Here's the question you clicked on:
baldymcgee6
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
Do you know how to find the derivative?
Yeah, do we just find the derivative and plug in the points?
Yes find the derivative of both sides with respect to x And then replace x with 2 and y with 1
What does this part of the question mean? y=f(x), near x=2, y =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!
in practical value, you can ignore that statement and proceed as myininaya said
Ahh, thank you satellite, makes more sense now.
But this point isn't even on the curve.....
@myininaya \[2xy+x^2y'+2y^3+2x3y^2y' = 0\] I got this when I differentiated both sides, but how do I get y'?
Nevermind, got it! Thanks guys