# Differentiating and Integrating the Lambert W function

1. Dec 30, 2012

### pierce15

Here was my thinking for differentiation (which, by the way, is wrong):

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

$$W(xe^x)=x$$

By the chain rule and product rule:

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

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. Dec 30, 2012

### chiro

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.

3. Dec 30, 2012

### lurflurf

you have found

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

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

to find

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

4. Dec 31, 2012

### JJacquelin

The expression :
$$\frac{dW}{dx}=(e^x+xe^x)^{-1}$$ is true, but ambiguous.

Do not confuse :
$$\frac{dW(xe^x)}{dx}=(e^x+xe^x)^{-1}$$
and
$$\frac{dW(x)}{dx}=\frac{W(x)}{x(W(x)+1)}$$

5. Dec 31, 2012

### pierce15

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?

6. Dec 31, 2012

### chiro

Have you tried using separation of variables?

7. Dec 31, 2012

### lurflurf

^right very nice

$\int W \text{ dx}=\int W \text{ d}(We^W)=\int (W+1)W e^W\text{ dW}$

Last edited: Dec 31, 2012
8. Jan 1, 2013

### pierce15

Sorry what did you do there?

9. Jan 1, 2013

### lurflurf

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

x=W(x)e^W(x)

10. Jan 2, 2013

### JJacquelin

The LambertW integral :

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