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Math Help - y' and y'' using implicit differentiation

  1. #1
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    y' and y'' using implicit differentiation

    Hi guys, having some trouble on implicit differentiation. The problem is

    find y'' given x^2 - y^2 = 1

    I got the answer (y^2-x^2)/y^3
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  2. #2
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    Quote Originally Posted by sleepiiee View Post
    Hi guys, having some trouble on implicit differentiation. The problem is

    find y'' given x^2 - y^2 = 1

    I got the answer (y^2-x^2)/y^3
    Remember that you can treat y's just like the chain rule with x's. So:

    \frac{d}{dx}y^2=2y(y')

    and so on. Taking our first derivative:

    \frac{d}{dx}[y^2-x^2]=\frac{d}{dx}1

    2yy'-2x=0

    \frac{d}{dx}[2yy'-2x]=\frac{d}{dx}0

    2yy''+2(y')^2-2=0

    To get the answer you had posted, you just substitute from above:

    2yy'-2x=0

    y'=\frac{x}{y}

    and you get:

    2yy''+2(\frac{x}{y})^2-2=0

    which can be rearranged thusly:

    y''=\frac{y^2-x^2}{y^3}
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  3. #3
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    thanks alot
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