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Math Help - 2nd implicit derivative

  1. #1
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    2nd implicit derivative

    I was just wondering if somebody could check over my math if they have a second.

    Find x^3+y^3 = 126 and find y''(x) @ (5 , 1)

    Took d/dx 3x^2+3y^2(dy/dx) = 0

    Solved for dy/dx by subtracting and dividing (dy/dx) = (-3x^2)/(3y^2)

    Took  d^2y/dx^2 using quotient rule \frac {(-3x^2*6y(dy/dx)-3y^2(-6x))}{(3y^2)^2}

    Plugged in dy/dx to the second derivative \frac {(-3x^2*6y(-3x^2/3y^2)-3y^2(-6x))}{(3y^2)^2} At this point does the procedure look correct? I plug in x = 5 and y = 1 and get 1260.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: 2nd implicit derivative

    I would simplify the first derivative to:

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

    and so:

    \frac{d^2y}{dx^2}=-\frac{2xy^2-2x^2y\frac{dy}{dx}}{y^4}=-\frac{2xy^2+2x^4y^{-1}}{y^4}=-\frac{\frac{2x}{y}(x^3+y^3)}{y^4}

    thus:

    \frac{d^2y}{dx^2}\left|_{(5,1)}=-1260
    Last edited by MarkFL; October 7th 2012 at 03:09 PM. Reason: correct sign error
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    Re: 2nd implicit derivative

    I would have avoided the quotient rule altogether.

    x^{3}+y^{3} = 126,

    3x^{2} + 3y^{2}\frac{dy}{dx}=0, (so \frac{dy}{dx}= -\frac{x^{2}}{y^{2}}),

    6x + 6y\left(\frac{dy}{dx}\right)^{2} + 3y^{2}\frac{d^{2}y}{dx^{2}} = 0, etc.
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    Re: 2nd implicit derivative

    Quote Originally Posted by BobP View Post
    I would have avoided the quotient rule altogether.





    6x + 6y\left(\frac{dy}{dx}\right)^{2} + 3y^{2}\frac{d^{2}y}{dx^{2}} = 0, etc.
    Why did you square  \frac {dy}{dx} and then add another  3y^2\frac{d^2y}{dx^2}?
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    Re: 2nd implicit derivative

    I didn't square it, I just differentiated 3y^{2}\frac{dy}{dx} as a product.

    Differentiating 3y^{2} gets you 6y\frac{dy}{dx} and differentiating \frac{dy}{dx} gets you \frac{d^{\2}y}{dx^{2}}.

    Put them together using the product rule.
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  6. #6
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    Lightbulb Re: 2nd implicit derivative

    Interesting results!

    Direct extraction of y(x):

    y''(x)=-\frac{2 x^4}{\left(126-x^3\right)^{5/3}}-\frac{2 x}{\left(126-x^3\right)^{2/3}}

    and:

    y''(5) = -1260
    Last edited by MaxJasper; October 7th 2012 at 02:09 PM.
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    Re: 2nd implicit derivative

    Quote Originally Posted by BobP View Post
    I didn't square it, I just differentiated 3y^{2}\frac{dy}{dx} as a product.

    Differentiating 3y^{2} gets you 6y\frac{dy}{dx} and differentiating \frac{dy}{dx} gets you \frac{d^{\2}y}{dx^{2}}.

    Put them together using the product rule.
    Ah, I see. I'm still kinda figuring out implicits.
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