# How do you find the second derivative of  ln(x^2+5x) ?

Sep 25, 2016

$- \frac{2 {x}^{2} + 10 x + 25}{{x}^{2} + 5 x} ^ 2$

#### Explanation:

First, we need to know the derivative of $\ln \left(x\right)$.

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

So, since we have a function embedded within $\ln \left(x\right)$, we must use the chain rule to differentiate it.

$\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

Thus:

$\frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 5 x\right) = \frac{1}{{x}^{2} + 5 x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 5 x\right)$

$= \frac{2 x + 5}{{x}^{2} + 5 x}$

Now, to differentiate this again, use the quotient rule:

$\frac{d}{\mathrm{dx}} f \frac{x}{g \left(x\right)} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Thus:

${d}^{2} / {\mathrm{dx}}^{2} \ln \left({x}^{2} + 5 x\right) = \frac{\left({x}^{2} + 5 x\right) \frac{d}{\mathrm{dx}} \left(2 x + 5\right) - \left(2 x + 5\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 5 x\right)}{{x}^{2} + 5 x} ^ 2$

$= \frac{\left({x}^{2} + 5 x\right) \left(2\right) - \left(2 x + 5\right) \left(2 x + 5\right)}{{x}^{2} + 5 x} ^ 2$

$= \frac{2 {x}^{2} + 10 x - \left(4 {x}^{2} + 20 x + 25\right)}{{x}^{2} + 5 x} ^ 2$

$= \frac{- 2 {x}^{2} - 10 x - 25}{{x}^{2} + 5 x} ^ 2$

Sep 25, 2016

$- \frac{2 {x}^{2} + 10 x + 25}{{x}^{2} + 5 x} ^ 2$

#### Explanation:

First derivative: $\left(\frac{1}{{x}^{2} + 5 x}\right) \cdot \left(2 x + 5\right) = \frac{2 x + 5}{{x}^{2} + 5 x}$

Second derivftive: $\frac{2 \left({x}^{2} + 5 x\right) - \left(2 x + 5\right) \left(2 x + 5\right)}{{x}^{2} + 5 x} ^ 2$

$= - \frac{2 {x}^{2} + 10 x + 25}{{x}^{2} + 5 x} ^ 2$

Sep 25, 2016

$\frac{- 2 {x}^{2} - 10 x - 25}{{x}^{2} + 5 x} ^ 2$

#### Explanation:

To find the first derivative use the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots . . \left(A\right)$

let $u = {x}^{2} + 5 x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2 x + 5$

then $y = \ln u \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$

substitute these values into (A) changing u back to x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times \left(2 x + 5\right) = \frac{2 x + 5}{{x}^{2} + 5 x}$
$\textcolor{b l u e}{\text{-----------------------------------------------------}}$

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

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

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

here $g \left(x\right) = 2 x + 5 \Rightarrow g ' \left(x\right) = 2$

and $h \left(x\right) = {x}^{2} + 5 x \Rightarrow h ' \left(x\right) = 2 x + 5$

$f ' \left(x\right) = \frac{\left({x}^{2} + 5 x\right) .2 - \left(2 x + 5\right) \left(2 x + 5\right)}{{x}^{2} + 5 x} ^ 2$

simplifying the numerator.

$= \frac{2 {x}^{2} + 10 x - 4 {x}^{2} - 20 x - 25}{{x}^{2} + 5 x} ^ 2$

The second derivative is therefore.

$= \frac{- 2 {x}^{2} - 10 x - 25}{{x}^{2} + 5 x} ^ 2 = \frac{- \left(2 {x}^{2} + 10 x + 25\right)}{{x}^{2} + 5 x} ^ 2$