Finding the nth Derivative of Cosine Function

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SUMMARY

The discussion focuses on finding the nth derivative of the cosine function, specifically \(\frac{d^{2n}}{dx^{2n}}\cos x\) where \(n\) is a natural number. The series expansion for cosine, \(\cos x=\sum^{\infty}_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}\), is utilized to derive the solution. Participants suggest evaluating the derivatives for small values of \(n\) (0, 1, and 2) to identify a pattern, and emphasize the recursive relationship \(\frac{d^{n+4}}{d x^{n+4}} \cos(x) = \frac{d^n}{d x^n} \cos(x)\) as a key to solving the problem.

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Homework Statement


[tex]\frac{d^{2n}}{dx^{2n}}\cos x[/tex]

[tex]n \in N[/tex]


Homework Equations


[tex]\cos x=\sum^{\infty}_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}[/tex]


The Attempt at a Solution


[tex]\frac{d^{2n}}{dx^{2n}}x^{2n}=(2n)![/tex]

But k is different that [tex]n[/tex]. I don't have a clue how to solve that.
 
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Here's a recommendation: Check the derivative for certain small values of n and see if you can find a pattern. I recommend n=0, n=1, and n=2. Then just remember that

[tex]\frac{d^{n+4}}{d x^{n+4}} cos(x) = \frac{d^n}{d x^n} cos(x)[/tex]

And that should finish your problem.
 

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