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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
 one year ago
 one year 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
 one year ago
 one year ago

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myininayaBest ResponseYou've already chosen the best response.1
Do you know how to find the derivative?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Yeah, do we just find the derivative and plug in the points?
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
Yes find the derivative of both sides with respect to x And then replace x with 2 and y with 1
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
What does this part of the question mean? y=f(x), near x=2, y =1.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
it 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!
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
in practical value, you can ignore that statement and proceed as myininaya said
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Ahh, thank you satellite, makes more sense now.
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
But this point isn't even on the curve.....
 one year ago

baldymcgee6Best 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'?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Nevermind, got it! Thanks guys
 one year ago
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