# How do you find the derivative of (1+4x)^5(3+x-x^2)^8 using the chain rule?

Nov 1, 2016

$y ' = 8 \left(1 - 2 x\right) {\left(1 + 4 x\right)}^{5} {\left(3 + x - {x}^{2}\right)}^{7} + 20 {\left(1 + 4 x\right)}^{4} {\left(3 + x - {x}^{2}\right)}^{8}$

#### Explanation:

$y = {\left(1 + 4 x\right)}^{5} {\left(3 + x - {x}^{2}\right)}^{8}$

Use product rule $\left(f g\right) ' = f g ' + g f '$ and chain rule (f(g(x))'=f'(g(x))*g'(x)

$f = {\left(1 + 4 x\right)}^{5} ,$ $g = {\left(3 + x - {x}^{2}\right)}^{8}$

$f ' = 5 {\left(1 + 4 x\right)}^{4} \cdot 4 ,$ $g ' = 8 {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left(1 - 2 x\right)$

$y ' = f g ' + g f '$

$y ' = 8 \left(1 - 2 x\right) {\left(1 + 4 x\right)}^{5} {\left(3 + x - {x}^{2}\right)}^{7} + 20 {\left(1 + 4 x\right)}^{4} {\left(3 + x - {x}^{2}\right)}^{8}$