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Thread: 2nd Derivative Implicit

  1. #1
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    2nd Derivative Implicit

    Find $\displaystyle \frac {d^2y}{dx^2} $

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

    I didn't get very far after differentiating with respect to x; here's my step 1): $\displaystyle 2\frac{dy}{dx}*e^{2y}\frac{dy}{dx} + 1 = \frac{dy}{dx} $.

    I have several questions. Can I multiply $\displaystyle \frac{dy}{dx} $ by another to get $\displaystyle \frac {d^2y}{dx^2} $?

    If not, I'll show the steps I went through with a brief explanation: I moved a dy/dx and the +1 to different sides and then factored out a dy/dx:
    2) $\displaystyle \frac{dy}{dx}(2e^{2y}-1) = -1 $

    Solved for dy/dx
    3) $\displaystyle \frac{dy}{dx}= \frac{-1}{2e^{2y}-1} $

    Differentiated again with respect to x using quotient rule / chain rule.
    4) $\displaystyle \frac {d^2y}{dx^2} = {\frac{(-2e^{2y}\frac{dy}{dx}*2\frac{dy}{dx})*-1}{(2e^{2y}-1)^2}$

    Simplified the top by distributing the negative one and multiplying the dy/dx terms (if that's legal).
    5) $\displaystyle \frac {d^2y}{dx^2} = {\frac{4e^{2y}\frac{dy}{dx}}{(2e^{2y}-1)^2}$

    Substituted dy/dx back into the game.
    6) $\displaystyle \frac {d^2y}{dx^2} = {\frac{4e^{2y}*\frac{-1}{2e^{2y}-1}}{(2e^{2y}-1)^2}$

    Multiplied lowest denominator across the middle denominator.
    7) $\displaystyle \frac {d^2y}{dx^2} = {\frac{4e^{2y}}{(2e^{2y}-1)^3}$
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    Re: 2nd Derivative Implicit

    $\displaystyle \frac{d}{dx} \left(e^{2y} + x = y \right)$

    $\displaystyle e^{2y} \cdot 2\frac{dy}{dx} + 1 = \frac{dy}{dx}$

    $\displaystyle 1 = \frac{dy}{dx} - e^{2y} \cdot 2\frac{dy}{dx}$

    $\displaystyle \frac{dy}{dx} = \frac{1}{1 - 2e^{2y}}$


    $\displaystyle \frac{d}{dx} \left[\frac{dy}{dx} = (1 - 2e^{2y})^{-1} }\right]$

    $\displaystyle \frac{d^2y}{dx^2} = -(1 - 2e^{2y})^{-2} \cdot (-4e^{2y}) \cdot \frac{dy}{dx}$

    $\displaystyle \frac{d^2y}{dx^2} = \frac{4e^{2y}}{(1 - 2e^{2y})^2} \cdot \frac{1}{1 - 2e^{2y}} = \frac{4e^{2y}}{(1 - 2e^{2y})^3}$

    I almost agree ...
    Thanks from AZach
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    Re: 2nd Derivative Implicit

    I didn't realize I had almost gotten that right. Thanks for showing me your work, Skeeter.
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    Re: 2nd Derivative Implicit

    I have another problem of a similar nature to add to this. $\displaystyle sin(x) + x^{2}y = 1 $

    First derivative with respect to x: 1) $\displaystyle cos(x) + 2xy+x^{2}*\frac{dy}{dx} = 0 $

    Solve for dy/dx: 2) $\displaystyle \frac{dy}{dx} = \frac{-cos(x)-2xy}{x^{2}} $

    Now things get hairy taking second d: 3) $\displaystyle \frac{(sin(x)-2y-2x\frac{dy}{dx})x^{2}-(-cos(x)-2xy)2x}{x^{4}} $

    Substituting dy/dx back into the game: 4) $\displaystyle \frac{(sin(x)-2y-2x(\frac{-cos(x)-2xy}{x^{2}}))x^{2}-(-cos(x)-2xy)2x}{x^{4}} $

    step 5), $\displaystyle \frac{(sin(x)-2y+(\frac{(2x)cos(x)-4x^{2}y}{x^{2}}))x^{2}+(2x)cos(x)-4x^{2}y}{x^{4}} $

    And then, $\displaystyle \frac{(sin(x)-2y+{(2x)cos(x)-4x^{2}y})x^{2}+(2x)cos(x)-4x^{2}y}{x^{6}} $

    $\displaystyle \frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)-4x^{4}y+(2x)cos(x)+4x^{2}y}{x^{6}} $


    I get stuck here now. I can't factor out the 2. $\displaystyle \frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)+(2x)cos(x)}{x^{6}} $
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    Re: 2nd Derivative Implicit

    Quote Originally Posted by skeeter View Post
    $\displaystyle \frac{d}{dx} \left(e^{2y} + x = y \right)$

    $\displaystyle e^{2y} \cdot 2\frac{dy}{dx} + 1 = \frac{dy}{dx}$

    $\displaystyle 1 = \frac{dy}{dx} - e^{2y} \cdot 2\frac{dy}{dx}$

    $\displaystyle \frac{dy}{dx} = \frac{1}{1 - 2e^{2y}}$
    Personally, I would not solve for dy/dx here. From
    $\displaystyle 2e^{2y}\frac{dy}{dx}+ 1= \frac{dy}{dx}$
    use "implicit differentiation" again
    $\displaystyle 4e^{2y}\frac{dy}{dx}+ 2e^{2y}\frac{d^2y}{dx^2}= \frac{d^2y}{dx^2}$
    Then $\displaystyle \frac{d^2y}{dx^2}= \frac{4e^{2y}\frac{dy}{dx}}{1- 2e^{2y}}$
    And, if you are required to write it in terms of x only, go back and and solve for $\displaystyle \frac{dy}{dx}$

    $\displaystyle \frac{d}{dx} \left[\frac{dy}{dx} = (1 - 2e^{2y})^{-1} }\right]$

    $\displaystyle \frac{d^2y}{dx^2} = -(1 - 2e^{2y})^{-2} \cdot (-4e^{2y}) \cdot \frac{dy}{dx}$

    $\displaystyle \frac{d^2y}{dx^2} = \frac{4e^{2y}}{(1 - 2e^{2y})^2} \cdot \frac{1}{1 - 2e^{2y}} = \frac{4e^{2y}}{(1 - 2e^{2y})^3}$

    I almost agree ...
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    Re: 2nd Derivative Implicit

    Quote Originally Posted by AZach View Post
    I have another problem of a similar nature to add to this. $\displaystyle sin(x) + x^{2}y = 1 $

    First derivative with respect to x: 1) $\displaystyle cos(x) + 2xy+x^{2}*\frac{dy}{dx} = 0 $
    Differentiating here, $\displaystyle -sin(x)+ 2y+ 2xy'+ 2xy'+ x^2+ x^2y''= 0$, y''= \frac{sin(x)- 2y- 4xy'}{x^2}[/tex]


    Solve for dy/dx: 2) $\displaystyle \frac{dy}{dx} = \frac{-cos(x)-2xy}{x^{2}} $

    Now things get hairy taking second d: 3) $\displaystyle \frac{(sin(x)-2y-2x\frac{dy}{dx})x^{2}-(-cos(x)-2xy)2x}{x^{4}} $

    Substituting dy/dx back into the game: 4) $\displaystyle \frac{(sin(x)-2y-2x(\frac{-cos(x)-2xy}{x^{2}}))x^{2}-(-cos(x)-2xy)2x}{x^{4}} $

    step 5), $\displaystyle \frac{(sin(x)-2y+(\frac{(2x)cos(x)-4x^{2}y}{x^{2}}))x^{2}+(2x)cos(x)-4x^{2}y}{x^{4}} $

    And then, $\displaystyle \frac{(sin(x)-2y+{(2x)cos(x)-4x^{2}y})x^{2}+(2x)cos(x)-4x^{2}y}{x^{6}} $

    $\displaystyle \frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)-4x^{4}y+(2x)cos(x)+4x^{2}y}{x^{6}} $


    I get stuck here now. I can't factor out the 2. $\displaystyle \frac{(x^{2})sin(x)-2yx^{2}+(2x^{3})cos(x)+(2x)cos(x)}{x^{6}} $
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    Re: 2nd Derivative Implicit

    Quote Originally Posted by HallsofIvy View Post
    Differentiating here, $\displaystyle -sin(x)+ 2y+ 2xy'+ 2xy'+ x^2+ x^2y''= 0$, $\displaystyle y''= \frac{sin(x)- 2y- 4xy'}{x^2}$
    I see you did the product rule, but I'm confused. You got $\displaystyle 2xy'+x^2 $ which I think is the result of differentiating $\displaystyle x^{2}y' $, but then there's also an $\displaystyle x^{2}y'' $ at the end of the left side of the expression. When you differentiate $\displaystyle \frac{dy}{dx} $ using the product rule does that term go to 1? Wait, is that single $\displaystyle x^2 $ term before solving for $\displaystyle y'' $ a typo? I don't see it moved over to the right hand side.

    Thanks!
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