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TechnoFictionBest ResponseYou've already chosen the best response.0
\[y=x ^{11}e ^{x}\] Much easier to look at.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
do you know the product rule
 one year ago

TechnoFictionBest ResponseYou've already chosen the best response.0
Yeah, I somehow worked out the problem by myself. Just want to make sure if I got it right or not.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
well tell me what you got :)
 one year ago

TechnoFictionBest ResponseYou've already chosen the best response.0
I look really ridiculous, but: \[11x ^{10}e ^{x}+x ^{11}e ^{x}\] Here you go.
 one year ago

TechnoFictionBest ResponseYou've already chosen the best response.0
hm....ok, if you say so. I just don't feel right somehow. It look too long and weird.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
It's right! \(u=x^{11} => u' = 11x^{10}\) \(v=e^{x} => v' = e^{x}\) \(=> 11x^{10}*e^{x}+ x^{11}e^{x}\)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
You can also write it like: \(e^{x} (11x^{10}+x^{11})\) if it looks too "weird" :)
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
You could also factor the \[x ^{10}\]and express the derivative as\[\frac{ dy }{ dx }=x ^{10}e ^{x}(11+x)\]
 one year ago
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