Question

I need help in calculus. how would you go about getting the 1000th derivative of xe^-x

Answer 1

\[f(x)=xe ^{-x}\]

Answer 2

I would find the first few derivatives and see if I could spot a pattern amongst them

Answer 3

\[f'(x)=e ^{-x}-xe ^{-x}\]

Answer 4

\[f^{(1)}=e^{-x}(1-x) \\ f^{(2)}=e^{-x}(-1)+-e^{-x}(1-x)=-e^{-x}(1+1-x)=-e^{-x}(2-x)\]
find the third one and put it in this form (factored form)

Answer 5

I think three will be all you need to see the pattern
but you can do the 4th one if you are still not sure

Answer 6

Did I make a mistake in the third derivative I got:
\[f'''(x)=2e ^{x}-e ^{-x}+xe ^{-x}\]

Answer 7

oh wait I made a mistake in my second derivative which caused my third derivative to incorrect

Answer 8

you don't have to multiple it out then find the derivative

Answer 9

ok one second

Answer 10

\[f'''(x)=e ^{-x}(3-x)\]

Answer 11

beauty

Answer 12

awesome

Answer 13

\[f^{(1)}=e^{-x}(1-x) \\ f^{(2)}=-e^{-x}(2-x) \\ f^{(3)}=e^{-x}(3-x) \\ .... \\ f^{(1000)}=?\]

Answer 14

so the 1000th derivative just be \[= e ^{-x}(1000-x)\]

Answer 15

almost!

Answer 16

you have one more thing to attach to that

Answer 17

oh negative since its even

Answer 18

yep

Answer 19

ah I see

Answer 20

like if you did go ahead and find the 4th derivative
you would see something similar happen that happened in the 2nd derivative

Answer 21

\[f^{(4)}=(e^{-x})'(3-x)+(3-x)'(e^{-x}) \\ f^{(4)}=(-e^{-x})(3-x)+(0-1)(e^{-x}) \\ f^{(4)}=(-1)e^{-x}(3-x)+(-1)e^{-x} \\ f^{(4)}=-e^{-x}(3-x+1) \\ f^{(4)}=-e^{-x}(4-x)\]

Answer 22

it should be pretty easy to come up with the nth derivative

Answer 23

right so theres a pattern, if the derivative is a even number then there is a negative infront of it.

Answer 24

\[f^{(n)}=(-1)^{n+1}e^{-x}(n-x)\]

Answer 25

yup that makes sense, Thank you! :)

Answer 26

np