# How do you find the first and second derivative of ln(x/(x^2+1))?

Nov 21, 2016

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

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

$f ' ' \left(x\right) = \frac{{x}^{4} - 2 {x}^{2} - 1}{{\left({x}^{3} + x\right)}^{2}}$

#### Explanation:

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

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

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

$f ' ' \left(x\right) = \frac{\left({x}^{3} + x\right) \left(- 2 x\right) - \left(- {x}^{2} + 1\right) \left(3 {x}^{2} + 1\right)}{{\left({x}^{3} + x\right)}^{2}}$
$f ' ' \left(x\right) = \frac{\left(- 2 {x}^{4} - 2 {x}^{2}\right) - \left(- 3 {x}^{4} - {x}^{2} + 3 {x}^{2} + 1\right)}{{\left({x}^{3} + x\right)}^{2}}$
$f ' ' \left(x\right) = \frac{- 2 {x}^{4} - 2 {x}^{2} + 3 {x}^{4} - 2 {x}^{2} - 1}{{\left({x}^{3} + x\right)}^{2}}$

$f ' ' \left(x\right) = \frac{{x}^{4} - 2 {x}^{2} - 1}{{\left({x}^{3} + x\right)}^{2}}$