Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 16

Determine the function $y = \ln \left( x^4 \sin^2 x \right)$

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \ln \left( x^4 \sin^2 x \right)\\
\\
y' &= \frac{1}{x^4 \sin^2 x} \cdot \frac{d}{dx} \left( x^4 \sin^2 x \right)\\
\\
y' &= \frac{1}{x^4 \sin^2 x} \left[ x^4 \frac{d}{dx} (\sin x)^2 + \sin^2 x \frac{d}{dx} (x^4) \right]\\
\\
y' &= \frac{1}{x^4 \sin^2 x} \left[ x^4 \cdot 2 \sin x \frac{d}{dx} (\sin x) + \sin^2x \cdot 4x^3 \right]\\
\\
y' &= \frac{1}{x^4 \sin^2 x} \left( 2x^4 \sin x \cos x + 4x^3 \sin^2 x\right)\\
\\
y' &= \frac{x^3 \sin x (2x \cos x + 4 \sin x)}{x^4 \sin^2 x}\\
\\
y' &= \frac{2\cancel{x}\cos x}{\cancel{x}\sin x} + \frac{4\cancel{\sin x}}{x \cancel{\sin x}}\\
\\
y' &= 2 \cot x + \frac{4}{x}
\end{aligned}
\end{equation}
$