Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 44

Find the derivative of the function $\displaystyle y = \sqrt[4]{\frac{x^2 + 1}{x^2 -1}}$, using log differentiation

$
\begin{equation}
\begin{aligned}
\ln y &= \ln \sqrt[4]{\frac{x^2+1}{x^2-1}}\\
\\
\ln y &= \ln \frac{\sqrt[4]{x^2+1}}{\sqrt[4]{x^2 - 1}}\\
\\
\ln y &= \ln \sqrt[4]{x^2 + 1} - \ln \sqrt[4]{x^2 - 1}\\
\\
\ln y &= \ln (x^2 + 1)^{\frac{1}{4}} - \ln (x^2 - 1)^{\frac{1}{4}}\\
\\
\ln y &= \frac{1}{4} \ln (x^2 + 1) - \frac{1}{4} \ln (x^2 -1)\\
\\
\frac{d}{dx} \ln y &= \frac{1}{4} \frac{d}{dx} \ln (x^2 +1 ) - \frac{1}{4} \frac{d}{dx} \ln (x^2 -1)\\
\\
\frac{1}{y} \frac{dy}{dx} &= \frac{1}{4} \cdot \frac{1}{x^2 +1} \frac{d}{dx} (x^2 + 1) - \frac{1}{4} \cdot \frac{1}{x^2 -1} \frac{d}{dx} (x^2 - 1)\\
\\
\frac{1}{y} y' &= \frac{1}{4(x^2+1)} \cdot 2x - \frac{1}{4(x^2 -1)} \cdot 2x\\
\\
\frac{y'}{y} &= \frac{x}{2(x^2+1)} - \frac{x}{2(x^2-1)}\\
\\
y' &= y \left[ \frac{x}{2(x^2+1)} - \frac{x}{2(x^2-1)} \right]\\
\\
y' &= \sqrt[4]{\frac{x^2+1}{x^2 -1 }} \left[ \frac{x}{2(x^2+1)} - \frac{x}{2(x^2-1)} \right]
\end{aligned}
\end{equation}
$