# Does the derivative of the Lambert W function's identity still hold equal?

As the title states, I want to differentiate the identity of the Lambert W function. (I have a tendency to use brackets)

Identity:

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

If you don't know what the Lambert W function is, it is as follows:

$W(x)=y$ and $x=ye^y$, which, using substitution and dividing by $W(x)$, produces the above identity.

Since $W(x)=f^{-1}(x)$

when $f(x)=xe^x$,

we can find $W'(x)$.

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

which using the identity...

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

By then using chain rule, I attempted to differentiate each part of the identity separately.

$\frac{x}{W(x)}dx=\frac{1}{W(x)}-\frac{x}{W^2(x)}$

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

By the identity, these two solutions should be equivalent to each other.

$\frac{1}{W(x)}-\frac{x}{W^2(x)}=\frac{1}{[W(x)+1]}$

I finish this off by solving for $W(x)$.

$\frac{1}{W(x)}-\frac{x}{W^2(x)}=\frac{1}{[W(x)+1]}$

I multiply by the GCF.

$[W(x)+1][W(x)-x]=W^2(x)$

Multiply.

$W^2(x)+(-x+1)W(x)-x=W^2(x)$

Subtract $W^2(x)$ and add $x$

$(-x+1)W(x)=x$

Divide by $-x+1$.

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

Then I realized this was completely untrue and that apparently, the Lambert $W$ function can't be expressed in terms of elementary functions, according to Wikipedia.

I was, in fact, trying to prove that the Lambert $W$ function might be solvable in terms of elementary functions with this method, but it appears to fail me.

Did I do something wrong or is this just plain weird? If I did everything right, is this just a mystery of the Lambert $W$ function or can someone explain it to me. No sources that I have found on the internet have ever even attempted this sort of an approach, so I am unsure if everything I have posted here. I am confident in my calculus skills, but I could have made a mistake. If I did, point out my mistake and solve for $W(x)$.

Thank you.

P.S. If this becomes famous or revolutionary, I call the credit for thinking of this unless someone can prove otherwise. If someone else solves this, they can have they credit of solving it.

• I didn't read through your entire post, but I can already tell you you made a mistake, because $W(x)=\frac{x}{-x+1}$ is not true, and you can check that by proving that $x\neq \frac{x}{-x+1}\cdot e^{\frac{x}{-x+1}}$... – 5xum Oct 29 '15 at 0:33

$\frac{x}{W(x)}dx=\frac{1}{W(x)}-\frac{x}{W^2(x)}$
$$\frac{d}{dx} (\frac{x}{W(x)}) = \frac{W(x) - xW'(x)}{W^2(x)} = \frac{1}{W(x)} - \frac{W'(x) x}{W^2(x)} \neq \frac{1}{W(x)}-\frac{x}{W^2(x)}$$