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First and second derivative of a parametric

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]x = t - e^{t}[/tex]

    [tex]y = t + e^{-t}[/tex]

    Find dy/dx and [tex]d^{2}y/dx.[/tex]


    2. Relevant equations
    Derivative equations.


    3. The attempt at a solution
    dy/dt = [tex]1 - e^{-t}[/tex]

    dx/dt = [tex]1 - e^{t}[/tex]

    The dy/dx I came up with is:
    dy/dx = [tex](1 - e^{-t}) / (1 - e^{t})[/tex]

    Second derivative I came up with is:
    [tex]d^{2}y/dx[/tex] =[tex] - (e^{t} - e^{2t})^{-2} (e^{t} - 2e^{2t})[/tex]
     
    Last edited: Sep 9, 2010
  2. jcsd
  3. Sep 8, 2010 #2

    Dick

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    Assuming you meant y=t+e^(-t) the first derivative looks ok. The second derivative doesn't. Mind showing us how you got it?
     
  4. Sep 9, 2010 #3
    I think I tried to do a chain rule but screwed up somewhere.
     
  5. Sep 9, 2010 #4

    Char. Limit

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    It's rather hard to find [itex]\frac{d^2 y}{dx}[/itex]. You should be looking for [itex]\frac{d^2 y}{d x^2}[/itex].
     
  6. Sep 10, 2010 #5
    Sorry, that's what I meant, the second derivative (just forgot the squared on the bottom).

    My main issue is I'm not sure how I was supposed to have found the second derivative. I tried to do a reverse chain rule but I am pretty sure I did it wrong.
     
  7. Sep 10, 2010 #6

    vela

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    It would help if you showed us your actual work. Just saying "I tried this and it didn't work" isn't very helpful.

    By the way, the first derivative simplifies quite a bit, down to dy/dx=-e-t.
     
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