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Thread: Differentials

  1. #1
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    Differentials

    If $\displaystyle x^2y + yx^2 = 6$, then what does $\displaystyle \frac{d^2y}{dx^2}$ equal at the point $\displaystyle (1,3)$?


    **I have not attempted this solution because I do not understand how $\displaystyle \frac{d^2y}{dx^2}$ works.
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  2. #2
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    Quote Originally Posted by Cursed View Post
    If $\displaystyle x^2y + yx^2 = 6$, then what does $\displaystyle \frac{d^2y}{dx^2}$ equal at the point $\displaystyle (1,3)$?


    **I have not attempted this solution because I do not understand how $\displaystyle \frac{d^2y}{dx^2}$ works.
    I am going to assume this was supposed to be
    $\displaystyle x^2y + xy^2 = 6$

    Start with your first derivative:
    $\displaystyle 2xy + 2x \frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$

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

    Now take the second derivative:
    $\displaystyle \frac{d^2y}{dx^2} = -\frac{\left ( 2y + 2x\frac{dy}{dx} \right ) (2x + 2xy) + (2xy + y^2) \left ( 2 + 2y + 2x \frac{dy}{dx} \right ) }{(2x + 2xy)^2}$

    Now plug in your expression for the derivative and simplify.

    -Dan
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  3. #3
    Eater of Worlds
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    Well, we can differentiate implicitly and then do it again.

    $\displaystyle x^{2}y+x^{2}y=6$

    $\displaystyle 2(x^{2}y)=6$

    $\displaystyle x^{2}y=3$

    Product rule:

    $\displaystyle x^{2}y'+2xy=0$

    $\displaystyle y'=\frac{-2xy}{x^{2}}=\frac{-2y}{x}$

    Do it again using quotient rule:

    $\displaystyle y''=\frac{-2xy'+2y}{x^{2}}=\frac{2(y-xy')}{x^{2}}$

    But, note what y' is equal to from before. Sub that in:

    $\displaystyle y''=\frac{2(y-x(\frac{-2y}{x}))}{x^{2}}=\frac{6y}{x^{2}}$

    Plug in your values.
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  4. #4
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    Thank you both for the help. I understand it now.
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