# How do you use the product rule to differentiate f(x)=(6x+5)(x^3-2)?

$f ' \left(x\right) = 24 {x}^{3} + 3$

#### Explanation:

This is actually a bit easy than the other ones, so make sure to understand this.

So, The Product Rule of Differentiation is:-

If $h \left(x\right) = f \left(x\right) \cdot g \left(x\right)$, then, $h ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$ .

So, Using the Rule,

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(6 x + 5\right) \cdot \left({x}^{3} - 2\right) + \left(6 x + 5\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} - 2\right)$

$= \left\{\frac{d}{\mathrm{dx}} \left(6 x\right) + \frac{d}{\mathrm{dx}} \left(5\right)\right\} \cdot \left({x}^{3} - 2\right) + \left(6 x + 5\right) \cdot \left\{\frac{d}{\mathrm{dx}} \left({x}^{3}\right) - \frac{d}{\mathrm{dx}} \left(2\right)\right\}$

$= \left(6 + 0\right) \cdot \left({x}^{3} - 2\right) + \left(6 x + 5\right) \cdot \left(3 \cdot {x}^{3 - 1} - 0\right)$

[Using the identities :-
$\frac{d}{\mathrm{dx}} \left(\text{constant}\right) = 0$, $\frac{d}{\mathrm{dx}} \left(m x\right) = m$ where $m$ is a constant and $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$
]

$= 6 \left({x}^{3} - 2\right) + 3 {x}^{2} \left(6 x + 5\right) = 6 {x}^{3} - 12 + 18 {x}^{3} + 15 = 24 {x}^{3} + 3$

So, $f ' \left(x\right) = 24 {x}^{3} + 3$.

Hope this helps.