# How do you find the derivative of y=ln((x-1)/(x+1))^(1/3)?

Nov 11, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3 {x}^{2} - 3}$

#### Explanation:

$y = \ln {\left(\frac{x - 1}{x - 1}\right)}^{\frac{1}{3}}$
$\therefore y = \frac{1}{3} \ln \left(\frac{x - 1}{x + 1}\right)$
$\therefore 3 y = \ln \left(\frac{x - 1}{x + 1}\right)$
$\therefore {e}^{3 y} = \left(\frac{x - 1}{x + 1}\right)$ ..... [1]

Differentiating the LHS implicity and the RHS using the quotient rule gives:

$\therefore 3 {e}^{3 y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x + 1\right) \left(1\right) - \left(x - 1\right) \left(1\right)}{x + 1} ^ 2$
$\therefore 3 {e}^{3 y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 1 - x + 1}{x + 1} ^ 2$
$\therefore 3 {e}^{3 y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x + 1} ^ 2$

$\therefore 3 \left(\frac{x - 1}{x + 1}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x + 1} ^ 2$ (using [1])
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \cdot \frac{1}{x + 1} ^ 2 \cdot \frac{x + 1}{x - 1}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \cdot \frac{1}{x + 1} \cdot \frac{1}{x - 1}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3} \cdot \frac{1}{{x}^{2} - 1}$

Hence,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{3 {x}^{2} - 3}$

Nov 17, 2016

$y = \ln {\left(\frac{x - 1}{x - 1}\right)}^{\frac{1}{3}}$

$= \frac{1}{3} \ln \left(\frac{x - 1}{x - 1}\right)$

$= \frac{1}{3} \left[\ln \left(x - 1\right) - \ln \left(x + 1\right)\right]$

Recall that $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$, so

$\frac{d}{\mathrm{dx}} \left(\ln \left(x - 1\right)\right) = \frac{1}{x - 1}$ and $\frac{d}{\mathrm{dx}} \left(\ln \left(x + 1\right)\right) = \frac{1}{x + 1}$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \left[\frac{1}{x - 1} - \frac{1}{x + 1}\right]$

$= \frac{2}{3 \left({x}^{2} - 1\right)}$