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TuringTest
 2 years ago
Let\[y=x^p(1+x)^q\]find\[y^{(p+q)}\]
TuringTest
 2 years ago
Let\[y=x^p(1+x)^q\]find\[y^{(p+q)}\]

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TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0that's the (p+q)th derivative by the way Sorry that I have to go soon, but I look forward to seeing your responses when I return.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1ln(y) = ln(x^p) + ln(1+x)^q y'/y = p/x + q/(1+x) hmm

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2i think the answer is just (p+q)!

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2since the highest power of x will be x^(p+q), and all the other powers of x will become 0 after P+q derivatives.

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1sounds plausible to me

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0hint:(Leibniz' theorem/ binomial stuff)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0at least in MIT's solution I'd like to see others

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1lol, id like to see mits solution :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I'll send it to you....

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.0im curious, what class is this problem from?

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.0Ahh, okay. Thank you. :)

Mr.Math
 2 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Leibniz_rule_(generalized_product_rule)
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