Finding the Power Series of f'(x) from f(x) = x^2cos(2x)

Magnawolf
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Homework Statement



Find the power series of f'(x), given f(x) = x2cos2(x)

Homework Equations



attachment.php?attachmentid=58196&stc=1&d=1366887854.png

Correct me if I'm wrong

The Attempt at a Solution



Can I just take the derivative of the solution I got previously? If so, what's a good way to write the sequence out so I can easily make a series representation. Or is there a better approach to the problem?
 

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Looks good. I would multiply the ##x^2/2## into the parentheses and include it under the summation. Then differentiate it.
 
My main problem is trying to figure out how to represent what I'm getting as a series. When I multiply in the x^2 and differentiate I get

x + x - 8x^3 + 4x^5 - (32/45)x^7...

I can't figure it out
 
When you multiply the ##\frac{x^2} 2## in you have$$
\frac {x^2} 2 +\sum_{n=0}^\infty\frac{(-1)^n2^{2n}x^{2n+2}}{(2n)!}$$Just differentiate the first term as you did and differentiate under the sum and you will have your formula.
 
When you say differentiate under the sum, do you mean differentiate each term in the sequence or do you mean you can actually take the derivative of the series equation? I tried googling and checking my textbook and I don't know how to take a derivative of a series in equation form, with the n's and such. If you can show me, that'd be great!
 
Yes, "differentiate under the sum" means "differentiate term by term". I don't know what you mean by "take the derivative of the equation"!
 
Magnawolf said:
When you say differentiate under the sum, do you mean differentiate each term in the sequence or do you mean you can actually take the derivative of the series equation? I tried googling and checking my textbook and I don't know how to take a derivative of a series in equation form, with the n's and such. If you can show me, that'd be great!

If you look at each term inside the sum, it just a constant times a power of ##x##. Use the power rule.
 
Oh okay, I got it now. Thanks for everything man!
 
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