# Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 43

$y' = 3x^{2} + 8x + 1$ $y'' = 6x +8$ $y''' = 6$ $y'''' = 0$ (and all higher derivatives will also be 0)

#### Work Step by Step

$y = (x-1)(x+2)(x+3)$ $y = (x^{2}+x-2)(x+3)$ $y = x^{3}+3x^{2}+x^{2}+3x-2x-6$ $y = x^{3}+4x^{2}+x-6$ $y' = 3x^{2} + 8x + 1$ $y'' = 6x +8$ $y''' = 6$ $y'''' = 0$ (and all higher derivatives will also be 0)

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