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Differentiating and Integrating the Lambert W function

  1. Dec 30, 2012 #1
    Here was my thinking for differentiation (which, by the way, is wrong):

    By the definition of the function, the following equations are equal:


    By the chain rule and product rule:

    $$\frac{dW}{dx}( e^x+xe^x ) =1$$

    What is the error here? What is the correct way to differentiate it? Also, how would I integrate it?

    P.S. Why is the fraction bar in the third equation bolder than in the second equation? They are typesetted the same way...
  2. jcsd
  3. Dec 30, 2012 #2


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    Hey piercebeatz.

    Basically you use the chain rule: d/dx f(g(x)) = g'(x)f'(g(x)).

    Note that W(x) in this case is W(x*e^x) or W(g(x)) where g(x) = x*e^x.
  4. Dec 30, 2012 #3


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    you have found

    w'(e^x+x e^x)=1/(e^x+x e^x)

    you might want to start with

    w(x) e^w(x)=x

    to find

    w'(x)=1/(e^w(x)+w(x) e^w(x))
  5. Dec 31, 2012 #4
    The expression :
    $$\frac{dW}{dx}=(e^x+xe^x)^{-1}$$ is true, but ambiguous.

    Do not confuse :
  6. Dec 31, 2012 #5
    Looking back at the top, that was a pretty lousy use of the chain rule. I see what I should have done now.

    Do you have any idea how to integrate it?
  7. Dec 31, 2012 #6


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    Have you tried using separation of variables?
  8. Dec 31, 2012 #7


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    ^right very nice

    [itex]\int W \text{ dx}=\int W \text{ d}(We^W)=\int (W+1)W e^W\text{ dW}[/itex]
    Last edited: Dec 31, 2012
  9. Jan 1, 2013 #8
    Sorry what did you do there?
  10. Jan 1, 2013 #9


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    ^As chiro suggested separation of variables, which is equivalent to the substitution

  11. Jan 2, 2013 #10
    The LambertW integral :

    Attached Files:

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