# How do you find the derivative of f(x) = (x-1)(x^2+2)^3?

Aug 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left({x}^{2} + 2\right)}^{3} \left[\left(6 {x}^{2} - 6 x\right) + 1\right]$

#### Explanation:

Given -

$y = \left(x - 1\right) {\left({x}^{2} + 2\right)}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left(x - 1\right) 3 {\left({x}^{2} + 2\right)}^{2} \left(2 x\right)\right] + \left[{\left({x}^{2} + 2\right)}^{3} \left(1\right)\right]$

dy/dx=[6x^2-6x)(x^2+2)^3]+[(x^2+2)^3]

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left({x}^{2} + 2\right)}^{3} \left[\left(6 {x}^{2} - 6 x\right) + 1\right]$