Locating roots by iteration

The function f(x)=xexf(x) = x - e^{-x} has a real root which can be expressed in terms of the Lambert-W/Product Log Function:

f(x)=0    x=W(1)0.5671\displaystyle f(x) = 0 \implies x = W(1) \approx 0.5671 \dots

And since this root is a solution to the equation x=exx = e^{-x}, we can conclude that, at the point we're interested in, ex=W(1)0.5e^{-x} = W(1) \approx 0.5. The gradiant (i.e. derivative) of exe^{-x} at this point is...?
 
The function f(x)=xexf(x) = x - e^{-x} has a real root which can be expressed in terms of the Lambert-W/Product Log Function:

f(x)=0    x=W(1)0.5671\displaystyle f(x) = 0 \implies x = W(1) \approx 0.5671 \dots

And since this root is a solution to the equation x=exx = e^{-x}, we can conclude that, at the point we're interested in, ex=W(1)0.5e^{-x} = W(1) \approx 0.5. The gradiant (i.e. derivative) of exe^{-x} at this point is...?

I don't think I've learnt about that yet, but thank you

The gradient of f+ g is the gradient of f plus the gradient of g: (f+ g)'= f'+ g' where I am using ' to indicate the gradient of a function.

This is so embarrassing I can't believe I didn't realise that
 
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