267. La función $f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-1 \right)$ tiene derivada

a) $3{{x}^{2}}+2x+1$

b) ${{x}^{2}}-2x+1$

c) $3{{x}^{2}}-2x-1$

Calculando primero el producto:

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

$=\quad {{x}^{3}}-x-{{x}^{2}}+1\quad =\quad {{x}^{3}}-{{x}^{2}}-x+1$

Y dado que la función se puede expresar:

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

la derivada pedida se calcula:

$f'\left( x \right)\quad =\quad 1\ \cdot 3\ \cdot {{x}^{3-1}}-1\ \cdot \ 2\ \cdot \ {{x}^{2-1}}-1\ \cdot \ 1\ \cdot \ {{x}^{1-1}}+0\quad =$

$=\quad 3{{x}^{2}}-2{{x}^{1}}-1{{x}^{0}}\quad =\quad 3{{x}^{2}}-2x-1$