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

Feb 2, 2016

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

#### Explanation:

To find the first derivative, use the chain rule in conjunction with the knowledge that $\frac{d}{\mathrm{dx}} \left[\ln x\right] = \frac{1}{x}$. The rule for the natural logarithm is $\frac{d}{\mathrm{dx}} \left[\ln u\right] = \frac{u '}{u}$. Here, $u = {x}^{2} + x$, so

$f ' \left(x\right) = \frac{\frac{d}{\mathrm{dx}} \left[{x}^{2} + 2\right]}{{x}^{2} + 2} = \frac{2 x}{{x}^{2} + 2}$

To find the second derivative, use the quotient rule. This gives:

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

These derivatives can be found through the power rule.

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

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