Differentiating and Integrating the Lambert W function

In summary: L}(x)=\int_{-\infty}^x W(x)\,d\xi$$The expression : $$\frac{dW}{dx}=(e^x+xe^x)^{-1}$$ is true, but ambiguous.In summary, the incorrect differentiation is chain rule based and the incorrect integration is separation of variables.
  • #1
pierce15
315
2
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...
 
Physics news on Phys.org
  • #2
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
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))
 
  • #4
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
JJacquelin said:
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?
 
  • #6
Have you tried using separation of variables?
 
  • #7
^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:
  • #8
lurflurf said:
^right very nice

[itex]\int W \text{ dx}=\int W \text{ d}(We^W)=\int (W+1)W e^W\text{ dW}[/itex]

Sorry what did you do there?
 
  • #9
^As chiro suggested separation of variables, which is equivalent to the substitution

x=W(x)e^W(x)
 
  • #10
The LambertW integral :
 

Attachments

  • Wintegral.JPG
    Wintegral.JPG
    3.5 KB · Views: 375

Related to Differentiating and Integrating the Lambert W function

1. What is the Lambert W function and what does it do?

The Lambert W function, also known as the product logarithm function, is a special function that represents the inverse of the function f(x) = xe^x. In other words, it is the solution to the equation x = ye^y. The function is useful in solving equations that involve both an exponential and a polynomial term.

2. How is the Lambert W function different from other special functions?

The Lambert W function is unique in that it cannot be expressed in terms of elementary functions such as polynomials, trigonometric functions, or logarithms. It also has multiple branches, which means that for certain inputs, the function has multiple possible outputs.

3. What are some applications of the Lambert W function?

The Lambert W function has applications in various fields of science and mathematics, including physics, engineering, and finance. It can be used to solve equations involving exponential growth, population models, and heat transfer. It is also useful in finding the equilibrium points of certain dynamical systems.

4. How do you differentiate and integrate the Lambert W function?

The derivative of the Lambert W function can be calculated using the chain rule, while the integral can be evaluated using integration by parts. However, due to the complexity of the function, these calculations can be quite involved and are usually done using numerical methods.

5. Are there any real-world examples of the use of the Lambert W function?

Yes, the Lambert W function has been used in various real-world applications such as modeling electrical circuits, studying enzyme kinetics, and analyzing the behavior of certain physical systems. It has also been used in finance to model stock prices and in economics to study economic growth.

Similar threads

Replies
2
Views
1K
  • Calculus
Replies
27
Views
2K
Replies
19
Views
3K
  • Calculus
Replies
6
Views
2K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
20
Views
3K
Replies
5
Views
2K
Replies
3
Views
741
Back
Top