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

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(A) Given $f(x) = \cot^{-1} \left(\frac{x^{x} - x^{-x}}{2}\right)$.Applying the chain rule,$f'(x) = -\frac{1}{1 + \left(\frac{x^x - x^{-x}}{2}\right)^

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(A) Given $f(x) = \cot^{-1} \left(\frac{x^{x} - x^{-x}}{2}\right)$.Applying the chain rule,$f'(x) = -\frac{1}{1 + \left(\frac{x^x - x^{-x}}{2}\right)^2} \cdot \frac{d}{dx} \left(\frac{x^x - x^{-x}}{2}\right)$.Simplifying the expression,$f'(x) = -\frac{4}{4 + (x^x - x^{-x})^2} \cdot \frac{1}{2} \cdot \frac{d}{dx} (x^x - x^{-x})$.Note that $(x^x + x^{-x})^2 = (x^x - x^{-x})^2 + 4$,so $f'(x) = -\frac{2}{(x^x + x^{-x})^2} \cdot \frac{d}{dx} (x^x - x^{-x})$.Using $\frac{d}{dx}(x^x) = x^x(1 + \log x)$ and $\frac{d}{dx}(x^{-x}) = -x^{-x}(1 + \log x)$,we get $\frac{d}{dx}(x^x - x^{-x}) = (x^x + x^{-x})(1 + \log x)$.Thus,$f'(x) = -\frac{2}{(x^x + x^{-x})^2} \cdot (x^x + x^{-x})(1 + \log x) = -\frac{2(1 + \log x)}{x^x + x^{-x}}$.At $x = 1$,$f'(1) = -\frac{2(1 + \log 1)}{1^1 + 1^{-1}} = -\frac{2(1 + 0)}{1 + 1} = -\frac{2}{2} = -1$.

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

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