1. **Problem Statement:** We analyze the sigmoid activation function $$\sigma(x) = \frac{1}{1+e^{-x}}$$ and its differential at specific points, then compare with the ReLU function. 2. **Formula for the differential of sigmoid:** The derivative of $$\sigma(x)$$ is $$\sigma'(x) = \sigma(x)(1 - \sigma(x))$$. This comes from differentiating the sigmoid function using the chain rule. 3. **(a) Find $$d\sigma$$ at $$x=0$$:** - Calculate $$\sigma(0) = \frac{1}{1+e^{0}} = \frac{1}{2}$$. - Then $$\sigma'(0) = \sigma(0)(1 - \sigma(0)) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$$. 4. **(b) Calculate $$d\sigma$$ at $$x=2$$ and $$x=-2$$:** - For $$x=2$$: - $$\sigma(2) = \frac{1}{1+e^{-2}} \approx 0.8808$$. - $$\sigma'(2) = 0.8808 \times (1 - 0.8808) \approx 0.8808 \times 0.1192 = 0.10499$$. - For $$x=-2$$: - $$\sigma(-2) = \frac{1}{1+e^{2}} \approx 0.1192$$. - $$\sigma'(-2) = 0.1192 \times (1 - 0.1192) \approx 0.1192 \times 0.8808 = 0.10499$$. 5. **(c) Vanishing gradient problem and sigmoid differential:** - The derivative $$\sigma'(x)$$ becomes very small when $$x$$ is large positive or large negative. - This means gradients vanish during backpropagation, slowing or stopping learning in deep networks. - At $$x=\pm 2$$, the derivative is about 0.105, much smaller than at 0. 6. **(d) ReLU function and its differential:** - ReLU is defined as $$f(x) = \max(0, x)$$. - Its derivative is: - $$f'(x) = 0$$ for $$x < 0$$ - $$f'(x) = 1$$ for $$x > 0$$ - Undefined at $$x=0$$ but often set to 0 or 1 in practice. - Implications: - ReLU avoids vanishing gradients for positive $$x$$ since derivative is 1. - However, for negative $$x$$, gradient is zero, which can cause "dead neurons". **Final answers:** - $$d\sigma(0) = 0.25$$ - $$d\sigma(2) = d\sigma(-2) \approx 0.105$$ - Sigmoid suffers from vanishing gradients at large $$|x|$$. - ReLU has piecewise derivative 0 or 1, helping mitigate vanishing gradients but with other trade-offs.