If f(x) = x^{operatorname{Sec}^{-1} x},then f | vedclass

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(C) Given $f(x) = x^{\operatorname{Sec}^{-1} x}$. Taking the natural logarithm on both sides,we get $\log f(x) = \operatorname{Sec}^{-1} x \cdot \log

Vedclass · Accepted Answer

$\frac{2^{\pi / 3}}{6}(\pi + \sqrt{3} \log 2)$

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(C) Given $f(x) = x^{\operatorname{Sec}^{-1} x}$. Taking the natural logarithm on both sides,we get $\log f(x) = \operatorname{Sec}^{-1} x \cdot \log x$. Differentiating both sides with respect to $x$,we have $\frac{1}{f(x)} f^{\prime}(x) = \frac{d}{dx} (\operatorname{Sec}^{-1} x) \cdot \log x + \operatorname{Sec}^{-1} x \cdot \frac{d}{dx} (\log x)$. Using the derivative formula $\frac{d}{dx} \operatorname{Sec}^{-1} x = \frac{1}{|x| \sqrt{x^2 - 1}}$,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{\log x}{x \sqrt{x^2 - 1}} + \frac{\operatorname{Sec}^{-1} x}{x}$. Thus,$f^{\prime}(x) = x^{\operatorname{Sec}^{-1} x} \left( \frac{\log x}{x \sqrt{x^2 - 1}} + \frac{\operatorname{Sec}^{-1} x}{x} \right)$. At $x = 2$,$\operatorname{Sec}^{-1} 2 = \frac{\pi}{3}$. So,$f^{\prime}(2) = 2^{\pi / 3} \left( \frac{\log 2}{2 \sqrt{2^2 - 1}} + \frac{\pi / 3}{2} \right) = 2^{\pi / 3} \left( \frac{\log 2}{2 \sqrt{3}} + \frac{\pi}{6} \right)$. Simplifying,$f^{\prime}(2) = 2^{\pi / 3} \left( \frac{\sqrt{3} \log 2}{6} + \frac{\pi}{6} \right) = \frac{2^{\pi / 3}}{6} (\pi + \sqrt{3} \log 2)$.

If $f(x) = x^{\operatorname{Sec}^{-1} x}$,then $f^{\prime}(2) =$

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