Sigmoid 函数导数推导

\[ \begin{align} g(z) = & \frac{1}{1+e^{-z}} \\ g(z)' = & (\frac{1}{1+e^{-z}})' \\ = & \Big((1+e^{-z})^{-1}\Big)' \\ = & (-1) \cdot (1 + e^{-z})^{-2} \cdot (1 + e^{-z})' \\ = & (-1) \cdot (1 + e^{-z})^{-2} \cdot (e^{-z})' \\ = & (-1) \cdot (1 + e^{-z})^{-2} \cdot ((e^z)^{-1})' \\ = & (-1) \cdot (1 + e^{-z})^{-2} \cdot (-1) \cdot (e^z)^{-2} \cdot (e^z)' \\ = & (-1) \cdot (1 + e^{-z})^{-2} \cdot (-1) \cdot (e^z)^{-2} \cdot e^z \\ = & (-1) \cdot (1 + e^{-z})^{-2} \cdot (-1) \cdot (e^z)^{-1} \\ = & (1 + e^{-z})^{-2} \cdot (e^{-z}) \\ = & \frac{e^{-z}}{(1 + e^{-z})^{2}} \\ = & \frac{1}{1 + e^{-z}} \cdot \frac{e^{-z}}{1 + e^{-z}} \\ = & \frac{1}{1 + e^{-z}} \cdot (1 - \frac{1}{1 + e^{-z}}) \\ = & g(z) \cdot \Big(1 - g(z)\Big) \\ \end{align} \]