Question

How do you use the product rule to differentiate `f(x)=(x^2-3x+1)(3x+2)`?

Answer 1

The product rule states that:

`(d(g(x)*h(x)))/dx = (dg(x))/dx*h(x) + g(x)*(dh(x))/dx`

Explanation:

Pose:

`g(x) = (x^2-3x+1) and h(x) =3x+2`

`(df(x))/dx = (d(x^2-3x+1))/dx*(3x+2)+(x^2-3x+1)*(d(3x+2))/dx = (2x-3)(3x+2)+3(x^2-3x+1) =`
`= 6x^2+4x-9x-6+3x^2-9x+3= 9x^2-14x-3`