15 may 2019

230. La función $f\left( x \right)={}^{3}\!\!\diagup\!\!{}_{\left( {{x}^{2}}-1 \right)}\;$ tiene derivada
a)$f'\left( x \right)={}^{3}\!\!\diagup\!\!{}_{{{\left( {{x}^{2}}-1 \right)}^{2}}}\;$
b)$f'\left( x \right)={}^{6{{x}^{2}}}\!\!\diagup\!\!{}_{{{\left( {{x}^{2}}-1 \right)}^{2}}}\;$
c)$f'\left( x \right)={}^{-6x}\!\!\diagup\!\!{}_{{{\left( {{x}^{2}}-1 \right)}^{2}}}\;$

Calculando dicha derivada como derivada de un cociente tendremos:

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


$=\quad \frac{0\cdot \left( {{x}^{2}}-1 \right)-3\left( 2x-0 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\quad =\quad \frac{0-6x}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\quad =\quad \frac{-6x}{{{\left( {{x}^{2}}-1 \right)}^{2}}}$

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