# How do you find the derivative of f(x) = [3(x)^2] - 4x?

Mar 31, 2018

$f ' \left(x\right) = 6 x - 4$

#### Explanation:

We have:

$f \left(x\right) = 3 {x}^{2} - 4 x$

Remember the following rules:

The power rule:
$\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ if $n$ is a constant.

The constant multiplication rule:

If a variable is being multiplied by a constant, you can always bring the constant outside the derivative. For example:

$\frac{d}{\mathrm{dx}} \left[3 x\right] = 3 \cdot \frac{d}{\mathrm{dx}} \left[x\right]$

Subtraction rule (Here is an example):

$\frac{d}{\mathrm{dx}} \left[x - 2 x\right] = \frac{d}{\mathrm{dx}} \left[x\right] - \frac{d}{\mathrm{dx}} \left[2 x\right]$

Therefore:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[3 {x}^{2} - 4 x\right]$

$\implies f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[3 {x}^{2}\right] - \frac{d}{\mathrm{dx}} \left[4 x\right]$

$\implies f ' \left(x\right) = 3 \cdot \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] - 4 \cdot \frac{d}{\mathrm{dx}} \left[{x}^{1}\right]$

$\implies f ' \left(x\right) = 3 \cdot 2 \cdot {x}^{2 - 1} - 4 \cdot 1 \cdot {x}^{1 - 1}$

$\implies f ' \left(x\right) = 6 \cdot {x}^{1} - 4 \cdot {x}^{0}$

$\implies f ' \left(x\right) = 6 \cdot x - 4 \cdot 1$

$\implies f ' \left(x\right) = 6 x - 4$