# How do you differentiate (x-1)(x^2+2)^3?

Aug 18, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6} - 6 {x}^{5} + 30 {x}^{4} - 24 {x}^{3} + 36 {x}^{2} - 24 x + 8$

#### Explanation:

Let $u = \left(x - 1\right) \to \frac{\mathrm{du}}{\mathrm{dx}} = 1$

Let $v = {\left({x}^{2} + 2\right)}^{3} \to \frac{\mathrm{du}}{\mathrm{dx}} = 3 {\left({x}^{2} + 2\right)}^{2} \left(2 x\right) = 6 x {\left({x}^{2} + 2\right)}^{2}$

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

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
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$\frac{\mathrm{dy}}{\mathrm{dx}} \text{ "=" } \left[{\left({x}^{2} + 2\right)}^{3} \times 1\right] + \left(x - 1\right) \left[6 x {\left({x}^{2} + 2\right)}^{2}\right]$......Eqn(1)

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

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \text{ "=" } {\left({x}^{2} + 2\right)}^{2} \left(7 {x}^{2} - 6 x + 2\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \text{ "=" } \left({x}^{4} + 4 {x}^{2} + 4\right) \left(7 {x}^{2} - 6 x + 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{b l u e}{7 {x}^{6} - 6 {x}^{5} + 2 {x}^{4}} \textcolor{g r e e n}{+ 28 {x}^{4} - 24 {x}^{3} + 8 {x}^{2}} \textcolor{p u r p \le}{+ 28 {x}^{2} - 24 x + 8}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 7 {x}^{6} - 6 {x}^{5} + 30 {x}^{4} - 24 {x}^{3} + 36 {x}^{2} - 24 x + 8$

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