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Thread: 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 $\displaystyle x^3+y^3 = 126$ and find y''(x) @ (5 , 1)

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

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

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

    Plugged in dy/dx to the second derivative $\displaystyle \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:

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

    and so:

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

    thus:

    $\displaystyle \frac{d^2y}{dx^2}\left|_{(5,1)}=-1260$
    Last edited by MarkFL; Oct 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.

    $\displaystyle x^{3}+y^{3} = 126,$

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

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

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





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

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

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

    Put them together using the product rule.
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  6. #6
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: 2nd implicit derivative

    Interesting results!

    Direct extraction of y(x):

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

    and:

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

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

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

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