# How do you differentiate (4x^-2) - 8x + 1?

$y ' = - 8 {x}^{-} 3 - 8$
$y ' = 4 \left({x}^{-} 2\right) ' - 8 \left(x\right) ' + \left(1\right) ' = 4 \cdot \left(- 2\right) \cdot {x}^{- 2 - 1} - 8 \cdot 1 + 0$
$y ' = - 8 {x}^{-} 3 - 8$