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Math Help - Implicit Derivative

  1. #1
    Member VitaX's Avatar
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    Implicit Derivative

    a. By differentiating x^2 - y^2 = 1 implicitly, show that \frac{dy}{dx} = \frac{x}{y}
    b. Then show that \frac{d^2y}{dx^2} = -\frac{1}{y^3}

    a. x^2 - y^2 = 1
    2x - 2yy' = 0
    2yy' = 2x
    y' = \frac{2x}{2y}
    y' = \frac{x}{y}

    b. y'' = \frac{(y)(1) - (x)(y')}{y^2}
    y'' = \frac{y - xy'}{y^2}
    y'' = \frac{y - x \left(\frac{x}{y}\right)}{y^2}
    y'' = \frac{\frac{y^2 - x^2}{y}}{y^2}
    y'' = \frac{y^2 -x^2}{y^3}

    There something wrong with my second derivative?
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  2. #2
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    skeeter's Avatar
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    Quote Originally Posted by VitaX View Post
    a. By differentiating x^2 - y^2 = 1 implicitly, show that \frac{dy}{dx} = \frac{x}{y}
    b. Then show that \frac{d^2y}{dx^2} = -\frac{1}{y^3}

    a. x^2 - y^2 = 1
    2x - 2yy' = 0
    2yy' = 2x
    y' = \frac{2x}{2y}
    y' = \frac{x}{y}

    b. y'' = \frac{(y)(1) - (x)(y')}{y^2}
    y'' = \frac{y - xy'}{y^2}
    y'' = \frac{y - x \left(\frac{x}{y}\right)}{y^2}
    y'' = \frac{\frac{y^2 - x^2}{y}}{y^2}
    y'' = \frac{\textcolor{red}{y^2 -x^2}}{y^3}

    There something wrong with my second derivative?

    nothing wrong ...

    if \textcolor{red}{x^2 - y^2 = 1} , then what is the value of \textcolor{red}{y^2 - x^2} ?
    ...
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  3. #3
    Member VitaX's Avatar
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    Eh I did not see that. Thanks
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  4. #4
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    I wrote y' = x(1/y) and used the product rule. I get

    y^{\prime\prime} = \frac{1}{y}-\frac{x}{y^2}\cdot y^\prime = \frac{1}{y}-\frac{x^2}{y^3}

    This is \frac{y^2-x^2}{y^3} = \frac{-(y^2-x^2)}{y^3} = -\frac{1}{y^3}
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