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

Jun 7, 2017

$- 6 x - 4$

#### Explanation:

$f \left(x\right) = \left(2 + x\right) \left(2 - 3 x\right)$

let say $u = \left(2 + x\right)$, then $u ' = 1$ and $v = \left(2 - 3 x\right)$, then $v ' = - 3$

$f ' \left(x\right) = u v ' + v u '$

$f ' \left(x\right) = \left(2 + x\right) \left(- 3\right) + \left(2 - 3 x\right) \left(1\right)$

$f ' \left(x\right) = - 6 - 3 x + 2 - 3 x = - 6 x - 4$

Jun 7, 2017

-6x-4

#### Explanation:

If we want to differentiate $f \left(x\right) = \left(2 + x\right) \left(2 - 3 x\right)$ using the product rule we use the following $\left(f '\right) \left(g\right) + \left(g '\right) \left(f\right)$. Where $f$ is your first term $\left(2 + x\right)$ and $g$ is your second term $\left(2 - 3 x\right)$. So, we take the $\frac{d}{\mathrm{dx}}$ of $f$ which is 1 using the power rule $n {x}^{n - 1}$ keep in mind that the $\frac{d}{\mathrm{dx}}$ of a constant is zero, while $g$ remains the same. At this point what we have is $1 \left(2 - 3 x\right)$ now we take the $\frac{d}{\mathrm{dx}}$ of $g$ which is $- 3$, $f$ remains the same.

After we have derived the equation we now have the following:

$1 \left(2 - 3 x\right) - 3 \left(2 + x\right)$

$2 - 3 x - 6 - 3 x = - 6 x - 4$
Our final answer is $- 6 x - 4$.