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Math Help - Implicit Differentation

  1. #1
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    Implicit Differentation

    I've been trying to figure it out, read quite a bit on it, but still struggling a little bit.

    2x^6+y^3-2x+2y=1062933

    and

    3125=(2y^4-4x^5)^5

    Here is what I worked out.

    2x^6+y^3-2x+2y=1062933
    12x^5+(3y^2)y'-2+2y'=0
    y'=\frac{-12x^5+2}{(3y^2)+2}

    I'm not too sure where to begin with the 2nd one. Also I'm pretty sure y' = dy/dx right?
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  2. #2
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    Quote Originally Posted by Crell View Post
    I've been trying to figure it out, read quite a bit on it, but still struggling a little bit.

    2x^6+y^3-2x+2y=1062933

    and

    3125=(2y^4-4x^5)^5

    Here is what I worked out.

    2x^6+y^3-2x+2y=1062933
    12x^5+(3y^2)y'-2+2y'=0
    y'=\frac{-12x^5+2}{(3y^2)+2}

    I'm not too sure where to begin with the 2nd one. Also I'm pretty sure y' = dy/dx right?
    right



    3125=(2y^4-4x^5)^5

    For this one you have to use the chain rule. Think of it like this:

    Let u=2y^4-4x^5

    0=\frac{d}{dx}(2y^4-4x^5)^5

    =\left(\frac{d}{du}u^5\right)\left(\frac{du}{dx}\r  ight)

    =(5u^4)(6y^3y'-20x^4)

    =5(2y^4-4x^5)^4(6y^3y'-20x^4)

    So now just solve for y'
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