$\frac{d}{d x} [(\cos x)^{\log x} + (\log x)^{x}] =$

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$(\log x)^{x} [\frac{1}{\log x} + \log (\log x)] + (\cos x)^{\log x} [\frac{1}{x} \log (\cos x) - \log x \cdot \tan x]$

$(\cos x)^{\log x} [\log (\cos x) - \cot x \cdot \log x] + (\log x)^{\log x} [1 + \log (\log x)]$

$((\cos x)^{\log x} + (\log x)^{x}) [\log x \cdot \cos x + x \log x]$

$(\log x)^{x} [\log x + \log (\log x)] (\cos x)^{\log x}$

answer is A.

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Detailed Solution

$u = (\cos x)^{\log x}$

$\begin{array}{l} \Rightarrow \log u = \log x \log (\cos x) \\ \Rightarrow \frac{1}{u} \frac{d u}{d x} = \frac{1}{x} \log (\cos x) + (\log x) \frac{1}{\cos x} (- \sin x) \\ \Rightarrow \frac{d u}{d x} = (\cos x)^{\log x} (\frac{\log (\cos x)}{x} - \tan x \log x) \end{array}$

$and v = (\log x)^{x}$

$\begin{array}{l} \Rightarrow \log v = x \log (\log x) \\ \frac{d v}{d x} = (\log x)^{x} (\frac{1}{\log x} + \log \log x) \\ \Rightarrow \frac{1}{v} \frac{d v}{d x} = (\log (\log x) + \frac{x}{\ln} \times \frac{1}{x}) \end{array}$