Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 60

Find the derivative of $\displaystyle y = \frac{(x^3 + 1)^4 \sin ^2 x}{\sqrt[3]{x}}$ by using logarithmic differentiation.

By taking logarithms of both sides..

$\displaystyle \ln y = \ln \left[ \frac{(x^3 + 1)^4 \sin^2 x}{\sqrt[3]{x}} \right]$

If we apply the Laws of logarithm, we have


$\begin{equation} \begin{aligned} \ln y =& \ln (x^3 + 1)^4 + \ln \sin^2 x - \frac{1}{3} \sqrt[3]{x} && \text{recall that } \ln (xy) = \ln x + \ln y \text{ and } \ln \left( \frac{x}{y} \right) = \ln x - \ln y \\ \\ \ln y =& 4 \ln (x^3 + 1) + 2 \ln \sin x - \frac{1}{3} \ln x && \text{recall that } \ln (x)^k = k \ln x \end{aligned} \end{equation}$


By taking the derivative implicitly, we have..


$\begin{equation} \begin{aligned} &\frac{\displaystyle \frac{d}{dx} (y)}{y} = 4 \left( \frac{\displaystyle \frac{d}{dx} (x^3 + 1)}{x^3 + 1} \right) + 2 \left(\frac{\displaystyle \frac{d}{dx} (\sin x)}{\sin x} \right) - \frac{1}{3} \left( \frac{\displaystyle \frac{d}{dx} (x)}{x } \right) \\ \\ & \frac{\displaystyle \frac{dy}{dx}}{y} = 4 \left( \frac{3x^2}{x^3 + 1} \right) + \frac{2 \cos x}{\sin x} - \frac{1}{3} \left( \frac{1}{x} \right) \\ \\ & \frac{\displaystyle \frac{dy}{dx}}{y} = \frac{12x^2}{x^3 + 1} + 2 \cot x - \frac{1}{3x} \\ \\ & \frac{dy}{dx} = y \left[ \frac{12x^2}{x^3 + 1} + 2 \cot x - \frac{1}{3x} \right] \\ \\ & \frac{dy}{dx} = \left[ \frac{(x^3 + 1)^4 \sin ^2 x}{\sqrt[3]{x}} \right] \left[ \frac{12x^2}{x^3 + 1} + 2 \cot x - \frac{1}{3x} \right] \end{aligned} \end{equation}$