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Thread: find where the slope is defined..

  1. #1
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    find where the slope is defined..

    (x^2)y - x(y^2) = 4

    I came out with...

    y' = [(y^2) - 2xy] / [(x^2)-2xy)]


    The answer in the back isnt this...
    How do you do this?
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  2. #2
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    Quote Originally Posted by Morgan82 View Post
    (x^2)y - x(y^2) = 4

    I came out with...

    y' = [(y^2) - 2xy] / [(x^2)-2xy)]


    The answer in the back isnt this...
    How do you do this?
    $\displaystyle x^2y-xy^2=4$

    So we have to differentiate implicitly,

    $\displaystyle
    2x(y)+x^2y'-1\cdot y^2-x(2y)y'=0
    $

    And solve for y' to get

    $\displaystyle x^2y'-2xyy'=y^2-2xy$

    $\displaystyle y'(x^2-2xy)=y^2-2xy$

    $\displaystyle y'=\frac{y^2-2xy}{x^2-2xy}$

    The slope is defined as long as the denominator doesn't equal 0.

    It is not defined when x=0, it is not defined when $\displaystyle y=\frac{x}{2}$
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  3. #3
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    Quote Originally Posted by Morgan82 View Post
    (x^2)y - x(y^2) = 4

    I came out with...

    y' = [(y^2) - 2xy] / [(x^2)-2xy)]


    The answer in the back isnt this...
    How do you do this?
    I agree with you ... what "answer" was in the back of the book?

    $\displaystyle x^2y' + 2xy - 2xy y' - y^2 = 0$

    $\displaystyle x^2y' - 2xy y' = y^2 - 2xy$

    $\displaystyle y'(x^2 - 2xy) = y^2 - 2xy$

    $\displaystyle y' = \frac{y^2-2xy}{x^2-2xy}$
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