# How do you use the Product Rule to find the derivative of y=(x^2+1) (x+2)^2 (x-3)^3?

Aug 17, 2015

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

#### Explanation:

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

The Product Rule, states that if $y = u v w$, then

$\frac{\mathrm{dy}}{\mathrm{dx}} = u v \frac{\mathrm{dw}}{\mathrm{dx}} + u w \frac{\mathrm{dv}}{\mathrm{dx}} + v w \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

Let $u = {x}^{2} + 1$, $v = {\left(x + 2\right)}^{2}$, and $w = {\left(x - 3\right)}^{3}$

Then

$\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$, $\frac{\mathrm{dv}}{\mathrm{dx}} = 2 \left(x + 2\right)$, and$\frac{\mathrm{dw}}{\mathrm{dx}} = 3 {\left(x - 3\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = u v \frac{\mathrm{dw}}{\mathrm{dx}} + u w \frac{\mathrm{dv}}{\mathrm{dx}} + v w \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

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

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

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