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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Given
    x^2 + 25y^2 = 100
    show that \frac{d^2y}{dx^2} = - \frac{4}{25y^3}

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  2. #2
    TD!
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    Differentiate and solve for y':

    <br />
x^2  + 25y^2  = 100 \Rightarrow 2x + 50yy' = 0 \Leftrightarrow y' =  - \frac{x}{{25y}}<br />

    Differentiate again and substitute y' with the previous expression:

    <br />
y'' = \frac{{xy' - y}}{{25y^2 }} = \frac{{x\left( { - \frac{x}{{25y}}} \right) - y}}{{25y^2 }} = \frac{{ - \frac{{x^2  + 25y^2 }}{{25y}}}}{{25y^2 }}<br />

    Now you can use the initial relation again in the numerator:

    <br />
y'' = \frac{{ - \frac{{x^2  + 25y^2 }}{{25y}}}}{{25y^2 }} = \frac{{ - \frac{{100}}{{25y}}}}{{25y^2 }} = \frac{{ - \frac{4}{y}}}{{25y^2 }} =  \boxed{- \frac{4}{{25y^3 }}}<br />
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    Given: .[1] x^2 + 25y^2 = 100, .show that: . \frac{d^2y}{dx^2} = - \frac{4}{25y^3}

    Differentiate: . 2x + 50y\left(\frac{dy}{dx}\right) \:=\:0\quad\Rightarrow\quad\frac{dy}{dx}\:=\:-\frac{1}{25}\,\frac{x}{y} [2]

    Differentiate again: . \frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{y - x(\frac{dy}{dx})}{y^2}\right]

    Substitute [2]: . \frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{y - x(-\frac{x}{25y})}{y^2}\right] \:=\:-\frac{1}{25}\left[\frac{y + \frac{x^2}{25y}}{y^2}\right]

    Multiply top and bottom by 25y\!:\;\;\frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{25y^2 + x^2}{25y^3}\right]

    From [1], the numerator equals 100: . \frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{100}{25y^3}\right]

    Therefore: . \boxed{\frac{d^2y}{dx^2} \:=\:-\frac{4}{25y^3}}

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