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anonymous
 5 years ago
find the derivative of x^2 y^2+x^3+y=8 at (2,0)
anonymous
 5 years ago
find the derivative of x^2 y^2+x^3+y=8 at (2,0)

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.22xy^2+x^2(2yy')+3x^2+y'=0 can you solve this for y'

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2y'(2yx^2+1)=3x^22xy^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2y'=[3x^22xy^2]/[2yx^2+1]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks so now do I just plug in 2 for x and 0 for y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative \[2xy^2 + 2x^2y(y') + 3x^2 + y' = 0\] So rearrange giving: \[y'(1+ 2x^2y) = 2xy^2  3x^2\] as myinaya said
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