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  1. #1
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    What is this called?

    2y^2-2x^2+0=0

    evaluate y' at given point (-2,-2)

    What is this called?

    How would I do it?
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  2. #2
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    Quote Originally Posted by Zanderist View Post
    2y^2-2x^2+0=0

    evaluate y' at given point (-2,-2)

    What is this called?

    How would I do it?
    Go to your notes or textbook and look for the heading Implicit Differentiation.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The relation y^{2} - x^{2}=0 defines in implicit form a function y= f(x) and its derivative in a point (y_{0},x_{0}) can be obtained from a well known explicit formula...

    Kind regards

    \chi \sigma
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  4. #4
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    So basically all I have to do is isolate y, then find the derivative?
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  5. #5
    MHF Contributor chisigma's Avatar
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    If You have a function y(x) implicity defined as...

    f(x,y)=0 (1)

    ... its derivative in a point (x_{0},y_{0}) is...

    y^{'} (x_{0}) = - \frac{f^{'}_{x} (x_{0},y_{0})}{f^{'}_{y} (x_{0},y_{0})} (2)

    In Your case is f(x,y)= y^{2} - x^{2}, x_{0}=-2, y_{0}=-2, so that is y^{'} (-1) = 1. A more simple solution for this particular case is to write (1) as y^{2}=x^{2} so that y(x)= \pm x. The 'right' y(x) passes from the point (-2,-2) so that is y(x)=x and y^{'} (x)=1...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by Zanderist View Post
    2y^2-2x^2+0=0

    evaluate y' at given point (-2,-2)

    What is this called?

    How would I do it?
    Implicit diferentiation gives you 4y \frac{dy}{dx} - 4x = 0. Substitute (-2, -2) and solve for dy/dx.
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  7. #7
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    I've manged to get the answer 1.

    But that was only a lucky guess steming from the research I did on this website.

    http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html
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  8. #8
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    Quote Originally Posted by Zanderist View Post
    I've manged to get the answer 1. [snip]
    Correct.
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