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Math Help - Implicit Differentiations

  1. #1
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    Implicit Differentiations

    I understand the concept behind this type of funciton, but am unable to simplify it into the for that my teacher wants:

    2x^2-3y^2 = 4, d^2y/dx^2 (second derivative)

    The answer is -8/9y^3.

    Also, how do I get my questions in the proper format? I notice mine are crude and basic compared to some here.

    Thanks for your help, and sorry to waste your time with such a simple question.
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  2. #2
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    Hello,

    I'll show you how to do. After that, it'd be nice if you tried to do some others
    Quote Originally Posted by Dack1000 View Post
    I understand the concept behind this type of funciton, but am unable to simplify it into the for that my teacher wants:

    2x^2-3y^2 = 4, d^2y/dx^2 (second derivative)

    The answer is -8/9y^3.

    Also, how do I get my questions in the proper format? I notice mine are crude and basic compared to some here.

    Thanks for your help, and sorry to waste your time with such a simple question.
    Well, your writing is very correct compared to some people
    To see how to make these, go to the Latex help section, or click on the images I'll write


    2x^2-3y^2=4

    The aim is to differentiate the whole equation with respect to x, noting that y is a function of x. This latter remark means that you will have to use the chain rule every time y appears.

    For more convenience, let y'=\frac{dy}{dx}

    (2x^2)'=2(x^2)'=2(2x)=4x

    (-3y^2)'=-3(y^2)'=-3 \left(2 \cdot y' \cdot y\right)=-6 \cdot y' \cdot y

    (4)'=0

    So after differentiating one time, we have :

    4x-6y'y=0 \implies 2x-3y'y=0

    Isolate y' :

    \boxed{y'=\frac{2x}{3y}}

    _____________________________________________

    Get back to 2x-3y'y=0 and do it the same way.

    (2x)'=2

    (-3y'y)'=-3y''y-3(y')^2, by using the product rule.

    So we have 2-3y''y-3(y')^2=0

    Isolate y''=\frac{d^2y}{dx^2} :

    y''=\frac{2-3(y')^2}{3y}


    Now substitute with the boxed equation :

    y''=\frac{2-3 \left(\tfrac{2x}{3y}\right)^2}{3y}=\frac{2-3 \left(\tfrac{4x^2}{9y^2}\right)}{3y}=\frac{6y^2-4x^2}{3y^2 \cdot y}=\frac{6y^2-4x^2}{3y^3}


    Now, use the very first equation 2x^2-3y^2=4. From this one, we get 2x^2=4+3y^2 \implies 4x^2=8+6y^2

    Substitute that in the latter expression for y'' and you'll find the result^^
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  3. #3
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    Ok... Thanks. I get it now. My teacher actually showed your way of doing it as well, but then said it was "too complicated" and shwed a different way, but I think your way is better.

    Thanks again.
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