Given $f(x) = \frac{1}{(1+e^{−x})}$, prove $f'(x) = f(1−f)$

Question

Coming from a Comp Sci, rather than Mathematics, background I am comfortable with simple differentiation and integration, but this problem is beyond my understanding.

Could someone please walk me through the proof that, given $f(x) = \frac{1}{(1+e^{−x})}$, then $f'(x) = f(1−f)$.

I can't seem to map on $f(1-f)$ to the derivative I (and google) produce.

This is probably really obvious to most of you, but thanks in advance!

Answer 1

You can start checking who is $$f\left( {1 - f} \right) = {1 \over {\left( {1 + {e^{ - x}}} \right)}}\left( {1 - {1 \over {\left( {1 + {e^{ - x}}} \right)}}} \right) = {1 \over {\left( {1 + {e^{ - x}}} \right)}} - {1 \over {{{\left( {1 + {e^{ - x}}} \right)}^2}}} = {{1 + {e^{ - x}} - 1} \over {{{\left( {1 + {e^{ - x}}} \right)}^2}}} = {{{e^{ - x}}} \over {{{\left( {1 + {e^{ - x}}} \right)}^2}}}$$

and then, using the chain rule,

$$\eqalign{ & f\left( x \right) = {1 \over {\left( {1 + {e^{ - x}}} \right)}} = {\left( {1 + {e^{ - x}}} \right)^{ - 1}} \cr & f'\left( x \right) = - {\left( {1 + {e^{ - x}}} \right)^{ - 2}} - {e^{ - x}} \cr & f'\left( x \right) = {{{e^{ - x}}} \over {{{\left( {1 + {e^{ - x}}} \right)}^2}}} \cr} $$

Answer 2

Use $\left(\frac1v\right)'(x)=-\frac{v'(x)}{v(x)^2}$ and notice that $1-f(x)=\frac{\mathrm e^{-x}}{1+\mathrm e^{-x}}$. Then $$ f'(x)=-\frac{-\mathrm e^{-x}}{(1+\mathrm e^{-x})^2}=\frac{1}{1+\mathrm e^{-x}}\frac{\mathrm e^{-x}}{1+\mathrm e^{-x}}=f(x)(1-f(x)). $$

Answer 3

The right hand side is a simple computation. The left hand side is where needs some work:

Rewrite $f$ to make differentiation more intuitive, it's $$f(x)=(1+e^{-x})^{-1} $$ So use chain rule to differentiate the power first: $$f'(x) = \frac{d}{dx} \left( (1+e^{-x})^{-1} \right) = (-1)(1+e^{-x})^{-2} \cdot \frac{d}{dx} \left( 1+e^{-x} \right) = (*)$$ And now you are left with something easier to differentiate: $$ (*) = -(1+e^{-x})^{-2} ( -e^{-x} ) = \frac{e^{-x}}{(1+e^{-x})^2} $$

Answer 4

I think this is an ambiguous notation problem: $$f'(x) = (1+e^{-x})^{-2}*e^{-x} = \frac{e^{-x}}{(1+e^{-x})(1+e^{-x})} =\frac{1}{(1+e^{-x})}\frac{e^{-x}}{(1+e^{-x})} = f(x)*(1-f(x)). $$ So when they say $f(1-f)$ they dont mean f applied to $1-f$ they mean times.