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

Apr 13, 2018

The answer is (3x^2-3)*(2x^2 + 3x + 5) + (x^3 − 3x)*(4x+3), which simplifies to $10 {x}^{4} + 12 {x}^{3} - 3 {x}^{2} - 18 x - 15$.

#### Explanation:

According to the product rule,
 ( f ⋅ g ) ′ = f ′ ⋅ g + f ⋅ g ′
This just means that when you differentiate a product, you do derivative of the first, leave second alone, plus derivative of the second, leave the first alone.

So the first would be (x^3 − 3x) and the second would be $\left(2 {x}^{2} + 3 x + 5\right)$.

Okay, now the derivative of the first is $3 {x}^{2} - 3$, times the second is $\left(3 {x}^{2} - 3\right) \cdot \left(2 {x}^{2} + 3 x + 5\right)$.

The derivative of the second is $\left(2 \cdot 2 x + 3 + 0\right)$, or just $\left(4 x + 3\right)$.
Multiply it by the first and get (x^3 − 3x)*(4x+3).

Add both portions together now: (3x^2-3)*(2x^2 + 3x + 5) + (x^3 − 3x)*(4x+3)

If you multiply it all out and simplify, you should get $10 {x}^{4} + 12 {x}^{3} - 3 {x}^{2} - 18 x - 15$.

Apr 13, 2018

$\frac{d}{\mathrm{dx}} f \left(x\right) = 10 {x}^{4} + 12 {x}^{3} - 3 {x}^{2} - 18 x - 15$

#### Explanation:

The product rule states that for a function, $f$ such that;

$f \left(x\right) = g \left(x\right) h \left(x\right)$
$\frac{d}{\mathrm{dx}} f \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

The function $f$ is given as $f \left(x\right) = \left({x}^{3} - 3 x\right) \left(2 {x}^{2} + 3 x + 5\right)$, which we can split into the product of two functions $g$ and $h$, where;

$g \left(x\right) = {x}^{3} - 3 x$
$h \left(x\right) = 2 {x}^{2} + 3 x + 5$

By applying the power rule, we see that;

$g ' \left(x\right) = 3 {x}^{2} - 3$
$h ' \left(x\right) = 4 x + 3$

Plugging $g$, $g '$, $h$, and $h '$ into our power rule function we get;

$\frac{d}{\mathrm{dx}} f \left(x\right) = \left(3 {x}^{2} - 3\right) \left(2 {x}^{2} + 3 x + 5\right) + \left({x}^{3} - 3 x\right) \left(4 x + 3\right)$

$\frac{d}{\mathrm{dx}} f \left(x\right) = 6 {x}^{4} + 9 {x}^{3} + 15 {x}^{2} - 6 {x}^{2} - 9 x - 15 + 4 {x}^{4} + 3 {x}^{3} - 12 {x}^{2} - 9 x$

$\frac{d}{\mathrm{dx}} f \left(x\right) = 10 {x}^{4} + 12 {x}^{3} - 3 {x}^{2} - 18 x - 15$