How do you differentiate f(x)=8x(2x+1)^2 + 3(2x+1)^3 using the product rule?

Dec 6, 2015

$f ' \left(x\right) = 168 {x}^{2} + 136 x + 26$

Explanation:

Find the derivative of each part.

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

$= 8 {\left(2 x + 1\right)}^{2} + 16 x \left(2 x + 1\right) \frac{d}{\mathrm{dx}} \left[2 x + 1\right]$

$= 8 {\left(2 x + 1\right)}^{2} + 32 x \left(2 x + 1\right)$

$= 96 {x}^{2} + 64 x + 8$

$\frac{d}{\mathrm{dx}} \left[3 {\left(2 x + 1\right)}^{3}\right] = 9 {\left(2 x + 1\right)}^{2} \frac{d}{\mathrm{dx}} \left[2 x + 1\right]$

$= 18 {\left(2 x + 1\right)}^{2}$

$= 72 {x}^{2} + 72 x + 18$

Add the two derivatives we've found:

$f ' \left(x\right) = 96 {x}^{2} + 64 x + 8 + 72 {x}^{2} + 72 x + 18$

$= 168 {x}^{2} + 136 x + 26$