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baldymcgee6 Group Title

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

  • 2 years ago
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  1. myininaya Group Title
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    Do you know how to find the derivative?

    • 2 years ago
  2. baldymcgee6 Group Title
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    Yeah, do we just find the derivative and plug in the points?

    • 2 years ago
  3. myininaya Group Title
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    Yes find the derivative of both sides with respect to x And then replace x with 2 and y with 1

    • 2 years ago
  4. baldymcgee6 Group Title
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    What does this part of the question mean? y=f(x), near x=2, y =1.

    • 2 years ago
  5. satellite73 Group Title
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    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!

    • 2 years ago
  6. satellite73 Group Title
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    in practical value, you can ignore that statement and proceed as myininaya said

    • 2 years ago
  7. baldymcgee6 Group Title
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    Ahh, thank you satellite, makes more sense now.

    • 2 years ago
  8. myininaya Group Title
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    But this point isn't even on the curve.....

    • 2 years ago
  9. baldymcgee6 Group Title
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    @myininaya \[2xy+x^2y'+2y^3+2x3y^2y' = 0\] I got this when I differentiated both sides, but how do I get y'?

    • 2 years ago
  10. baldymcgee6 Group Title
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    @satellite73 ?

    • 2 years ago
  11. baldymcgee6 Group Title
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    Nevermind, got it! Thanks guys

    • 2 years ago
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