Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 36

Differentiate $y = x^{\cos x}$


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$