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

Mar 26, 2017

$\frac{d}{\mathrm{dx}} g \left(x\right) = - 6 x - 4$

(In the answer, I use h(x) where g(x) is traditionally used to avoid confusion since the question already defines g(x)).

#### Explanation:

Product rule states that:

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

In this case, $f \left(x\right) = 2 + x$, and $h \left(x\right) = 2 - 3 x$. If we differentiate each of these separately, we get that:

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left(2 + x\right) = 1$

$\frac{d}{\mathrm{dx}} h \left(x\right) = \frac{d}{\mathrm{dx}} \left(2 - 3 x\right) = - 3$

Therefore, using product rule, we get:

$\frac{d}{\mathrm{dx}} \left(2 + x\right) \left(2 - 3 x\right) = \left(2 + x\right) \left(- 3\right) + \left(1\right) \left(2 - 3 x\right)$

Now, all we have left to do is simplify.

$\textcolor{w h i t e}{\text{XX}} \left(2 + x\right) \left(- 3\right) + \left(1\right) \left(2 - 3 x\right)$

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

$= - 6 x - 4$