How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!

Thanks in advance!

Printable View

- May 16th 2010, 02:43 PMs3aCompute the 9th derivative
How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!

Thanks in advance! - May 16th 2010, 03:10 PMJester
- May 16th 2010, 03:13 PMskeeter
- May 16th 2010, 03:49 PMs3a
For arctan(x) I get = sigma from 0 to inf of (-x^(2n)) which gives - x - x^3 /3 - x^5 /5 etc without the alternating. What did I do wrong so far? I did d/dx arctan(x) = 1/(1-(-x^2)) then integrate the sum.

- May 16th 2010, 03:56 PMskeeter
- May 16th 2010, 09:35 PMs3a
Ok I see how to get the summation of arctan(x^3 /4) but then I don't know what to do. Do I just write down the polynomials up to the x^9 power and do something there? What about f^(n) (a) x^n /n! formula ? What I was trying to do before my current attempt was to just do the summation of f^(n) (a) x^n /n! and to replace f^(n) with it's appropriate algebraic expression but it was tough to compute the second derivative and on.

Sorry for being stupid :( - May 17th 2010, 12:37 AMdrumist
So you're OK with this, correct?

Now, consider what happens when you take 9 derivatives of this expression.

The first term will eventually become zero (beyond 3 derivatives).

The second term will become a constant. What constant does it become?

The remaining terms will turn into the form where is a constant and is a positive integer. When you substitute into a term of this form, does it make sense that these terms will all become zero?

Therefore, there is only one nonzero term, which is the second term. - May 17th 2010, 09:03 AMs3a
OMG! Yes, I get it now, thanks!