Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 38

Differentiate $\displaystyle y = (\cos x)^x$


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$