From: evan 20070719 07:20 pm (UTC)
possibly embarrassing myself here  (Link)

I forget calculus syntax. Is that asking you to just evaluate at x=2pi? In that case, the sin is +1 and the cos is 1 at that point, so it's just 7  4 = 3.
 From: erik 20070719 07:54 pm (UTC)
Re: possibly embarrassing myself here  (Link)

You mean I was only off by 4?! Not bad!
 From: zarex 20070719 08:22 pm (UTC)
Re: possibly embarrassing myself here  (Link)

You have to take the derivative first, then evaluate it at the point x=2pi. Anyway, sin(N*2pi)=0, cos(N*2pi+pi/2)=0.
From: evan 20070719 09:22 pm (UTC)
Re: possibly embarrassing myself here  (Link)

I meant "evaluate the derivative", and "the sin is +1" I meant "the derivative at that point is +1". Nikolas did it more thoroughly below.
 From: zarex 20070719 09:45 pm (UTC)
Re: possibly embarrassing myself here  (Link)

He messed up too. :) Must be a lot of programmers hanging around here, and not enough engineers.
 From: nibot 20070719 10:18 pm (UTC)
Re: possibly embarrassing myself here  (Link)

I think you both forgot to apply the chain rule. The derivative of 7Sin[6x] is 42 Cos[6x], etc. (Deleted comment)
From: evan 20070719 09:23 pm (UTC)
Re: Spelling it out ...  (Link)

Alternatively, what I did was visualize what the derivatives at those wellknown points are: the sin is maximally sloping upwards (+1) and the cos maximally downwards (1), which got me to your secondtolast step without computing any derivatives directly. :)
 From: zarex 20070719 09:44 pm (UTC)
Re: Spelling it out ...  (Link)

You missed a factor on the derivative. D[ sin(ax) ] = a*cos(ax) . Remember D[ sin(u) ] = cos(u)du . See my solution above.
 From: nibot 20070719 10:15 pm (UTC)
you forgot the chain rule  (Link)

zarex has the correct answer above, 14.
From: evan 20070719 10:23 pm (UTC)
Re: Spelling it out ...  (Link)

Haha, we suck. Chain rule? 