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Derivatives, chain rule 2

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of y=xe-kx

    2. Relevant equations

    Chain rule

    3. The attempt at a solution

    y = xeu

    [itex]\frac{dy}{du}[/itex] = xeu+eu

    u = -kx

    [itex]\frac{du}{dx}[/itex] = -k


    [itex]\frac{dy}{dx}[/itex] = (xe-kx+e-kx)(-k)

    = e-kx(x+1)(-k)

    = e-kx(-kx-k)

    The answer is e-kx(-kx+1). What did I do wrong?
     
  2. jcsd
  3. Nov 9, 2013 #2

    LCKurtz

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    There is a subtle error here. The problem is that if ##u=-kx## then ##x=-\frac u k## and you have to include ##\frac{dx}{du} = -\frac 1 k## in your second term. But it's best to avoid this error completely by noting that you have a product and should use the product rule before the chain rule. Try$$
    \frac d {dx} xe^u = e^u\frac d {dx} x + x \frac d {dx}e^u$$ and use your method on that last term.
     
  4. Nov 9, 2013 #3
    I'm sorry, but I don't get what you mean.

    [itex]\frac{d}{dx}[/itex] xeu = eu[itex]\frac{d}{dx}[/itex]x+x[itex]\frac{d}{dx}[/itex]eu

    is still equal to eu + xeu.
     
  5. Nov 9, 2013 #4
    Okay, I get it. Thanks!
     
  6. Nov 9, 2013 #5

    LCKurtz

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    No it isn't. By your own steps in your original post you showed$$
    \frac d {dx} e^u = -ke^{-kx}$$and if you put that in you get the correct answer.
     
  7. Nov 9, 2013 #6
    I get it now. Thanks!
     
  8. Nov 9, 2013 #7
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