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

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