How do you differentiate f(x)=(2x-1)(3x-2)(5x+1) using the product rule?

Dec 13, 2015

Well, you don't have to do it with the product rule, but I guess I can do it both ways.

Product Rule:

$f \left(x\right) = g \left(x\right) h \left(x\right)$
$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)$

$f \left(x\right) = k \cdot {x}^{n}$
$f ' \left(x\right) = k n \cdot {x}^{n - 1}$

Easier way:

$f \left(x\right) = \left(6 {x}^{2} - 4 x - 3 x + 2\right) \left(5 x + 1\right)$

$= 30 {x}^{3} + 6 {x}^{2} - 20 {x}^{2} - 4 x - 15 {x}^{2} - 3 x + 10 x + 2$

$= 30 {x}^{3} - 29 {x}^{2} + 3 x + 2$

$\textcolor{b l u e}{f ' \left(x\right) = 90 {x}^{2} - 58 x + 3}$

Here you just multiply everything out first and then use the power rule. Personally I find the power rule easier to remember via muscle memory.

Harder way:

$f \left(x\right) = \left(6 {x}^{2} - 4 x - 3 x + 2\right) \left(5 x + 1\right)$

$\textcolor{b l u e}{f ' \left(x\right)} = \left(6 {x}^{2} - 7 x + 2\right) \left(5\right) + \left(5 x + 1\right) \left(12 x - 7\right)$

$= 30 {x}^{2} - 35 x + 10 + \left(5 x + 1\right) \left(12 x - 7\right)$

$= 30 {x}^{2} - 35 x + 10 + 60 {x}^{2} - 35 x + 12 x - 7$

$= \textcolor{b l u e}{90 {x}^{2} - 58 x + 3}$

Here you partially multiply things out, and then use the product rule, and then continue multiplying things out.