# How do you find the first and second derivative of y=ln [ x / (x^2 - 1) ]?

Mar 18, 2017

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

$f ' ' \left(x\right) = \frac{{x}^{4} + 4 {x}^{2} - 1}{{x}^{2} {\left(1 - {x}^{2}\right)}^{2}}$

#### Explanation:

Since
$f \left(x\right) = \ln \left(g \left(x\right)\right) \to f ' \left(x\right) = \frac{1}{g} \left(x\right) \cdot g ' \left(x\right)$

and

$g \left(x\right) = \frac{n \left(x\right)}{d \left(x\right)} \to g ' \left(x\right) = \frac{n ' \left(x\right) d \left(x\right) - n \left(x\right) d ' \left(x\right)}{d \left(x\right)} ^ 2$,

the first derivative is:

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

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

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

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

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

The second derivative is:

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

=(2x^2-2x^4-1-x^2+3x^2+3x^4)/(x^2(1-x^2)^2

$= \frac{{x}^{4} + 4 {x}^{2} - 1}{{x}^{2} {\left(1 - {x}^{2}\right)}^{2}}$