# Compute the 9th derivative

• May 16th 2010, 02:43 PM
s3a
Compute the 9th derivative
How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!
• May 16th 2010, 03:10 PM
Jester
Quote:

Originally Posted by s3a
How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!

Use the hint. The Taylor series for $\tan^{-1}x$ is

$
\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.
$

Subs in for $x \rightarrow \frac{x^3}{4}$. You really only need to consider powers up to $x^9$.
• May 16th 2010, 03:13 PM
skeeter
Quote:

Originally Posted by s3a
How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!

did you use the hint?

$f(x) = \arctan\left(\frac{x^3}{4}\right) = \frac{x^3}{4} - \frac{x^9}{4^3 \cdot 3} + \frac{x^{15}}{4^5 \cdot 5} - \frac{x^{21}}{4^7 \cdot 7} + ...$

$f^{(9)}(0) = -\frac{9!}{4^3 \cdot 3}$
• May 16th 2010, 03:49 PM
s3a
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 PM
skeeter
Quote:

Originally Posted by s3a
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.

$\frac{1}{1-r} = 1 + r + r^2 + r^3 + ...$

$\frac{1}{1-(-x^2)} = 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + ... = 1 - x^2 + x^4 - x^6 + ...$
• May 16th 2010, 09:35 PM
s3a
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 AM
drumist
So you're OK with this, correct?

$f(x) = \arctan \left(\frac{x^3}{4}\right) = \frac{x^3}{4} - \frac{x^9}{4^3 \cdot 3} + \frac{x^{15}}{4^5 \cdot 5} - \frac{x^{21}}{4^7 \cdot 7} + \cdots$

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

The first term $\frac{x^3}{4}$ will eventually become zero (beyond 3 derivatives).

The second term $-\frac{x^9}{4^3 \cdot 3}$ will become a constant. What constant does it become?

The remaining terms will turn into the form $C \cdot x^n$ where $C$ is a constant and $n$ is a positive integer. When you substitute $x=0$ 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 AM
s3a
OMG! Yes, I get it now, thanks!