# How do you find the first and second derivative of ln[lnx^2+1)]?

Jul 24, 2017

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

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

#### Explanation:

Suppose we let:

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

Using the chain rule we have:

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

To find the second derivative we will need to apply the quotient rule:

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

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

And using the product rule, combined with the chain rulewe have:

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

Injecting this result into our earlier findings we get:

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

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

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

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

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