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?
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.
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!
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, $$g'(x) = \frac{1}{\frac{x}{W(x)}(W(x)+1)}$$ $$= \frac{W(x)}{x(1+W(x))}$$
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
$$W'(x)=\frac{W(x)}{x+W(x)x}$$
$$W'(x)=\frac{W(x)}{x(1+W(x))}$$
