# How do you find the derivative of f(x) = 3x^2 ln 2x?

Mar 17, 2018

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x \ln 2 x + 3 x$

#### Explanation:

we need to use the product rule

$\frac{d}{\mathrm{dx}} \left(\textcolor{red}{u} v\right) = v \textcolor{red}{\frac{\mathrm{du}}{\mathrm{dx}}} + \textcolor{red}{u} \frac{\mathrm{dv}}{\mathrm{dx}}$

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln 2 x \times \textcolor{red}{6 x} + \textcolor{red}{3 {x}^{2}} \times \frac{1}{2 x} \times 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 x \ln 2 x + 3 x$