Question

compute the 9th derivative of
f(x)=arctan(x^(3)/4) at x=0

Answer 1

arc tan is a trouble as variable factory while taking n'th derivative.

Answer 2

f1(x) = 1/(1+(x^3/4)^2) * 1/4 * 3x
f2(x) = 3x/4( 1 /(1+(x^3/4)^2)^2*-1*3x^2/4 + 3/4(1+(x^3/4)^2) =

Answer 3

looks it's a long way ... until we find a general pattern.

Answer 4

I tried at first spending so much time listing them out
and I could not find the pattern

Answer 5

I think finding pattern and writing the equation is the most difficult part for me

Answer 6

it's quite difficult .... it's going to be series.

Answer 7

yeah. you dont see it as well?

Answer 8

i on second derivative.

Answer 9

let's take help form wolfram.

Answer 10

http://www.wolframalpha.com/input/?i=d%5E9%2Fdx%5E9%28arctan%28x%5E%283%29%2F4%29%29

this is something terribly ugly ... i think we have to look for another method.

Answer 11

Okay..

Answer 12

couldn't you just do this since you want to evaluate at x=0

http://www.wolframalpha.com/input/?i=d^9%2Fdx^9%28arctan%28x^%283%29%2F4%29%29+at+x%3D0

Answer 13

aren't you glad we have software now so you don't have to do these by hand :|

Answer 14

isn't there another method ... it would take a week by hand.

Answer 15

probably but i don't know of it off top of my head

Answer 16

You could write arctan as a series expansion to get:\[\arctan(\frac{x^3}{4})=\frac{x^3}{4}-\frac{x^9}{192}+\frac{x^{15}}{5120}+...\]and then notice that the 9th derivative at x=0 will be zero for all terms except the term involving \(x^9\) which will end up as:\[-\frac{9!}{192}=-1890\]

Answer 17

this seems correct I think.

Answer 18

this was genius

Answer 19

yep. thank you for your help!! experimentx

Answer 20

Ah yes only up to third derivative
http://www.wolframalpha.com/input/?i=d%5E3%2Fdx%5E3%28arctan%28x%5E%283%29%2F4%29%29+at+x%3D0