## University Calculus: Early Transcendentals (3rd Edition)

$y' = 2x^{3} - 3x -1$ $y''= 6x ^{2} -3$ $y''' = 12x$ $y'''' = 12$ $y''''' = 0$ (and all higher derivatives will also be 0)
$y = \frac{x^{4}}{2} -\frac{3x^{2}}{2} - x$ $y' = 2x^{3} - 3x -1$ $y''= 6x ^{2} -3$ $y''' = 12x$ $y'''' = 12$ $y''''' = 0$ (and all higher derivatives will also be 0)
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