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

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Question

(C) Let $u = (\log x)^{x}$.Taking logarithm on both sides,$\log u = x \log(\log x)$.Differentiating with respect to $x$,we get $\frac{1}{u} \frac{du}{

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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Assertion $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$
Reason $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$

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