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Homework Help: Derivative exponential problem

  1. Feb 25, 2012 #1


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    1. The problem statement, all variables and given/known data

    Compute the derivative of the following function


    2. Relevant equations

    Product Rule and Quotient rule

    3. The attempt at a solution

    My problem here is that I come up with two different answers when I use the quotient rule vs. the product rule.

    Trying it with the product rule

    f(x)= (1-2x)e^-x
    f'(x)= -2e^-x + e^-x(1-2x)
    f'(x)= e^-x(-2+(1-2x))
    [itex]f'(x)= e^{-x}(-1-2x)[/itex] or [itex]\frac{-1-2x}{e^x}[/itex]

    With the quotient rule

    f'(x)= (e^-x(1-2x) - (-2e^-x)) / [e^-x]^2
    f'(x)= (e^-x-2e^-x+2e^-x) / [e^-x]^2
    f'(x)= (e^-x(-2x+2)) / [e^-x]^2
    [itex]f'(x)= \frac{-2x+2}{e^{-x}}[/itex] or [itex]e^x(-2x+2))[/itex]

    As you can see, these are two different answers. I would think that I should have the same solution either way I do this, so what am I doing wrong?
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2
    I don't get either of your answers when I take the derivative.

    (1-2x)e^{-x} = e^{-x} -2xe^{-x} = -e^{-x} -2(product rule) = ??
  4. Feb 25, 2012 #3


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    Put the whole exponent in braces { } .

    [itex ]e^{-x}[/itex ] gives   [itex]e^{-x}[/itex]
  5. Feb 25, 2012 #4


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    Thanks SammyS. I cleaned up some of the above. Hopefully that helps.
  6. Feb 25, 2012 #5


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    Here's the problem. It should be a minus above, not a plus. (Remember, chain rule.)
    Whoa. If you want to use the quotient rule, then you need to rewrite f(x) as
    ... so there shouldn't be any e-x in the first step:
    [itex]f'(x) = \frac{e^x(1-2x) - (-2e^x)}{(e^x)^2}[/itex]
    = ...
  7. Feb 25, 2012 #6


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    Thanks eumyang. I'm actually seeing numerous errors that I've typed above, and I'm working on cleaning it up with Latex.

    Please stand by until I can clean up this mess. :smile:
  8. Feb 25, 2012 #7


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    Ahh, I didn't need to completely retype that. You are perfectly correct, eumyang, I was making those mistakes on both. I wasn't considering that the derivative of [itex]e^{-x}[/itex] is [itex]-e^{-x}[/itex]. And I wasn't playing with the signs right on the quotient rule.

    Thank you very much, I appreciate the help.
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