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I want to differentiate the Lambert W function (the inverse of $y = xe^x$), I didn't think it would be that difficult a problem but it's causing me some problems. I tried this method:

(1.) Implicitly differentiating $f(g(x)) = x$ and solving for $g'(x)$ yields $g'(x) = \frac{1}{f'(g(x))}$, so substituting $f = xe^x$ and $g = W(x)$ gives us $W'(x) = \frac{1}{x + e^{W(x)}}$.

I then got paranoid and tried a second method,

(2.) Implicitly differentiating $W(x)e^{W(x)} = x$ directly gives us $W'(x)e^{W(x)} + W(x)e^{W(x)}W'(x) = 1$, or $W'(x)(e^{W(x)} + x) = 1$. Solving for $W'$ gives us the exact same answer as (1.)

My issue arises from the fact that WolframAlpha tells me that $W'(x) = \frac{W(x)}{xW(x) + x}$, which is nothing like what I got. What is wrong with my method?

user3002473
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1 Answers1

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The two answers are in fact equivalent. Remember that $W(x)e^{W(x)} = x$.

Therefore, $\dfrac{1}{x+e^{W(x)}} = \dfrac{1}{x+e^{W(x)}} \cdot \dfrac{W(x)}{W(x)} = \dfrac{W(x)}{xW(x)+W(x)e^{W(x)}} = \dfrac{W(x)}{xW(x)+x}$.

JimmyK4542
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