# How do you find the derivative of (2x+1)^5 (x^3-x+1)^4?

Sep 26, 2015

Use the product, power and chain rules.

#### Explanation:

$f \left(x\right) = {\left(2 x + 1\right)}^{5} {\left({x}^{3} - x + 1\right)}^{4}$

Use the product, power and chain rules. I use the product rule in the order, $\left(u v\right) ' = u ' v + u v '$

$f ' \left(x\right) = \left[5 {\left(2 x + 1\right)}^{4} \frac{d}{\mathrm{dx}} \left(2 x + 1\right)\right] {\left({x}^{3} - x + 1\right)}^{4} \textcolor{red}{+} {\left(2 x + 1\right)}^{5} \left[4 {\left({x}^{3} - x + 1\right)}^{3} \frac{d}{\mathrm{dx}} \left({x}^{3} - x + 1\right)\right]$

$= \left[5 {\left(2 x + 1\right)}^{4} \left(2\right)\right] {\left({x}^{3} - x + 1\right)}^{4} \textcolor{red}{+} {\left(2 x + 1\right)}^{5} \left[4 {\left({x}^{3} - x + 1\right)}^{3} \left(3 {x}^{2} - 1\right)\right]$

We have finished differentiating, but we can "clean up" the expression. First rewrite each of the two terms (with the red $\textcolor{red}{+}$ between them.

$= 10 {\left(2 x + 1\right)}^{4} {\left({x}^{3} - x + 1\right)}^{4} \textcolor{red}{+} 4 {\left(2 x + 1\right)}^{5} {\left({x}^{3} - x + 1\right)}^{3} \left(3 {x}^{2} - 1\right)$

Now remove the common facotrs of the two terms.

$= 2 {\left(2 x + 1\right)}^{4} {\left({x}^{3} - x + 1\right)}^{3} \left[5 \left({x}^{3} - x + 1\right) \textcolor{red}{+} 2 \left(2 x + 1\right) \left(3 {x}^{2} - 1\right)\right]$

Finally simplify what was left behind in the brackets.

$= 2 {\left(2 x + 1\right)}^{4} {\left({x}^{3} - x + 1\right)}^{3} \left[5 {x}^{3} - 5 x + 5 \textcolor{red}{+} 12 {x}^{3} + 6 {x}^{2} - 4 x - 2\right]$

$= 2 {\left(2 x + 1\right)}^{4} {\left({x}^{3} - x + 1\right)}^{3} \left(17 {x}^{3} + 6 {x}^{2} - 9 x + 3\right)$