How do you differentiate f(x)=(x+4x^2)(x^2-3x) using the product rule?

Jan 9, 2016

$f ' \left(x\right) = 16 {x}^{3} - 33 {x}^{2} - 6 x$

Explanation:

The product rule states that for a function $f \left(x\right) = g \left(x\right) h \left(x\right)$,

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + h ' \left(x\right) g \left(x\right)$

Thus,

$f ' \left(x\right) = \left({x}^{2} - 3 x\right) \frac{d}{\mathrm{dx}} \left(x + 4 {x}^{2}\right) + \left(x + 4 {x}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x\right)$

These derivatives are simple to find through the power rule:

$\frac{d}{\mathrm{dx}} \left(x + 4 {x}^{2}\right) = 1 + 8 x$

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

Plug these back in.

$f ' \left(x\right) = \left({x}^{2} - 3 x\right) \left(1 + 8 x\right) + \left(x + 4 {x}^{2}\right) \left(2 x - 3\right)$

Now, distribute and simplify.

$f ' \left(x\right) = 8 {x}^{3} - 23 {x}^{2} - 3 x + 8 {x}^{3} - 10 {x}^{2} - 3 x$

$f ' \left(x\right) = 16 {x}^{3} - 33 {x}^{2} - 6 x$