Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 62

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

By taking logarithms of both sides..

$\displaystyle \ln y = \ln \sqrt[4]{\frac{x^2 + 1}{x^2 - 1}}$

If we apply the Laws of logarithm, we have


$
\begin{equation}
\begin{aligned}

& \ln y = \frac{1}{4} \ln \frac{x^2 + 1}{x^2 - 1}
&& \text{recall that } \ln x^k = k \ln x
\\
\\
& \ln y = \frac{1}{4} [\ln (x^2 + 1) - \ln (x^2 - 1)]
&& \text{recall that } \ln \frac{x}{y} = \ln x - \ln y
\\
\\
& \ln y = \frac{1}{4} [\ln (x^2 + 1) - \ln (x - 1)(x + 1)]
\\
\\
& \ln y = \frac{1}{4} [\ln (x^2 + 1) - (\ln (x -1) + \ln (x + 1))]
\\
\\
& \ln y = \frac{1}{4} \ln (x^2 + 1) - \frac{1}{4} \ln (x - 1) - \frac{1}{4} \ln (x + 1)


\end{aligned}
\end{equation}
$


By taking the derivative implicitly, we have..


$
\begin{equation}
\begin{aligned}

& \frac{\displaystyle \frac{d}{dx} (y) }{y} = \frac{1}{4} \left( \frac{\displaystyle \frac{d}{dx} (x^2 + 1)}{x^2 + 1} \right) - \frac{1}{4} \left( \frac{\displaystyle \frac{d}{dx} (x - 1)}{x - 1} \right) - \frac{1}{4} \left( \frac{\displaystyle \frac{d}{dx} (x + 1)}{x + 1} \right)
\\
\\
& \frac{\displaystyle \frac{dy}{dx}}{y} = \frac{1}{4} \left( \frac{2x}{x^2 + 1} \right) - \frac{1}{4} \left( \frac{1}{x - 1} \right) - \frac{1}{4} \left( \frac{1}{x + 1} \right)
\\
\\
& \frac{dy}{dx} = \frac{y}{4} \left( \frac{2x}{x^2 + 1} - \frac{1}{x - 1} - \frac{1}{x + 1} \right)
\\
\\
& \frac{dy}{dx} = \frac{\displaystyle \sqrt[4]{\frac{x^2 + 1}{x^2 - 1}}}{4} \left( \frac{2x}{x^2 + 1} - \frac{1}{x - 1} - \frac{1}{x + 1} \right)



\end{aligned}
\end{equation}
$