# How do you calculate the fourth derivative of f(x)=2x^4+3sin2x+(2x+1)^4?

Jul 11, 2016

$y ' ' ' ' = 432 + 48 \sin \left(2 x\right)$

#### Explanation:

Application of the chain rule makes this problem easy, though it still requires some legwork to get to the answer:
$y = 2 {x}^{4} + 3 \sin \left(2 x\right) + {\left(2 x + 1\right)}^{4}$
$y ' = 8 {x}^{3} + 6 \cos \left(2 x\right) + 8 {\left(2 x + 1\right)}^{3}$
$y ' ' = 24 {x}^{2} - 12 \sin \left(2 x\right) + 48 {\left(2 x + 1\right)}^{2}$
$y ' ' ' = 48 x - 24 \cos \left(2 x\right) + 192 \left(2 x + 1\right)$
$= 432 x - 24 \cos \left(2 x\right) + 192$

Note that the last step allowed us to substantially simplify the equation, making the final derivative much easier:
$y ' ' ' ' = 432 + 48 \sin \left(2 x\right)$