# help 2nd derivative

• July 10th 2007, 09:27 AM
Samantha
help 2nd derivative
Find $\frac {d^2y} {dx^2} for y = cos^2 4x$
• July 10th 2007, 09:30 AM
Jhevon
Quote:

Originally Posted by Samantha
Find $\frac {d^2y} {dx^2} for y = cos^2 4x$

$\frac {d^2 y}{dx^2}$ means the second derivative. so you take the derivative of y and then take the derivative of the result, and that's your answer.

is it the derivative you need help with, or did you not know that was what you should do?
• July 10th 2007, 09:38 AM
Samantha
Quote:

Originally Posted by Jhevon
$\frac {d^2 y}{dx^2}$ means the second derivative. so you take the derivative of y and then take the derivative of the result, and that's your answer.

is it the derivative you need help with, or did you not know that was what you should do?

I don't know how to do it...

I have a hard time with derivatives...
• July 10th 2007, 10:57 AM
janvdl
Quote:

Originally Posted by Samantha
Find $\frac {d^2y} {dx^2} for y = cos^2 4x$

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

Use the chain rule.

$y = cos^2 4x$
$\frac{dy}{dx} = 2cos(4x) . (-sin(4x)) . 4$
$= -8cos(4x)(sin(4x))$

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

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

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

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
• July 10th 2007, 11:13 AM
Samantha
So the answer is - 32 cos 8x?
• July 10th 2007, 11:14 AM
servantes135
Quote:

Originally Posted by janvdl
$\frac {d^2y} {dx^2} for y = cos^2 4x$

Use the chain rule.

$y = cos^2 4x$
$\frac{dy}{dx} = 2cos(4x) . (-sin(4x)) . 4$
$= -8cos(4x)(sin(4x))$

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

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

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

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

Verry close.

$(-32*-sin^2(4x))+(-32*cos^2(4x))$

$-32*(cos^2(4x)-sin^2(4x))$

with the double angle form
$cos^2(x)-sin^2(x)= cos(2x)$

we can simplify this
$-32cos(8x)$
• July 10th 2007, 11:15 AM
janvdl
Quote:

Originally Posted by servantes135
Verry close.
$(-32*-sin^2(4x))+(-32*cos^2(4x))$
$-32*(cos^2(4x)-sin^2(4x))$
$cos^2(x)-sin^2(x)= cos(2x)$
$-32cos(8x)$