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baldymcgee6
 4 years ago
The equation x^2y+2xy^3=8 defines y as a function of x, y=f(x), near x=2, y =1.
Find the slope of the curve x^2y+2xy^3=8 when x=2, y=1
baldymcgee6
 4 years ago
The equation x^2y+2xy^3=8 defines y as a function of x, y=f(x), near x=2, y =1. Find the slope of the curve x^2y+2xy^3=8 when x=2, y=1

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1Do you know how to find the derivative?

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, do we just find the derivative and plug in the points?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1Yes find the derivative of both sides with respect to x And then replace x with 2 and y with 1

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0What does this part of the question mean? y=f(x), near x=2, y =1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it means that the curve itself might not represent a function, because it does not pass the "vertical line test" but "locally" it is a function, that is, near the point it does represent a function hello @myininaya!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in practical value, you can ignore that statement and proceed as myininaya said

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0Ahh, thank you satellite, makes more sense now.

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1But this point isn't even on the curve.....

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0@myininaya \[2xy+x^2y'+2y^3+2x3y^2y' = 0\] I got this when I differentiated both sides, but how do I get y'?

baldymcgee6
 4 years ago
Best ResponseYou've already chosen the best response.0Nevermind, got it! Thanks guys
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