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Math Help - Find the second derivative of this ellipse

  1. #1
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    Question Find the second derivative of this ellipse

    Find the second derivative of the ellipse
     \frac{x^2}{a^2} + \frac{y^2}{b^2}=1

    I know how to do the first derivative
    it would be

     \frac{2x}{a^2} + \frac{2y}{b^2}(\frac{dy}{dx}) = 0

    but what comes next if I want to find the second derivative?

    could someone show me full working out?

    thanks in advance
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  2. #2
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    Quote Originally Posted by differentiate View Post
    Find the second derivative of the ellipse
     \frac{x^2}{a^2} + \frac{y^2}{b^2}=1

    I know how to do the first derivative
    it would be

     \frac{2x}{a^2} + \frac{2y}{b^2}(\frac{dy}{dx}) = 0

    but what comes next if I want to find the second derivative?

    could someone show me full working out?

    thanks in advance
    Solve for \frac{dy}{dx}.

    I think you can see that

    \frac{2y}{b^2}\,\frac{dy}{dx} = -\frac{2x}{a^2}

    \frac{dy}{dx} = -\frac{b^2x}{a^2y}.


    So \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{b^2x}{a^2y}\right)

     = -\frac{d}{dx}\left(\frac{b^2x}{a^2y}\right)

     = -\frac{a^2b^2y - a^2b^2x\frac{dy}{dx}}{a^4y^2}

     = \frac{a^2b^2x\frac{dy}{dx} - a^2b^2y}{a^4y^2}

     = \frac{b^2x\frac{dy}{dx} - b^2y}{a^2y^2}.


    Now substitute \frac{dy}{dx}.
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