y=fxgx then dydx=fxgxgx1fx.f'x+logfx.g'x

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$\frac{d}{d x} \{x^{\frac{1}{x}}\} =$

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$x^{\frac{1}{x} - 2} (1 + \log x)$

$(1 + \log x)$

$1 + \log x^{\frac{1}{x} - 2}$

$x^{\frac{1}{x} - 2} (1 - \log x)$

answer is B.

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Detailed Solution

$y = f (x) g (x)$ then $\frac{d y}{d x} = f (x) g (x)$

$[g (x) \frac{1}{f (x)} . f' (x) + \log f (x) . g' (x)]$

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