$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{1}{1+e^{-2x}}) = \frac{ (1+e^{-2x}) \frac{\mathrm{d} }{\mathrm{d} x}(1) - (1) \frac{\mathrm{d} }{\mathrm{d} x} (1+e^{-2x}) }{ (1+e^{-2x})^2 } = \frac{ 0 - (1) \frac{\mathrm{d} }{\mathrm{d} x} (1+e^{-2x}) }{ (1+e^{-2x})^2 } = \frac{ - (1) (e^{-2x}) (-2) }{ (1+e^{-2x})^2 } = \frac{ 2e^{-2x} }{ (1+e^{-2x})^2 }$
Now Put x=0
$\frac{ 2e^{-2(0)} }{ (1+e^{-2(0)})^2 } = \frac{ 2(1) }{ (1+(1))^2 } =\frac{ 2}{ (2)^2 } =\frac{1}{2}$
therefore answer is B