}-1}{2x^{x}}\right)$
Put$ x^{x} = tan \,\theta$
$\therefore f \left(x\right)=cot^{-1}\left\{\frac{tan^{2}\,\theta-1}{2\,tan\,\theta}\right\}=cot^{-1}\left(-cot\,2\theta\right)$
$=\pi-cot^{-1}\left(cot\,2\theta\right)\, \left(\because tan 2\theta =\frac{2 tan \theta}{1-tan^{2}\theta}\right)$
$\Rightarrow f \left(x\right)=\pi-2\theta=\pi-2\,tan^{-1}\left(x^{x}\right)$
Differentiate w.r.t. x, we get
$f 'x=-\frac{2}{1+x^{2x}}.x^{x}\left(1+log\,x\right)$
$\therefore $ At $x=1$
$f '\left(1\right)=\frac{-2}{1+1}\left(1+0\right)=-1.$