• Prove that the derivative of the sigmoid function with respect to the input \(z\) is:

\[\frac{d}{d z} \sigma(z)=\sigma(z)(1-\sigma(z))\]

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• Recall that the sigmoid function is given by:

\[\sigma(z)=\frac{1}{1+e^{-z}}\]

• Here’s how we obtain its derivative:

\[\begin{aligned} \frac{d}{d z} \sigma(z) &=\frac{d}{d z}\left[\frac{1}{1+e^{-z}}\right] \\ &=\frac{d}{d z}\left(1+\mathrm{e}^{-z}\right)^{-1} \\ &=-\left(1+e^{-z}\right)^{-2}\left(-e^{-z}\right) \\ &=\frac{e^{-z}}{\left(1+e^{-z}\right)^{2}} \\ &=\frac{1}{1+e^{-z}} \cdot \frac{e^{-z}}{1+e^{-z}} \\ &=\frac{1}{1+e^{-z}} \cdot \frac{\left(1+e^{-z}\right)-1}{1+e^{-z}} \\ &=\frac{1}{1+e^{-z}} \cdot\left(\frac{1+e^{-z}}{1+e^{-z}}-\frac{1}{1+e^{-z}}\right) \\ &=\frac{1}{1+e^{-z}} \cdot\left(1-\frac{1}{1+e^{-z}}\right) \\ &=\boxed{\sigma(z) \cdot(1-\sigma(z))} \end{aligned}\]