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

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(A) Given $f(x) = \cot^{-1} \left( \frac{x^x - x^{-x}}{2} ight)$.Let $x^x = 	an 	heta$. Then $x^{-x} = \frac{1}{	an 	heta} = \cot 	heta$.Substi

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$-1$

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(A) Given $f(x) = \cot^{-1} \left( \frac{x^x - x^{-x}}{2} ight)$.Let $x^x = 	an 	heta$. Then $x^{-x} = \frac{1}{	an 	heta} = \cot 	heta$.Substituting this into the expression for $f(x)$:$f(x) = \cot^{-1} \left( \frac{	an 	heta - \cot 	heta}{2} ight) = \cot^{-1} \left( \frac{	an^2 	heta - 1}{2 	an 	heta} ight)$.Using the identity $\cot 2	heta = \frac{1 - 	an^2 	heta}{2 	an 	heta}$,we have $\frac{	an^2 	heta - 1}{2 	an 	heta} = -\cot 2	heta$.So,$f(x) = \cot^{-1} (-\cot 2	heta) = \pi - \cot^{-1} (\cot 2	heta) = \pi - 2	heta$.Since $	an 	heta = x^x$,we have $	heta = 	an^{-1}(x^x)$.Thus,$f(x) = \pi - 2 	an^{-1}(x^x)$.Differentiating with respect to $x$:$f'(x) = -2 \cdot \frac{1}{1 + (x^x)^2} \cdot \frac{d}{dx}(x^x)$.We know $\frac{d}{dx}(x^x) = x^x(1 + \log x)$.So,$f'(x) = -2 \cdot \frac{x^x(1 + \log x)}{1 + x^{2x}}$.At $x = 1$,$f'(1) = -2 \cdot \frac{1^1(1 + \log 1)}{1 + 1^2} = -2 \cdot \frac{1(1 + 0)}{2} = -1$.

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

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