Find 9th Derivative of f(x) = cos(6x^4)-1 at x=0 | Maclaurin Series Homework

demersal
Messages
41
Reaction score
0

Homework Statement




(1 pt) Compute the 9th derivative of:
f(x) = \frac{cos(6x^4)-1}{x^7}
at x=0.



Homework Equations



Hint: Use the MacLaurin series for f(x).

The Attempt at a Solution



I have tried many weird ways and cannot come up with the correct numerical answer. I've gotten 0 each time and it still comes up as wrong. Please help point me in the correct direction, even if it is only a verbal explanation!

Thanks so much
 
Physics news on Phys.org
The only term in the expansion that would contribute to a 9th derivative would be the term containing x^9. If you look up the Maclaurin series of cos you'll it can be written as a sum of even powers of it's argument. The fourth power of 6x^4 contains an x^16. 16-7=9. Is that enough of a hint.
 
Yes, thank you very much. I think I can work it out now!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
48
Views
5K
Replies
4
Views
3K
Replies
16
Views
3K
Replies
16
Views
2K
Replies
10
Views
2K
Replies
2
Views
5K
Back
Top