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Second Differential

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data


    y(x) = exp (-([tex]\sqrt{ms}/2t[/tex]) x[tex]^{2}[/tex])

    Find the y"(x)

    3. The attempt at a solution

    y'(x) = (-([tex]\sqrt{ms}/2t[/tex]) 2x) exp (-([tex]\sqrt{ms}/2t[/tex]) x[tex]^{2}[/tex])

    y"(x) = (-([tex]\sqrt{ms}/t[/tex])) exp (-([tex]\sqrt{ms}/2t[/tex]) x[tex]^{2}[/tex]) + (([tex]ms/4t^{2}[/tex])4x[tex]^{2}[/tex]) exp (-([tex]\sqrt{ms}/2t[/tex]) x[tex]^{2}[/tex])


    This is correct? Sorry if it looks a bit messy... Thanks.
     
    Last edited: Aug 25, 2010
  2. jcsd
  3. Aug 25, 2010 #2
    Could you please try to clean up the equation of the problem a little bit? Just for clarification.
     
  4. Aug 25, 2010 #3
    I hope that makes it easier.
     
  5. Aug 25, 2010 #4

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    I can't see any that yet.

    I think Jorge's point is, contrary to what many students imagine, there is no virtue in complicated-looking expressions, in carrying through all unnecessary complication in a problem until, maybe at the end, it matters. You have, if y'(x) means dy/dx, a function of ( x2 multiplied by a constant). You don't need to know how the constant is made up of this, that and the other when you differentiate. So just call it a. Or you can call it -a. Then you can see what you are doing easier and make fewer mistakes.

    As you go through phys and math you will see all the time where where where. I.e. w = some function of, say, [au + sin2(bv)] where a and b each = some other jumble of constants stuff (sometimes quite complicated stuff, like 'where a is the solution of this horrible equation' - something you could never carry through with everything explicit all the time). At the end of a calculation you might need to unravel or put back what is in the a and b to see how, e.g. a physical behaviour depends on the parameters inside them.
     
    Last edited: Aug 25, 2010
  6. Aug 25, 2010 #5

    Mark44

    Staff: Mentor

    Excellent point. After all, [tex]\sqrt{ms}/2t[/tex] is just a constant as far as differentiation with respect to x is concerned.

    BTW, you (the OP) are trying to find the second derivative, not the second differential. Also, this is hardly a precalculus problem.
     
  7. Aug 26, 2010 #6
    Thank you for the advice.

    @Mark44: sorry about that.
     
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