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anonymous
 4 years ago
find the partial derivatives with respect to x and y, if f(x,y) = \[\sum_{n=0}^{}\] (xy)^n
anonymous
 4 years ago
find the partial derivatives with respect to x and y, if f(x,y) = \[\sum_{n=0}^{}\] (xy)^n

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, wrt to one variable, you treat the other as a constant. So how would you differentiate (constant*x)^n?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(n)x^(n1)? but what about the sum?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@First question  Yep. @Second question  What's the upper limit? Also, what's the xy < 1 for? Sorry, I'm not quite understanding the question fully.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the upper limit is infinity. xy<1 is the condition that goes along with the question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah! Makes sense. So, basically, for x, it would be: \[ \displaystyle\sum_{n=0}^{\infty} nx^{n1}. \]At this point, I'll just wait for another response. I really have no idea myself. Sorry. :( All the best!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0patial derivative wrt to x \[\sum_{n=0}^{n=\infty}nx ^{n1}y ^{n}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and wrt to y \[\sum_{n=0}^{n=\infty}x ^{n}(ny ^{n1})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont think that is quite right. I dont get how you take a partial of a sum and how you would account for xy < 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok first tell me the partial derivative of xy^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just find out and tell me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0find 2 different partials, one with relation to x and one with relation to y
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