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

May 23, 2018

$f ' \left(x\right) = 3 {x}^{2} - 1$

Explanation:

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

for this question, $g \left(x\right) = {x}^{2} + x$ and $h \left(x\right) = x - 1$

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

May 23, 2018

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

$f ' \left(x\right) = \left({x}^{2} + x\right) \left(1\right) + \left(2 x + 1\right) \left(x - 1\right)$

$f ' \left(x\right) = {x}^{2} + x + \left(2 x + 1\right) \left(x - 1\right)$

$f ' \left(x\right) = {x}^{2} + x + 2 {x}^{2} - 2 x + x - 1$

$f ' \left(x\right) = 3 {x}^{2} - 1$