Derivative of Lambert W function.

Question

I'm trying to find the derivative of the Lambert W function which is defined such that:

$$W(x)e^{W(x)}=x$$

Through implicit differentiation I get:

$$W(x)e^{W(x)}W'(x)+W'(x)e^{W(x)}=1$$

$$W'(x)(W(x)e^{W(x)}+e^{W(x)})=1$$

And using $W(x)e^{W(x)}=x$ I get:

$$W'(x)=\frac{1}{x+e^{W(x)}}$$

However the answer should be:

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

Where did I go wrong?

Answer 1

Nowhere. But note that $$ \exp W(x) = \frac{W(x)\exp W(x)}{W(x)} = \frac{x}{W(x)}$$ Now use this in your expression to get the other expression.

Answer 2

You arrived correctly at

$$W'(x)=\frac{1}{x+e^{W(x)}}$$

Now, recalling that $x=W(x)e^{W(x)}$, then clearly $e^{W(x)}=x/W(x)$. Therefore,

$$\begin{align} W'(x)&=\frac{1}{x+e^{W(x)}}\\\\ &=\frac{1}{x+x/W(x)}\\\\ &=\frac{W(x)}{xW(x)+x}\\\\ &=\frac{W(x)}{x(W(x)+1)} \end{align}$$

as expected!

Answer 3

A bit late but I found another way to find the derivative.

First

$$W(ye^y)=y$$

so getting the derivative on both sides, we get

$$W'(ye^y)(e^y+ye^y)=1$$

$$W'(ye^y)=\frac{1}{e^y+ye^y}$$

multiply by y in the numerator and denominator

$$W'(ye^y)=\frac{y}{ye^y+y*ye^y}$$

let $x=ye^y$

$$W'(x)=\frac{y}{x+y*x}$$

and recall that $y=W(ye^y)$ which is W(x)

so

$$W'(x)=\frac{W(x)}{x+W(x)x}$$

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

Answer 4

Another way to find the derivative of the Lambert function:

1) recall that W(x) is the inverse of $f(x) = xe^x$.

2) recall that if g(x) is the inverse of f(x), then $$g'(x)= \frac{1}{f'(g(x))}$$

For $f(x) = xe^x$, $f'(x) = xe^x + e^x = e^x(x+1)$

So, if $f(x) = xe^x$, and $g(x) = W(x)$ is its inverse, then $$g'(x) = \frac{1}{e^{W(x)}(W(x)+1)}$$

3) recall that $$e^{W(x)} = \frac{x}{W(x)}$$

Therefore, $$g'(x) = \frac{1}{\frac{x}{W(x)}(W(x)+1)}$$ $$= \frac{W(x)}{x(1+W(x))}$$