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Math Help - dy/dx

  1. #1
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    dy/dx

    I got a question:
    Find dy/dx of:

    (x^3)+(y^3)-(2x)=3

    Now what i got is:
    ((-3x^2)-2) / (3y^2)

    By doing this:
    (3x^2)+(3y^2)dy/dx-2=0

    what am i doing wrong?
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  2. #2
    Senior Member DivideBy0's Avatar
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    It looks right, I think you just messed up the negative signs at the end a bit.

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

    3x^2+3y^2\frac{dy}{dx}-2=0

    3y^2\frac{dy}{dx}=2-3x^2

    \frac{dy}{dx}=\frac{2-3x^2}{3y^2}
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  3. #3
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    and this one:

    ((x^2)/4) + ((y^2)/16) = 1

    i did this:
    ((8x-(x^2))/16) + ((32y-(y^2))/256 dy/dx

    Now from here am confused if its right?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by taurus View Post
    ((x^2)/4) + ((y^2)/16) = 1
    To derive this, you gotta use the Implicit Differentiation.

    \frac14x^2+\frac1{16}y^2=1

    Since we want \frac{dy}{dx}, we use the chain rule for y, this yields

    \frac12x+\frac18yy'=0

    Now just make y' the subject of the equation.
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  5. #5
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    that doesnt yield the correct answer? could you run through the steps pls
    Last edited by taurus; September 17th 2007 at 10:22 AM.
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