Question

What is the fourth derivative of

Answer 1

with respect to x

Answer 2

\[\frac{1}{\sqrt{a}e^{\frac{1}{2a}x^2}}\]

Answer 3

ugh

Answer 4

\[\frac{1}{\sqrt{a}} e^{\frac{-1}{2a}}x^2 \] I meant this

Answer 5

Is that stuff in front all a coefficient?

Answer 6

If so, the fourth derivative would be 0.

Answer 7

yeah I'm supposed to take the fourth derivative of the laplace transform of this thing

Answer 8

\[x = \xi \]

Answer 9

OH, I hate me some Laplace.

Answer 10

I'm doing fourier transforms.. this is the guy that needs to be taken to the fourth derivative in terms of that xi thing which I replaced to x because it was a bit easier.

Answer 11

Bleh, I haven't worked with Fouriers yet. ;-; I'm sorry I can't help.

Answer 12

I'm sure @wio can helps.

Answer 13

\[i^4 \frac{d}{d /\xi^4}(\frac{1}{\sqrt{a}}e^{\frac{-1}{2a} \xi^2}\]

Answer 14

fourth derivative of this transform x.x

Answer 15

\[i^4 \frac{d^4}{d \xi^4}(\frac{1}{\sqrt{a}}e^{\frac{-1}{2a} \xi^2}\]

Answer 16

It's not fun. You just have to do it.

Answer 17

there now it's fixed.. I know the i^4 is just i^2 split so a negative and negative is a positive.

Answer 18

However, you might want to start out considering: \[
e^{f(x)}
\]first.

Answer 19

who do I treat as a constant? the a? and then take derivatives for the \[/\xi \]

Answer 20

maybe mathematica would work XD

Answer 21

We should treat \(a\) as a constant unless we have reason to believe otherwise. If we don't believe it is a constant, then we would just need to write \(da/d\xi\).

Answer 22

f[x_] := 1/Sqrt[a] e^{-1/(2 a) x^2} is my code but I have to replace the x with xi .. the thing is I don't kinow if mathematica would understand it

Answer 23

so...let me guess... product rule 4 times?

Answer 24

http://www.wolframalpha.com/input/?i=fourth+derivative+%28+sqrt%28a%29e%5E%28-x%5E2%2F%282a%29%29%29%5E%28-1%29