Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 42

Differentiate $y = (\ln x)^{\cos x}$


$
\begin{equation}
\begin{aligned}

\ln y =& \ln (\ln x)^{\cos x}
\\
\\
\ln y =& \cos x \ln (\ln x)
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\\
\frac{d}{dx} \ln y =& \frac{d}{dx} [\cos x \ln (\ln x)]
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\frac{1}{y} \frac{dy}{dx} =& \cos x \frac{d}{dx} \ln (\ln x) + \ln (\ln x) \frac{d}{dx} (\cos x)
\\
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\frac{1}{y} y' =& \cos x \cdot \frac{1}{\ln x} \frac{d}{dx} (\ln x) + \ln (\ln x) (-\sin x)
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\frac{y'}{y} =& \frac{\cos x}{\ln x} \cdot \frac{1}{x} - \sin x \ln (\ln x)
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y' =& y \left( \frac{\cos x}{x \ln x} - \sin x \ln (\ln x) \right)
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y' =& (\ln x)^{\cos x} \left[ \frac{\cos x}{x \ln x} - \sin x \ln (\ln x) \right]



\end{aligned}
\end{equation}
$