# What is the derivative of ln((x+1)/(x-1))?

May 14, 2015

In three moments: first, remember ln derivative rule. Second, chain rule. Third and finally, quocient rule.

Derivative rule: be $y = \ln \left(f \left(x\right)\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right)}{f \left(x\right)}$

Now, let's consider $u = \left(\frac{x + 1}{x - 1}\right)$

Deriving $\ln \left(u\right)$, we get, then, $\frac{u '}{u}$.

But the derivative of $u$ is the derivative of $\left(\frac{x + 1}{x - 1}\right)$, which is calculated as follows:

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

Remembering the values of $u$ and $u '$ and substituting them into the derivative of $\ln \left(u\right)$, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{- 2}{x - 1} ^ 2}{\frac{x + 1}{x - 1}}$

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

May 14, 2015

There are several ways to get to the correct answer. Here is one:

Use properties of logarithm to rewrite:

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

Now use $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$ to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x + 1} - \frac{1}{x - 1}$

If you prefer to write the result as a single fraction, do so.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{{x}^{2} - 1}$