Question

If f(x) is 1/(1 + sin (x^2)) how do you find d^6f/dx^6?

Answer 1

?

Answer 2

6th derivative?
Hmm I guess you start by finding first derivative,
then second,
and then look for a pattern.

If you can't find one, you'll have to take 6 full derivatives D:
That won't be fun.

Answer 3

Do you understand how to find f'(x) ?

Answer 4

i did try to take the 6 derivatives but it's awfully ugly and it takes a lot of time and since this a MC question in an exam, i think there should be a quicker way to do it?

Answer 5

Hmm :d

Answer 6

is there perhaps anything that when you integrate, you get 1/(1+sin(x^2)) or something of the sort?

Answer 7

It is really ugly!!

Answer 8

i've been trying to expand it as a power series too for a long time and in different ways as well but it doesn't seen to be correct

Answer 9

there's got to be some easy way to do it right? an MC question isn't supposed to be so ugly???

Answer 10

Pfft, pansies. It doesn't look like it'd take that long:
http://www.wolframalpha.com/input/?i=d%5E6y%2Fdx%5E6+for+y%3D1%2F(1+%2B+sin+(x%5E2))

Answer 11

@agent0smith i hope you are joking about it not taking too much time

Answer 12

and according to the solutions, the answer is supposed to be -600 but there's no damn explanation

Answer 13

He certainly WAS joking :D hence the link lol

-600? Then the 6th derivative must've been evaluated at some x...
Hmm

Answer 14

Post the full question, screenshot.

Answer 15

:D alright!

it must have which is why i've been trying to expand it but nothing is working at all. it's not even making sense and my teacher is not even responding to my emails >.<

Answer 16

first, i apologise for my crappy handwriting and the highlighter that bled through. i don't know if this actually helps but i tried to expand as a power series using these. although it didn't work for me, it might work for you and you might be able to show me the way?

Answer 17

is there no one at all who can help me?

Answer 18

Ooo ok I've got something for you!

Answer 19

I'm a little rusty on my taylor series,
so I can't exactly tell you how to generate this expansion,
https://www.wolframalpha.com/input/?i=power+series+of+1%2F(1%2Bsin(x%5E2))
I would need a few minutes to figure that out.

But anyway, click the "more terms" button to expand it out a little ways.
Notice that if you differentiate that series 6 times...
You'll be left with the x^6 term and everything after it.

But... when we evaluate this at x=0,
we'll only be left with the x^6 term because everything else zeros out!
So then...\[\large\rm \frac{d}{dx}\left(-\frac{5x^6}{6}\right)=\frac{-5\cdot6!}{6}=-600\]

Answer 20

Hopefully that can get you on the right track.
If you need help with the Taylor Expansion lemme know.

Answer 21

OH MY GOD! THANK YOU SO MUCH!!! I'll try this right away!!!

Answer 22

cool \c:/

Answer 23

@zepdrix ok, would you be able to explain how you go from the sin to the taylor series, please?

Answer 24

The sine is stuck in the denominator, making things tricky for us...
I wonder if we can do something like this,\[\large\rm \frac{1}{1+\sin x^2}\left(\frac{1-\sin x^2}{1-\sin x^2}\right)\quad=\frac{1-\sin x^2}{\cos^2 x^2}\]And then we can split it into two fractions,\[\large\rm =\sec^2x^2-\sec x^2 \tan x^2\]Hmm no.. that doesn't seem much better... hmmm

Answer 25

I'm not 100% sure, but this may be one way to do it
\[\Large \frac{1}{1-x} = 1+x+x^2+x^3+\cdots\]
\[\Large \frac{1}{1-(x)} = 1+x+x^2+x^3+\cdots\]
\[\Large \frac{1}{1-({\color{red}{x}})} = 1+({\color{red}{x}})+({\color{red}{x}})^2+({\color{red}{x}})^3+\cdots\]
\[\Large \frac{1}{1-({\color{red}{-x}})} = 1+({\color{red}{-x}})+({\color{red}{-x}})^2+({\color{red}{-x}})^3+\cdots\]
\[\Large \frac{1}{1+x} = 1-x+x^2-x^3+\cdots\]



\[\Large \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots\]
\[\Large \sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \frac{(x^2)^7}{7!}+\cdots\]
\[\Large \sin(x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \frac{x^{14}}{5040}+\cdots\]


Then you'll replace each 'x' in
\[\Large \frac{1}{1+x} = 1-x+x^2-x^3+\cdots\]
with sin(x^2), then replace each "sin(x^2)" on the right side with its corresponding power series expansion. As you can see, things get very messy. This is especially true when you get to x^2 and -x^3

Answer 26

okay, so when i tried to expand it again, i got the first 3 terms alright but for the term in x^6 the coefficient is not 5/6 =/
what i did was as in the attachment... >.< where did i go wrong?

Answer 27

@jim_thompson5910 wait, i'll give your method a go!

Answer 28

@jim_thompson5910 hmm... i basically did the same thing but i probably missed a term?

Answer 29

\[\large (1+x)^n = 1+nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3+\cdots\]
\[\large (1+x)^{-1} = 1+(-1)x + \frac{-1(-1-1)}{2}x^2 + \frac{-1(-1-1)(-1-2)}{6}x^3+\cdots\]
\[\large (1+x)^{-1} = 1-x + \frac{2}{2}x^2 + \frac{-6}{6}x^3+\cdots\]
\[\large (1+x)^{-1} = 1-x + x^2 -x^3+\cdots\]

Answer 30

based on what I see in your scratch work, it looks like you got the same steps

Answer 31

OKAY!!!! I GOT IT!!!!! Thank you guys so very much, especially @jim_thompson5910 and @zepdrix

I had forgotten the term in t^3 while expanding 1/(1+t) and when I added it, I did get the coefficient for x^6 correctly, i.e. -5/6. Thanks a LOT!!!

Answer 32

Refer to the attachment from Mathematica 9.

Answer 33

I still think this doesn't look all that hard to derive ;P

Answer 34

It's only about 27 terms! Walk in the park.

Answer 35

@agent0smith yeah, it's a piece of cake! way too easy, you see. i wanted something more complicated to show off so the teacher could give me extra marks in the exam! XD

Answer 36

Refer to the Mathematica attachment.