Find $\displaystyle \frac {d^2y} {dx^2} for y = cos^2 4x$

Printable View

- Jul 10th 2007, 09:27 AMSamanthahelp 2nd derivative
Find $\displaystyle \frac {d^2y} {dx^2} for y = cos^2 4x$

- Jul 10th 2007, 09:30 AMJhevon
- Jul 10th 2007, 09:38 AMSamantha
- Jul 10th 2007, 10:57 AMjanvdl
$\displaystyle \frac {d^2y} {dx^2} for y = cos^2 4x$

Use the chain rule.

$\displaystyle y = cos^2 4x$

$\displaystyle \frac{dy}{dx} = 2cos(4x) . (-sin(4x)) . 4$

$\displaystyle = -8cos(4x)(sin(4x))$

You must derive twice. But use the product rule now.

$\displaystyle \frac{dy}{dx} (-8cos(4x)(sin(4x))) = -8cos(4x)' . (sin(4x)) + -8cos(4x) . (sin(4x))' $

$\displaystyle = (-32(-sin(4x))) . (sin(4x)) + (-8(cos(4x)) . (4(cos(4x))) $

I hope that's right...

I don't have much experience with this. Jhevon's still helping me get the hang of these things... :D - Jul 10th 2007, 11:13 AMSamantha
So the answer is - 32 cos 8x?

- Jul 10th 2007, 11:14 AMservantes135
- Jul 10th 2007, 11:15 AMjanvdl