Question

Let F(x) =x^4e^x. Determine the nth derivative of f at x=0

Answer 1

taylor series?

Answer 2

no

Answer 3

\[f(x)=x^4e^x\]right?

Answer 4

\(\huge F(x) =x^4e^x\)
or
\(\huge F(x) =x^{4e^x}\)

???

Answer 5

Yes turingtest that is correct

Answer 6

really? the first one that agentx5 has?

Answer 7

because the second makes a lot more sense

Answer 8

Product rule! eh? :-)

Answer 9

It is the first one that agentx5 has

Answer 10

First d Second + Second d First

^_^ (this is the way I say it)

Answer 11

hmm just for the first part
\[ f'(x)=4x^3e^x+e^xx^4 \]
If we derive more and more the first term will become
\[ n! e^x \] right?

Answer 12

\[\large \frac{d}{dx}(x^4)(e^x) = (x^4)\frac{d}{dx}(e^x) + (e^x)\frac{d}{dx}(x^4)\]
\[\large f'(x) = (x^4)(e^x) + (e^x)(4x^3)\]

Answer 13

\[f'(x)=4x^3e^x+x^4e^x\]\[f''(x)=12x^2e^x+4x^3e^x+4x^3e^x+x^4e^x=12x^2e^x+8x^3e^x+x^4e^x\]\[f'''(x)=24xe^x+12x^2e^x+24x^2e^x+8x^3e^x+4x^3e^x+x^4e^x\]\[f^{(4)}(x)=24e^x+O(x)\]so only from the 4th derivative on do we get a non zero value for the nth derivative of f at 0

Answer 14

I wonder if we can apply\[(uv)'=\sum_{k=0}^n\binom nku^{(n-k)}v^{(k)}\]to get a more general formula...

Answer 15

You can write the first derivative also as:
\(e^x x^3 (x+4)\)

In which case the second is...
\(e^x x^2 (x^2+8 x+12)\)

Third is...
\(e^x x^2 (x^2+8 x+12)\)

Fourth is...
\(e^x x (x^3+12 x^2+36 x+24)\)

Fifth is...
\(e^x (x^4+16 x^3+72 x^2+96 x+24)\)

Sixth is...
\(e^x (x^4+20 x^3+120 x^2+240 x+120)\)

etc...

Answer 16

Oh for crying out loud... I wrote the second and third twice... err

Answer 17

But yeah @TuringTest , chain rule and product rule says your polynomial stuff doesn't go away... Right?

Answer 18

You keep cycling an x back into it, almost like recursion

Answer 19

go away? not sure what you mean...
you mean the O(x) part
yes it's only true for the fourth derivative, for the others it will come back into play

Answer 20

Yeah what's with the O(x) ?

Answer 21

It doesn't go to zero, does it?

Answer 22

Wait '0' or 'O'?

Answer 23

that means that they are "junk" terms
i.e. they vanish when you plug in x=0 because they all are being multiplied by x

Answer 24

Junk terms? :-D I wonder if I can do that on an exam :-3

Answer 25

but I didn't mean to imply that they can be ignored from that point on; they cannot and will need to be accounted for int the higher derivatives

Answer 26

"Sry, this section of the test is junk"
"Divergent by the Turing Junk Theorem" :-3

Answer 27

sure that notation is done a lot in calc proofs using the difference quotient
(proof of the power rule for example involves a bunch of junk terms that have an h in them, which dissappear as \(h\to0\) and are usually written as \(O(x)\) to save paper)

Answer 28

haha not quite that I don't think :P
I have a good like where an MIT prof uses it of I can find it...

Answer 29

Hmm after 4 derivatives the first term becomes \(n!e^x\) therefore a number, at x=0, after 5 derivates the first two terms become \(n!e^x\) and therefore a number for x=0, or are my derivatives wrong?

Answer 30

but yeh, it depends on more then just n.

Answer 31

i guess we use the method Turing suggested.

Answer 32

@agentx5 check out around 47:00 on here in the proof of the power rule to see how it is used properly

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/

Answer 33

...the Junk think I mean
@experimentX which method? the leibniz rule that I wrote up there? (I think that's what it's called at least....)

Answer 34

Err which one of the 4? (also: yikes!)

Answer 35

yes ...it's pretty useless to expand.

Answer 36

all those stupid 24's keep piling up. it's better this way.