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

Oct 22, 2015

$f ' \left(x\right) = 6 {\left(3 x + 1\right)}^{3} {\left(2 x - 3\right)}^{2} \left(6 x - 5\right)$

#### Explanation:

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

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

$f ' \left(x\right) = {\left(3 x + 1\right)}^{3} {\left(2 x - 3\right)}^{2} \left(12 \left(2 x - 3\right) + 6 \left(3 x + 1\right)\right)$

$f ' \left(x\right) = {\left(3 x + 1\right)}^{3} {\left(2 x - 3\right)}^{2} \left(42 x - 30\right)$

$f ' \left(x\right) = 6 {\left(3 x + 1\right)}^{3} {\left(2 x - 3\right)}^{2} \left(6 x - 5\right)$