1000th derivative of a function.?

Question

How do i go about finding the 1000th derivative of a function: xe^-x

Answer 1

u must find a patern

first derivative: e^-x - xe^-x

second der. : -e^-x - e^-x + xe^-x

third: e^-x + e^-x + e^-x - xe^-x

forth: -e^-x - e^-x - e^-x - e^-x + xe^-x

u got the pattern ?

nth derivative: (-1)^(n+1) * (n)(e^-x) + (-1)^n * xe^-x

find the 1000th derivative

which is -1000(e^-x) + xe^-x

Answer 2

The first derivative of (Ax + B) e^(Cx) is, by the product rule,

(Ax + B) C e^(Cx) + A e^(Cx) = (A C x + (A + B C)) e^(Cx)

As you differentiate again and again, you can see that the exponent in the e^... part will always be Cx, so the form will be

(something x + something else) e^(Cx)

with the original triple A, B, C at n=0 becomes

A C, A + B C, C at n=1

(A C) C, (A C) + (A + B C) C, C at n=2
= A C^2, 2 A C + B C^2, C at n=2

(A C^2) C, (A C^2) + (2 A C + B C^2) C, C at n=3
= A C^3, 3 A C^2 + B C^3, C at n=3

You can prove using the principle of mathematical induction that the n-th derivative of (A x + B) e^(Cx) is:

( (A C^n) x + (n A C^(n-1) + B C^n) ) e^(Cx)

For the 1000-th derivative of x e^-x just plug in A = 1, B = 0, C = -1, and n = 1000 to get: (x - 1000) e^-x

Answer 3

h2 has the correct answer, now here is an explanation to answer your question:

From what i remember, and its been a long time, its the first term times the derivitive of the second, plus the second times the derivitive of the 1st.

1st
d(xe^-x) = -xe^-x + e^-x

2nd
d(-xe^-x + e^-x) = xe^-x - e^-x - e^-x = xe^-x -2e^-x

3rd
d(xe^-x -2e^-x) = -xe^-x + e^-x +2e^-x = -xe^-x + 3e^-x

And so the even derivitives have an odd multiplier on the e^-x term., and the even derivities have a positive xe^-x term, also the order of the derivitive is the same as the constant multiplier, and so by inspection the answer is:


xe^-x -1000e^-x

Answer 4

HYPOTHESIS: the n-th derivative is (x - n)e^(-x)

PROOF BY INDUCTION:
it is clearly true that the zeroeth derivative is xe^(-x) = (x - 0)e^(-x)

suppose the statement is true for a certain value of n
then the (n+1)-th derivative is the derivative of

(x - n)e^(-x)

which is

(x - n)e^(-x) - e^(-x) = (x - [n+1]) e^(-x)

proving that the statement is also true for n+1.

END OF PROOF

Therefore the answer is (x - 1000) e^(-x)

Answer 5

Don't fret over 1000th derivative

Just open your book and know Leibnitz theorem

Answer 6

plug in values of p and h take organic log of the two aspects use residences of logs to seperate the proper component with ln e ln e = a million so which you will subtract the ln fee w/o the ok and divide the ok fee and bam

Answer 7

differentiate it three times. You should see a pattern.

Answer 8

I'm sure you will get a lot of answeres.

Answer 9

- 1000 e ^ -x + x e ^ -x