# What is the second derivative of f(x)= ln (x^2+2x)?

Jan 13, 2016

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

#### Explanation:

Use the substitution $t = {x}^{2} + 2 x$
Then $f \left(x\right) = \ln t$

$\frac{\mathrm{df}}{\mathrm{dt}} = \frac{1}{t}$
$\frac{\mathrm{dt}}{\mathrm{dx}} = 2 x + 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \left(\frac{1}{{x}^{2} + 2 x}\right) \cdot \left(2 x + 2\right)$
$f ' \left(x\right) = \frac{2 \left(x + 1\right)}{{x}^{2} + 2 x}$

We can now use the quotient rule to find the second derivative

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

$f ' ' \left(x\right) = \frac{2 {x}^{2} + 4 x - 2 {x}^{2} - 8 x - 4}{{x}^{2} + 2} ^ 2$
$f ' ' \left(x\right) = - 4 \frac{x + 1}{{x}^{2} + 2} ^ 2$