Question

Find a sixth-degree Maclaurin polynomial for the function \(g(x)=\frac{1}{(x^2+1)^\frac{5}{2}}\)

Answer 1

any easy way to do this, or do I just have to grind out 6 derivatives?

Answer 2

ii dont think there is any easy way.. u have to find six derivatives..

Answer 3

@nubeer after the second derivative it gets really nasty because the 2x comes out from the chain rule, and then I have to use the product rule...

Answer 4

well when i use to do those i had to normlaly find up to 4th derivative because teachers new it get nasty as move on to higher derivatives.

Answer 5

since it's maclaurin series though whenver the term in front has an x in it it becomes 0, which makes things easier but it's still a pain.

Answer 6

well simplification is easy but i think this kind of question normally would carry marks for showing the derivatives.

Answer 7

\(g'(0)=(0^2+1)^{-7/2}*\frac{-5}{2}*2*0\)

Answer 8

Yeah @nubeer these are practice problems for the AP test (free response section) so I am assuming that if I don't do it right I"ll get docked

Answer 9

lol maybe.. but it will certainly would mkae u good at taking derivatives ;)

Answer 10

the first term i think is 0 but later on i think when u start taking product rule u migh have terms which wont be turning into 0

Answer 11

yeah for g''(x) I have a term that isn't 0

Answer 12

calculating g'''(x) right now >.>

Answer 13

maybe fine 1/u^(5/2) and subin the innards afterwards

Answer 14

$235 fine should suffice .. itll crack :/

Answer 15

btw it's a "no calculator" problem -_-

Answer 16

thats why i keep a slide rule handy

Answer 17

LOL what are you talking about I've only heard those in tall tales

Answer 18

those old dusty journals of mathematics .... i know we are told they are just folk lore and childrens dreams .. but i feel deep down in my gut that there is some truth to them ;)

Answer 19

maclaurin is f^(n)(0) right?

Answer 20

yes.

Answer 21

i got one at a yard sale. looks great but i have no idea how to use it. uses logs somehow

Answer 22

on the bright side I just realized that u'''(x)=0, and that u'(0), and u''(0)=0. Unfortunately, I am still trying to find g'''(x)

Answer 23

I see a pattern in the derivative though

Answer 24

this is just a pain ....

Answer 25

yeah if I had this question on my AP test I would not be able to finish it in the entire time...

Answer 26

so far, letting k=-5/2
\[f=1\]
\[f'=0\]
\[f''=2k\]
\[f'''=0\]
\[f''''=12k(k-1)\\
\]

Answer 27

you think f''''' = 0?

Answer 28

the good news is its fitting so far :)
http://www.wolframalpha.com/input/?i=1-5x%5E2%2F2%2B35x%5E4%2F2%2C+y%3D1%2F%28x%5E2%2B1%29%5E%285%2F2%29

Answer 29

sorry I was in school and had to leave...