How do you find the first and second derivative of ln(x^2-4)?

Oct 2, 2016

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

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

Explanation:

let $f \left(x\right) = \ln \left({x}^{2} - 4\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and more generally. $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\ln \left(g \left(x\right)\right)\right) = \frac{g ' \left(x\right)}{g} \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

To find the second derivative, use the $\textcolor{b l u e}{\text{quotient rule}}$

Given $f \left(x\right) = g \frac{x}{h \left(x\right)} \text{ then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots \left(A\right)$

here $g \left(x\right) = 2 x \Rightarrow g ' \left(x\right) = 2$

and $h \left(x\right) = {x}^{2} - 4 \Rightarrow h ' \left(x\right) = 2 x$

substitute these values into (A)

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

$= \frac{2 {x}^{2} - 8 - 4 {x}^{2}}{{x}^{2} - 4} ^ 2 = \frac{- 2 {x}^{2} - 8}{{x}^{2} - 4} ^ 2 = \frac{- 2 \left({x}^{2} + 4\right)}{{x}^{2} - 4} ^ 2$