# How do you find the derivative of y=(2x^3+4)(x^2-3x+1)?

Jan 29, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 10 {x}^{4} - 24 {x}^{3} + 6 {x}^{2} + 8 x - 12$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{product rule}}$

$\text{Given "y=f(x).g(x)" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ product rule}$

$\text{here } f \left(x\right) = 2 {x}^{3} + 4 \Rightarrow f ' \left(x\right) = 6 {x}^{2}$

$\text{and } g \left(x\right) = {x}^{2} - 3 x + 1 \Rightarrow g ' \left(x\right) = 2 x - 3$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 {x}^{3} + 4\right) \left(2 x - 3\right) + \left({x}^{2} - 3 x + 1\right) .6 {x}^{2}$

distributing brackets.

$= 4 {x}^{4} - 6 {x}^{3} + 8 x - 12 + 6 {x}^{4} - 18 {x}^{3} + 6 {x}^{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 10 {x}^{4} - 24 {x}^{3} + 6 {x}^{2} + 8 x - 12$