# How do you find the derivative of ln(x^3+3x)?

Oct 17, 2016

$\frac{d}{\mathrm{dx}} \ln \left({x}^{3} + 3 x\right) = \frac{3 {x}^{2} + 3}{{x}^{3} + 3 x}$

#### Explanation:

We us the chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Let $y = \ln \left({x}^{3} + 3 x\right)$, and
Let $u = {x}^{3} + 3 x$

$y = \ln \left({x}^{3} + 3 x\right)$
$\therefore y = \ln u$
$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$

$u = {x}^{3} + 3 x$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2} + 3$

So using the chain rule we have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \left(3 {x}^{2} + 3\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 3}{{x}^{3} + 3 x}$