Question

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Answer 1

@jim_thompson5910 @iambatman

Answer 2

Find the first few derivatives and observe a pattern to find nth derivative of g

Answer 3

I think I would write as a product for fun
\[g(x)=xe^{-x}\]
then differentiate

Answer 4

1st:e^-x-xe^-x
2nd:e^-x-2xe^-x
3rd:3e^-x-xe^-x

Answer 5

it has something to do with -1(x-n) right?

Answer 6

with an exponent

Answer 7

\[g^{(1)}(x)=1e^{-x}-xe^{-x} \\ g^{(2)}(x)=-2e^{-x}+xe^{-x}=(-1)(2e^{-x}-xe^{-x}) \\ g^{(3)}(x)=3e^{-x}-xe^{-x}\]

Answer 8

Do you notice it is alternating

Answer 9

also the part in front of e^{-x} is the number of the n-th derivative

Answer 10

like what i mean is :
the 1st derivative we have a 1 in front of e^{-x}
the 2nd derivative we have a 2 in front of e^{-x}
and so on...

Answer 11

so its e^(-nx)?

Answer 12

nothing is changing in the exponent area

Answer 13

ok

Answer 14

by the pattern above i can say the 4th derivative would be
\[g^{(4)}=(-1)(4e^{-x}-xe^{-x})\]

Answer 15

the 5-th derivative would be
\[g^{(5)}=5e^{-x}-xe^{-x}\]

Answer 16

so can you take a guess at what the n-th derivative would look like

Answer 17

not yet. it has something to do with even and odd functions though

Answer 18

\[g^{(1)}(x)=(-1)^{1+1}(1e^{-x}-xe^{-x}) \\ g^{(2)}(x)=(-1)^{2+1}(2e^{-x}-xe^{-x}) \\ g^{(3)}(x)=(-1)^{3+1}(3e^{-x}-xe^{-x}) \\ g^{(4)}(x)=(-1)^{4+1}(4e^{-x}-xe^{-x})\]

Answer 19

g^(n)(x)=(-1)^(n+1)(e^(-x)-xe^(-x))

Answer 20

what about the number in front of e^(-x)?

Answer 21

ne^(-x)

Answer 22

ok

Answer 23

thank you so much

Answer 24

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