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Mathematics 14 Online
OpenStudy (anonymous):

i need help finding derivatives

OpenStudy (anonymous):

\[f(x)=xe^{-9x}\]

OpenStudy (anonymous):

i need to find 3 derivatives

OpenStudy (loser66):

what do you mean by find 3 derivatives?

OpenStudy (anonymous):

f' f'' and f'''

OpenStudy (loser66):

so? go ahead, f'=?? product rule

OpenStudy (anonymous):

\[-9x*e^{-9x}\]

OpenStudy (anonymous):

i mean -9 e^-9x

OpenStudy (loser66):

\(f'(x) = x'*e^{-9x}+x (e^{-9x})' = e^{-9x} -9xe^{-9x}\)

OpenStudy (anonymous):

does that = \[e^{-9x}(x-9)\]

OpenStudy (anonymous):

F'=(1-9x)*e^-9x

OpenStudy (loser66):

check again, my answer is ok:)

OpenStudy (anonymous):

f'' = \[(x-1)(1-9x)e^{-9x}\]

OpenStudy (loser66):

Please, tell me whether you get f' yet? and DON'T factor, it turns very complicated next in next steps

OpenStudy (anonymous):

I understand now that (1-9x)e^{-9x}= f'

OpenStudy (loser66):

Don't factor, please.

OpenStudy (loser66):

\(f'(x) = x'*e^{-9x}+x (e^{-9x})' = e^{-9x} -9xe^{-9x}\) \(f''(x) = (e^{-9x} -9xe^{-9x})'\)

OpenStudy (anonymous):

this is why i am so confused.. the factoring part is too much

OpenStudy (loser66):

It's ok when factoring, but why do we make the life harder by product rule again for f" ? now, just take derivative term by term and the second term is exactly what we did for f' * (-9) ,right?

OpenStudy (loser66):

\(f''(x) = (e^{-9x})' -9(xe^{-9x})'\)

OpenStudy (loser66):

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