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

Nov 1, 2016

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

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

*See logarithmic rules .

Now, if $g \left(x\right) = \ln \left(h \left(x\right)\right)$, $g ' \left(x\right) = \frac{h ' \left(x\right)}{h \left(x\right)}$, which means that...

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

So far so good... Now from here, what we get is...

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

Thanks to the quotient rule.

If we simplify the above we get...

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1}{{x}^{2}} - \frac{2 {x}^{2} + 2 - 4 {x}^{2}}{{x}^{2} + 1} ^ 2$

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

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

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