Answer:
The answer is
[tex] \frac{d}{dx} = 8 {x}^{3} + 3 {x}^{2} - 2[/tex]
Explanation:
You have to apply Product Rule,
[tex] \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} [/tex]
[tex]let \: u = 2x + 1[/tex]
[tex] \frac{du}{dx} = 2[/tex]
[tex]let \: v = {x}^{3} - 1[/tex]
[tex] \frac{dv}{dx} = 3 {x}^{2} [/tex]
Finally :
[tex] \frac{d}{dx} = (2x + 1)(3 {x}^{2} ) + ( {x}^{3} - 1)(2)[/tex]
[tex] \frac{d}{dx} = 6 {x}^{3} + 3 {x}^{2} + 2 {x}^{3} - 2[/tex]
[tex] \frac{d}{dx} = 8{x}^{3} + 3 {x}^{2} - 2[/tex]
