Derivative of sigmoid function

Sigmoid function is defined as $$\frac{1}{1+e^{-x}}$$ I tried to calculate the derivative and got $$\frac{e^{-x}}{(e^{-x}+1)^2}$$ Wolfram|Alpha however give me the same function but with exponents on $e$ changed of sign

Someone could explain this to me?

Multiply both numerator and denominator by $e^{2x}$ and you will get Wolfram|Alpha result.
Using the fact $$e^{-x}=\frac{1}{e^x}$$ we have that $$\frac{e^{-x}}{(1+e^{-x})^2}=\dfrac{\dfrac{1}{e^x}}{(1+\dfrac{1}{e^x})^2}=\dfrac{\dfrac{1}{e^x}}{(\dfrac{e^x+1}{e^x})^2}=$$ $$=\frac{\dfrac{1}{e^x}}{\dfrac{(e^x+1)^2}{e^x\cdot e^x}}=\frac{1}{\dfrac{(e^x+1)^2}{e^x}}=\frac{e^x}{(1+e^x)^2}$$
Use the formula: $\left(\frac{1}{f(x)}\right)'=-\frac{f'(x)}{f^2(x)}$ and we have:
$$\left(\frac{1}{1+e^{-x}}\right)'=\frac{-(1+e^{-x})'}{(1+e^{-x})^2}=\frac{-1'-(e^{-x})'}{(1+e^{-x})^2}=\frac{0-(-x)'(e^{-x})}{(1+e^{-x})^2}=\frac{-(-1)(e^{-x})}{(1+e^{-x})^2}=$$ $$=\frac{e^{-x}}{(1+e^{-x})^2}$$