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Math Help - Find dy/dx Need help again

  1. #1
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    Find dy/dx Need help again

    x^4+3x^2y^3-y^2=5

    Thanks!
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  2. #2
    Senior Member Pinkk's Avatar
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    Just follow the same thought process as the last problem you posted.
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  3. #3
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    Is this right?

    x^4+3x^2y^3-y^2=5
    (4x^3)+[(3x^2)(3y^2dy/dx)+(y^3)(6x)]-2ydy/dx=0
    (4x^3)+(9x^2y^2)dy/dx+(6xy^3)-(2y)dy/dx=0
    (4x^3)+(6xy^3)=(2y)dy/dx-(9x^2y^2)dy/dx
    (4x^3+6xy^3)/2y-9x^2y^2)=dy/dx

    Thanks for the help Pink!
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  4. #4
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    sorry i havnt figured out how to put a set of numbers over another set, so the / is the divide line.
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  5. #5
    Senior Member Pinkk's Avatar
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    Correct.

    To have a fraction like \frac{a}{b}, use \frac{a}{b} inside the [tex] tags.
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  6. #6
    MHF Contributor matheagle's Avatar
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    I use {a\over b} instead of frac
    This one can be solved by explicit differentiation.
    Let  w=x^2 and then solve the quadractic equation in w.
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