The derivative of (log x)^{x} with respect to | vedclass

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Question

(B) Let $u = (\log x)^x$. Taking logarithm on both sides,we get $\log u = x \log(\log x)$.Now,differentiating with respect to $x$,we have $\frac{1}{u}

Vedclass · Accepted Answer

$x(\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$

Vedclass · Answer

(B) Let $u = (\log x)^x$. Taking logarithm on both sides,we get $\log u = x \log(\log x)$.Now,differentiating with respect to $x$,we have $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1 = \frac{1}{\log x} + \log(\log x)$.Thus,$\frac{du}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.Let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\log x)^x [\frac{1}{\log x} + \log(\log x)]}{1/x} = x(\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.

The derivative of $(\log x)^{x}$ with respect to $\log x$ is

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