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

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(B) Let $y = f(x) = \cot^{-1} \left( \frac{x^x - x^{-x}}{2} \right)$.Using the derivative formula for $\cot^{-1}(u)$,where $u = \frac{x^x - x^{-x}}{2}

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$-1$

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(B) Let $y = f(x) = \cot^{-1} \left( \frac{x^x - x^{-x}}{2} \right)$.Using the derivative formula for $\cot^{-1}(u)$,where $u = \frac{x^x - x^{-x}}{2}$,we have $f'(x) = -\frac{1}{1+u^2} \cdot \frac{du}{dx}$.First,find $\frac{d}{dx}(x^x)$. Let $g(x) = x^x = e^{x \log x}$. Then $g'(x) = x^x (1 + \log x)$.Similarly,let $h(x) = x^{-x} = e^{-x \log x}$. Then $h'(x) = x^{-x} (-1 - \log x) = -x^{-x} (1 + \log x)$.Thus,$\frac{du}{dx} = \frac{1}{2} [g'(x) - h'(x)] = \frac{1}{2} [x^x(1 + \log x) + x^{-x}(1 + \log x)] = \frac{1}{2} (1 + \log x)(x^x + x^{-x})$.At $x = 1$,$u = \frac{1^1 - 1^{-1}}{2} = 0$.Also,at $x = 1$,$\frac{du}{dx} = \frac{1}{2} (1 + \log 1)(1^1 + 1^{-1}) = \frac{1}{2} (1)(2) = 1$.Therefore,$f'(1) = -\frac{1}{1+0^2} \cdot (1) = -1$.

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

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