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Hard derivative

  1. Dec 12, 2013 #1
    The problem statement, all variables and given/known data

    ψ2 = A(2αx2- 1)e-αx2/2

    First, calculate dψ2/dx, using A for A, x for x, and a for α.

    Second, calculate d2ψ2/dx2.

    3. The attempt at a solution
    so I got the first derivative correct, it was

    A((4*a*x)*exp((-a*x^2)/2) +(2*a*x^2 -1)*(-a*x*exp((-a*x^2)/2)))

    but i can seem to calculate the second derivative correctly I'm getting
    A((4*a)*(exp((-ax^2)/2)) + (4*a*x)*(-a*x*exp((-ax^2)/2)) +(4*a*x)*(-a*x*exp((-ax^2)/2)) +(2*a*x^2 -1)*(a^2*x^2*exp((-ax^2)/2)))

    but this incorrect, am I missing something?
  2. jcsd
  3. Dec 12, 2013 #2


    Staff: Mentor

    This is just about impossible to read. I could take a guess at what you're trying to say, but I shouldn't have to. Take a look at this, especially #2: https://www.physicsforums.com/showthread.php?t=617567.
  4. Dec 12, 2013 #3
    My bad this is the second derivative
    [itex] A((4a)(exp(( - ax^2 ) / 2)) + (4ax) * ( - ax * exp(( - ax^2) /2)) + (4ax) * ( - a xexp(( - ax^2 ) / 2)) + (2ax - 1) (a^2x^2 *exp(( - ax^2 ) / 2))) [/itex]
  5. Dec 12, 2013 #4


    Staff: Mentor

    That's a lot better.

    Is this ψ2?

    Here's what you have, cleaned up a little more, using more LaTeX and fewer parentheses.
    $$A((4a)(e^{ - (a/2)x^2}) + (4ax) * ( - ax * e^{-(a/2)x^2}) + (4ax) * ( - a xe^{- (a/2)x^2}) + (2ax - 1) (a^2x^2 *e^{- (a/2)x^2}) $$

    This ought to be at least close to what you have.
    Last edited: Dec 12, 2013
  6. Dec 12, 2013 #5


    User Avatar
    Gold Member

    Take a look at your last term, and consider the Product Rule. You missed something.
  7. Dec 12, 2013 #6
    This is the first derivative, correct?


    If so, in your computation of the second derivative, you need to perform a product rule within a product rule.
  8. Dec 12, 2013 #7
    yes that's my first derivative
  9. Dec 12, 2013 #8
    So my first derivative was simplified to [itex]\frac{d}{dx}\psi_2(x)=A\left[4\alpha x e^{-\frac{1}{2}\alpha x^2} - (2\alpha x^2 - 1)\alpha x e^{-\frac{1}{2}\alpha x^2}\right] [/itex]

    and if i factor out the [itex] \alpha x \exp(-\frac{1}{2}\alpha x^2)[/itex]

    i get that

    [itex]\frac{d}{dx}\psi_2(x) = A(\alpha x \exp(-\frac{1}{2}\alpha x^2))(5-2ax^2) [/itex]

    so my second derivative should be
    [itex] A(((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2)) +(4ax)) [/itex]

    is this right?
    Last edited: Dec 12, 2013
  10. Dec 12, 2013 #9
    Looks good to me. Can you differentiate that now?
  11. Dec 12, 2013 #10
    so my second derivative should be
    [itex] A(((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2)) +(4ax)) [/itex]

    is this right or am i still missing a product rule?
  12. Dec 12, 2013 #11
    Still missing a bit. Try doing this in two steps. Let ##f(x) = \alpha x \exp(-\frac{1}{2}\alpha x^2)## and ##g(x) = 5-2ax^2##

    What is the derivative of ##\frac{d}{dx}\psi_2(x) = Af(x)g(x)##? (simple application of product rule)

    After that, calulate ##f'(x)## and ##g'(x)## and then plug everything into the the second derivative formula you got.

    I know this is a lot of tedious work but I hope you'll be able to see why you were leaving out the terms you did.
  13. Dec 12, 2013 #12
    Hmmmm i may be forgetting to multiply the [itex] F' (x) by G(x) [/itex] and vice versa

    Is the derivative closer to

    [itex] A((((aexp((-1/2)ax^2) + (-a^2x^2exp((-1/2)ax^2))(5-2ax^2) + (4ax)( \alpha x \exp(-\frac{1}{2}\alpha x^2))[/itex]
  14. Dec 13, 2013 #13
    That looks better. You were just forgetting to put in that very last term. One minor issue with a sign error, the derivative of ##5-2\alpha x^2## is ##-4\alpha x## so you need a minus sign in the one place.

    And there's a bunch of parentheses; I'll just assume that they line up correctly. Just double check them if you are submitting them for homework.
  15. Dec 13, 2013 #14
    Thanks for your help sir!
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