If f(x) = cot^{-1}left(frac{x^x - x^{-x}}{2}r | vedclass

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(A) Given $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$.Let $u = x^x$. Then $f(x) = \cot^{-1}\left(\frac{u - u^{-1}}{2}\right) = \cot^{-1}\lef

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-$1$

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(A) Given $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}ight)$.Let $u = x^x$. Then $f(x) = \cot^{-1}\left(\frac{u - u^{-1}}{2}ight) = \cot^{-1}\left(\frac{u^2 - 1}{2u}ight)$.Using the identity $\cot^{-1}(z) = 	an^{-1}(1/z)$,we have $f(x) = 	an^{-1}\left(\frac{2u}{u^2 - 1}ight)$.Recall that $2	an^{-1}(u) = 	an^{-1}\left(\frac{2u}{1 - u^2}ight)$,so $f(x) = -2	an^{-1}(u) + \frac{\pi}{2}$ (for appropriate range).Differentiating with respect to $x$: $f'(x) = -2 \cdot \frac{1}{1 + u^2} \cdot \frac{du}{dx}$.Since $u = x^x$,$\frac{du}{dx} = x^x(1 + \ln x)$.At $x = 1$,$u = 1^1 = 1$ and $\frac{du}{dx} = 1^1(1 + \ln 1) = 1$.Thus,$f'(1) = -2 \cdot \frac{1}{1 + 1^2} \cdot 1 = -2 \cdot \frac{1}{2} = -1$.

If $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$,then the value of $f'(1)$ is equal to

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