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How do you find derivative of xe^(x^2)

  1. Apr 13, 2013 #1
    1. This may seem like a dumb question, but I don't really know the steps of how to find this derivative...

    f(x)= xex2

    2. The answer is supposed to be: e-x2*(1-2x2)

    I thought to do u-substitution.

    u= x2

    du= 2xdx

    1/2du = xdx

    (1/2)eudu

    (1/2)ueu-1

    (1/2)(x2)(ex2-1)

    hhhh =_= How is the real answer supposed to be: e-x2(1-2x2)???

    Thanks :)
     
  2. jcsd
  3. Apr 13, 2013 #2

    jedishrfu

    Staff: Mentor

    you need to apply the product rule y=uv dy/dx = du/dx * v + u * dv/dx

    and for the dv/dx you need to use the dy/dx = e^u du/dx

    There's also some confusion here: y = x * e ^ (x^2) or is it y = x * e ^ (-x^2)
     
    Last edited: Apr 13, 2013
  4. Apr 13, 2013 #3
    The product rule for this would be:

    (1)(ex2) + (x)(2xex2)

    = ex2 + 2x2ex2

    If I factor out ex2, I would get:

    ex2(1 + 2x2)

    Similar, but they have negatives! How is that? :?
     
  5. Apr 13, 2013 #4
    Is it possible that the problem was supposed to have [itex] e^{-x^2} [/itex]? That is the only explanation I can see.
     
  6. Apr 13, 2013 #5
    Hey, you were right! I copied it down wrong! #O_O

    It was xe-x2

    e-x2(1-2x2)
     
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