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Using the chain rule I am led to believe that the following can be differentiated nicely using the chain rule:

$f(x)=\dfrac{1}{(1+e^{-x})}$

It has been 3 years since I have used calculus though. If someone could show me how it's done and what $u$ and $v$ are then I would really appreciate it.

Thanks very much.

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    guy Yes, I realised it must have been real after I wrote that comment and deleted it afterwards. I hope the answer I gave below helped.2011-04-07

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guy. You can calculate the derivative as follows:

Let $u = 1+e^{-x}$. Note that u'(x) = -e^{-x}. Then $f(x) = 1/u(x)$ and f'(x) = -\frac{1}{u^2(x)}\cdot u'(x) = -\frac{1}{(1+e^{-x})^2}\cdot(-e^{−x}) = \frac{e^{−x}}{(1+e^{-x})^2}. Note that f'(x) = \frac{e^{−x}}{(1+e^{-x})^2} = f(x)\frac{e^{−x}}{(1+e^{-x})} = f(x)\frac{1+e^{−x}-1}{(1+e^{-x})} = f(x)\Big(1-\frac{1}{(1+e^{-x})}\Big), so f'(x) = f(x)(1-f(x)). I hope that helps.

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    @vivid-colours Absolutely no problem.2011-04-07
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Let $u(z)=1+e^{-z}$. Then, $f(u)=1/u$

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    Hey picakhu, I made a mistake in my post. Please see edit.2011-04-07