Question

When computing the nth derivative of atan(x^3/6), I've tried approaching the problem so that if atan(x) = 1/(1+x^2), then atan(x^3/6) = 1/(1+(x^3/6)^2). That's ok for the first few, but by the time you get to say, the ninth or tenth derivative, things get hairy in a hurry. Should I just be looking for the coefficient of the 9th term in the MacLaurin expansion of 1/(1+x^2)?

Answer 1

is this arctangent?

Answer 2

this statement
\[\arctan(x)=\frac{1}{1+x^2}\] cannot be right

Answer 3

that is the derivative

Answer 4

Ah yes, sorry that's what I meant to say. That I used that as the first derivative, so that \[\frac{ d }{ dx }\tan^{-1}(\frac{x^{6}}{3}) = (1+x^{2})^{-1}\]

Answer 5

you need the chain rule for this derivative

Answer 6

alternatively if what you are looking for is the mclaurin series
write the series for \(\arctan(x)\) and replace \(x\) by \(\frac{x^6}{3}\)

Answer 7

So, I have \[\frac{d}{dx}\tan^{-1}(\frac{x^{6}}{3}) = (1+(\frac{x^{6}}{3})^{2})^{-1}\] Where I think I'm freaking out is in taking that out to the 9th derivative to find \[f^{9}(0) \]

Answer 8

you are doing too much work and in any case that is not right

Answer 9

the derivative requires the chain rule

Answer 10

if you are looking for the maclaurin series
find the maclaurin series for \(\arctan(x)\) first
then substitute

Answer 11

The problem says to find the 9th derivative of \[f(x)=\tan^{-1}(\frac{x^{6}}{3}))\] so that at \[f^{9}(0)=?\]

Answer 12

yikes really? you need an expression for the ninth derivative? the second derivative is hard enough

Answer 13

I know, right? If I didn't know better, I'd swear he's singling me out. ;)

Answer 14

I thought there might be a way to generalize the power series and get to it that way.

Answer 15

i can't imagine how to do that sorry
finding \(f^{9}(0)\) may not be that hard though

Answer 16

Bummer. Yeah, I've seen typos in the homework before, but then again, many of the assigned problems have bordered on sadistic.

Answer 17

i am fairly sure it is zero

Answer 18

Yeah, I was, too, but Webwork didn't think so.

Answer 19

but the maclaurin series will have only terms with powers of 6

Answer 20

http://www.wolframalpha.com/input/?i=9th+derivative+arctan%28x^6%2F3%29

lolol

Answer 21

but for sure \(f^{9}(0)=0\)

Answer 22

Actually, I get \[(\frac{x^3}{6})-[\frac{(\frac{x^3}{6})^{3}}{3}]-...\]

Answer 23

Make that "+" on the end.

Answer 24

By the godlike Mathematica, it is \(f^9(0)=560\). Not sure how to get that though.

Answer 25

looks good to me

Answer 26

crap i had it backwards!
it is
\[\frac{x^3}{6}\] not \[\frac{x^6}{3}\] in which case there will be a degree 9 term

Answer 27

Thomas5267, I've seen that somewhere else, as well, but not with any steps...though. Satellite73, I feel your pain. I do that all the time. :)

Answer 28

So should I be taking the derivative of the second term?

Answer 29

i get the idea i think

Answer 30

find the maclaurin series for
\[\arctan(\frac{x^3}{6})\] out to three terms
the third term will have \(x^9\)

Answer 31

the coefficient will be
\[\frac{f^{9}(0)}{9!}\]

Answer 32

Ah, I think that's it! Working through it now. *Brutal*.

Answer 33

not that bad actually

Answer 34

I think this has something to do with this:
\[
\sum_{n=0}^\infty x^n=\frac{1}{1-x}
\]

Answer 35

cheat
the maclaurin series for arctangent is well known
use it

Answer 36

you can derive it using what @thomas5267 said but it is kind of annoying
note that the derivative is
\[\frac{1}{1+x^2}\] take \[\sum_{n=0}^\infty x^n=\frac{1}{1-x}\] replace \(x\) by \((-x^2)\) write the series
integrate term by term
too much work, look it up

Answer 37

Aren't we differentiating?

Answer 38

no

Answer 39

you find the maclaurin series for \[\frac{1}{1+x^2}\] then integrate term by term

Answer 40

MacLaurin for arctan is where I got the x^9 in the first place. Arctan = x-(x^3/3)+(x^5/5)...

Answer 41

in any case it is
\[x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...\]

Answer 42

replace \(x\) by \(\frac{x^3}{6}\)

Answer 43

you get \(x^9\) in the second term

Answer 44

if my arithmetic is correct you get
\[\frac{x^9}{648}\] which means
\[\frac{f^{9}(0)}{9!}=\frac{1}{648}\]

Answer 45

Weird. Mathematica returned -560.

Answer 46

oooh an check it out!!
\[\frac{9!}{648}=560\] just as mathematica said!!

Answer 47

right i forgot the minus sign

Answer 48

Good job!

Answer 49

that mathematica is very smart!!

Answer 50

thnx
trial an error strikes again

Answer 51

It is a software for mathematics, created by those guys behind Wolfram Alpha.

Answer 52

yeah i used to use maple, now since i don't do any math i just use wolfram alpha on line

Answer 53

Wow. Tough problem. So I want to be sure I understand here. You got the 648 from 3*216. The 9! is from it being the ninth term in the series, though the other powers don't appear. Is that right?

Answer 54

I was so close, but I don't think I was raising 6^3.

Answer 55

Thanks, both of you for your help!

Answer 56

You differentiate the second term of the series 9 times so you get 9!.
\[
\frac{d^9}{dx^9}x^9=9!
\]
Then the other terms with powers of x vanish because the question is asking the value of \(f^9\) at \(x=0\).