# How do you differentiate f(x)=x^3lnx  using the product rule?

Dec 12, 2016

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

#### Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

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

So with $f \left(x\right) = {x}^{3} \ln x$ we have;

$\left\{\begin{matrix}\text{Let "u = x^3 & => & (du)/dx = 3x^2' \\ "And } v = \ln x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x} '\end{matrix}\right.$

Applying the product rule we get:

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{3} \ln x\right) = \left({x}^{3}\right) \left(\frac{1}{x}\right) + \left(3 {x}^{2}\right) \left(\ln x\right)$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{3} \ln x\right) = {x}^{2} + 3 {x}^{2} \ln x$