# How do you differentiate f(x)= (2x^2-5)(x+1)  using the product rule?

Mar 2, 2017

$f ' \left(x\right) = 6 {x}^{2} + 4 x - 5$

#### Explanation:

The product rule is $k ' \left(x\right) = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)$

Written informally, this is:

(derivative of product)=(first)x(derivative of second)+(second)x(derivative of the first).

In this case the 'first' is $\left(2 {x}^{2} - 5\right)$ and the 'second' is $\left(x + 1\right)$.

The derivative of the 'first' is calculated as follows:
$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 5\right) = \frac{d}{\mathrm{dx}} 2 {x}^{2} + \frac{d}{\mathrm{dx}} - 5$

$= 2 \frac{d}{\mathrm{dx}} {x}^{2} - \frac{d}{\mathrm{dx}} 5$

$= 2 \cdot 2 x - 0$

$= 4 x$

The derivative of the 'second' is calculated:

$\frac{d}{\mathrm{dx}} \left(x + 1\right) = \frac{d}{\mathrm{dx}} x + \frac{d}{\mathrm{dx}} 1$

$= 1 + 0 = 1$

Using the product rule (and the derivatives worked out above) gives:

$f ' \left(x\right) = \left(2 {x}^{2} - 5\right) \left(1\right) + \left(x + 1\right) \left(4 x\right)$

$= 2 {x}^{2} - 5 + 4 {x}^{2} + 4 x$

$= 6 {x}^{2} + 4 x - 5$