# Any easy way to do this?

1. Nov 21, 2007

### chrono210

Compute the 9th derivative of cos(6x^4) - 1 / x^7 at x = 0.

Is there an easy way to do this that I'm not seeing? After taking a couple of derivatives, I realized how long this will take if I do it by taking more.

2. Nov 21, 2007

### Office_Shredder

Staff Emeritus
I would be a bit concerned by the 1/x^7, considering the derivative doesn't exist there, and the cos part doesn't offer any sort of cancellation you might be looking for.

My guess is you're supposed to say it's trivially not defined, or you wrote the wrong function

3. Nov 21, 2007

### chrono210

I should probably clarify: (cos(6x^4) - 1) / x^7 at x = 0.

The x^7 is the only thing in the denominator.

Last edited: Nov 21, 2007
4. Nov 21, 2007

### Xevarion

Expand the cosine using its Taylor series.

5. Nov 21, 2007

### rock.freak667

Try finding a general formula for the nth derivative n see if you get a pattern

6. Nov 21, 2007

### Dick

This is the only really practical suggestion. And it's not even that hard.

7. Nov 22, 2007

### Kreizhn

I'm not terribly sure if this will work in this case, but to extend the idea of expanding the cosine, we can exploit the definition of the Taylor series coefficients. This may be what Xevarion/Dick meant, but by finding the Taylor series of the entire function about a neighbourhood of zero then the coefficient of $x^9$ will be your solution. I know that this works with much simpler functions, but this one is a bit complicated so I might make an error.

You can get the series representation by expanding the cosine, and treating everything else like polynomials. That is, when you do the expansion, you should get something along the lines of

$$cos(6x^4) = 1-18x^8+54x^{16} -\ldots$$

Thus $$\frac{cos(6x^4) -1 }{x^7} = -18x + 54x^9-\ldots$$

Now since $$f(x) = \displaystyle \sum_{n=o}^\infty \frac{f^n(0)}{n!} x^n$$
then by equation coefficients, you can show that

$$f^9 (0) = 54\times 9!$$

Edit: This is the exact same answer you'll get if you just expand the cosine series, since the procedure is precisely the same; however, I find that this is a useful technique that can be used with much nastier functions, so just thought I'd throw in my two-cents

8. Nov 22, 2007

### GleefulNihilism

Aww, I wanted to be the one to sweep in with the breath-of-fresh-air answer. I agree.