# How do you find the derivative for 2x^4-1?

$f \left(x\right) = 2 {x}^{4} - 1 = g \left(x\right) + h \left(x\right)$, where $g \left(x\right) = 2 {x}^{4}$ and $h \left(x\right) = - 1$
$f ' \left(x\right) = g ' \left(x\right) + h ' \left(x\right) = \left(2 {x}^{4}\right) ' + \left(- 1\right) ' = 4 \cdot 2 {x}^{4 - 1} + 0 = 8 {x}^{3}$.