Differentiating and Integrating the Lambert W function

[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...

 

[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.

 

[lurflurf]

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))

 

[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
$$\frac{dW(x)}{dx}=\frac{W(x)}{x(W(x)+1)}$$

 

[pierce15]

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
$$\frac{dW(x)}{dx}=\frac{W(x)}{x(W(x)+1)}$$

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?

 

[chiro]

Have you tried using separation of variables?

 

[lurflurf]

^right very nice

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

 

[pierce15]

^right very nice

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

Sorry what did you do there?

 

[lurflurf]

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

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

 

[JJacquelin]

The LambertW integral :