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Mathematics 18 Online
OpenStudy (anonymous):

compute the derivative y=(e/e^x)^3

OpenStudy (anonymous):

\[y=(x/e^x)^3\]

OpenStudy (anonymous):

\[ y' = 2 \frac x {e^x} \frac d {dx}\left( x e^{-x} \right) \]

OpenStudy (anonymous):

Can you finish it now?

OpenStudy (anonymous):

kind of lost

OpenStudy (anonymous):

y= u^2 y'= 2 u u' Chain rule

OpenStudy (anonymous):

\[u= x e^{-x} \]

OpenStudy (anonymous):

\[ u'= e^{-x} - x e^{-x}\\ 2 u u'= 2 x e^{-x} \left( e^{-x} - x e^{-x} \right)= 2 x e^{-2x} - 2 x^2 e^{-2x} \]

OpenStudy (anonymous):

That is it.

OpenStudy (anonymous):

is both chain rule and quotant rule

OpenStudy (anonymous):

you can use product rule by writing \[ (x e^{-x})^2 \]

OpenStudy (anonymous):

ok eliassaab I had the first part right.Ok Andersfon12

OpenStudy (anonymous):

@hooverst did you get it?

OpenStudy (anonymous):

I have the final answer above.

OpenStudy (anonymous):

yes Sir I got it !!

OpenStudy (anonymous):

Did you understand it?

OpenStudy (anonymous):

yes SIr

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