# How do you use the product rule to differentiate f(x)=(x^2-3x+1)(3x+2)?

Nov 28, 2016

#### Answer:

The product rule states that:

$\frac{d \left(g \left(x\right) \cdot h \left(x\right)\right)}{\mathrm{dx}} = \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \cdot h \left(x\right) + g \left(x\right) \cdot \frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}}$

#### Explanation:

Pose:

$g \left(x\right) = \left({x}^{2} - 3 x + 1\right) \mathmr{and} h \left(x\right) = 3 x + 2$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{d \left({x}^{2} - 3 x + 1\right)}{\mathrm{dx}} \cdot \left(3 x + 2\right) + \left({x}^{2} - 3 x + 1\right) \cdot \frac{d \left(3 x + 2\right)}{\mathrm{dx}} = \left(2 x - 3\right) \left(3 x + 2\right) + 3 \left({x}^{2} - 3 x + 1\right) =$
$= 6 {x}^{2} + 4 x - 9 x - 6 + 3 {x}^{2} - 9 x + 3 = 9 {x}^{2} - 14 x - 3$