# Second order derivative

#### JayWalk

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!

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#### SenatorArmstrong

Please see the attachment. Be mindful of the double product rule and the -2 that must be distributed.

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#### Joppy

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 .

1 person

#### Country Boy

Math Team
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).

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 a moderator:
1 person

#### SenatorArmstrong

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. :ninja:

1 person