# How do you use the Product Rule to find the derivative of f(x)=(6x-4)(6x+1)?

Jul 17, 2015

$f ' \left(x\right) = 72 x - 18$

#### Explanation:

In general, the product rule states that if $f \left(x\right) = g \left(x\right) h \left(x\right)$ with $g \left(x\right)$ and h(x) some functions of $x$, then $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$.

In this case $g \left(x\right) = 6 x - 4$ and $h \left(x\right) = 6 x + 1$, so $g ' \left(x\right) = 6$ and $h ' \left(x\right) = 6$. Therefore $f \left(x\right) = 6 \left(6 x + 1\right) + 6 \left(6 x - 4\right) = 72 x - 18$.

We can check this by working out the product of $g$ and $h$ first, and then differentiating. $f \left(x\right) = 36 {x}^{2} - 18 x - 4$, so $f ' \left(x\right) = 72 x - 18$.

Jul 17, 2015

You can either multiply this out and then differentiate it, or actually use the Product Rule. I'll do both.

$f \left(x\right) = 36 {x}^{2} + 6 x - 24 x - 4 = 36 {x}^{2} - 18 x - 4$

Thus, $\textcolor{g r e e n}{\frac{\mathrm{dy}}{\mathrm{dx}} = 72 x - 18}$

or...

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

$= \left(6 x - 4\right) \cdot 6 + \left(6 x + 1\right) \cdot 6$

$= 36 x - 24 + 36 x + 6$

$= \textcolor{b l u e}{72 x - 18}$