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July 10th, 2017, 03:41 PM   #1
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Smile Second order derivative

So the question asks first to find y' of sin(x)cos(2x) which I worked out to be

cos(x)cos(2x) - 2sin(x)sin(2x)

Then question asks for y'' for which the answer is supposed to be

-5sin(x)cos(2x) - 4cos(x)sin(2x)

I am having trouble getting to that point, my answer somehow ends up a lot longer than that. Would anybody be able to help me with this problem?

Thanks!


Last edited by skipjack; July 11th, 2017 at 12:00 AM.
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July 10th, 2017, 04:10 PM   #2
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Please see the attachment. Be mindful of the double product rule and the -2 that must be distributed.
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July 10th, 2017, 05:40 PM   #3
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Originally Posted by SenatorArmstrong View Post
Please see the attachment. Be mindful of the double product rule and the -2 that must be distributed.
Your $d$'s start looking like $\partial$'s .
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July 10th, 2017, 06:09 PM   #4
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Quote:
Originally Posted by JayWalk View Post
So the question asks first to find y' of sin(x)cos(2x) which I worked out to be

cos(x)cos(2x) - 2sin(x)sin(2x)

Then question asks for y'' for which the answer is supposed to be

-5sin(x)cos(2x) - 4cos(x)sin(2x)
The second derivative of y is the derivative of that, which involves two applications of the product rule:
-sin(x)cos(2x) - 2cos(x)sin(2x) - 2cos(x)sin(2x) - 4 sin(x)cos(2x)

Now, -sin(x)cos(2x) - 4 sin(x)cos(2x) = -5 sin(x)cos(2x)
and -2cos(x)cos(2x) - 2 cos(x)sin(2x) = -4 sin(x)cos(2x).

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Originally Posted by JayWalk View Post
I am having trouble getting to that point, my answer somehow ends up a lot longer than that. Would anybody be able to help me with this problem?

Thanks!

Thanks from SenatorArmstrong

Last edited by skipjack; July 11th, 2017 at 12:03 AM.
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July 10th, 2017, 07:00 PM   #5
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Quote:
Originally Posted by Joppy View Post
Your $d$'s start looking like $\partial$'s .
Hahaha! No joke. I was doing partial derivatives all day long so that must be what happens to my d's. My brain just wants to go back to multivariable calc I guess haha.
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