1. **Problem Statement:** We are given the activation function $g(x) = \tanh(x)$ used in neural networks. (a) Find the first derivative $g'(x)$ and the second derivative $g''(x)$. (b) Determine the intervals where $\tanh(x)$ is concave up and concave down. (c) Explain how concavity affects vanishing gradients. 2. **Recall the formulas and rules:** - The derivative of $\tanh(x)$ is $\frac{d}{dx} \tanh(x) = 1 - \tanh^2(x)$. - The second derivative is the derivative of the first derivative. - Concavity is determined by the sign of the second derivative: if $g''(x) > 0$, the function is concave up; if $g''(x) < 0$, it is concave down. 3. **Find $g'(x)$:** $$ g'(x) = 1 - \tanh^2(x) $$ This comes from the identity $\frac{d}{dx} \tanh(x) = \operatorname{sech}^2(x) = 1 - \tanh^2(x)$. 4. **Find $g''(x)$:** Take the derivative of $g'(x)$: $$ g''(x) = \frac{d}{dx} (1 - \tanh^2(x)) = -2 \tanh(x) \cdot g'(x) = -2 \tanh(x) (1 - \tanh^2(x)) $$ 5. **Determine concavity intervals:** - Since $g''(x) = -2 \tanh(x) (1 - \tanh^2(x))$, note that $1 - \tanh^2(x) > 0$ for all real $x$ because $|\tanh(x)| < 1$. - The sign of $g''(x)$ depends on $-2 \tanh(x)$. - For $x < 0$, $\tanh(x) < 0$, so $-2 \tanh(x) > 0$ and $g''(x) > 0$; the function is concave up. - For $x > 0$, $\tanh(x) > 0$, so $-2 \tanh(x) < 0$ and $g''(x) < 0$; the function is concave down. - At $x=0$, $g''(0) = 0$. 6. **Concavity and vanishing gradients:** - Vanishing gradients occur when derivatives become very small, slowing learning in neural networks. - The first derivative $g'(x) = 1 - \tanh^2(x)$ approaches zero as $|x|$ becomes large because $\tanh(x)$ approaches $\pm 1$. - Concavity indicates how the slope changes: near $x=0$ (inflection point), the slope changes sign, affecting gradient flow. - Concave up region ($x<0$) means the slope is increasing, concave down ($x>0$) means slope is decreasing. - Understanding concavity helps in analyzing gradient behavior and designing activation functions to mitigate vanishing gradients. **Final answers:** (a) $$g'(x) = 1 - \tanh^2(x), \quad g''(x) = -2 \tanh(x) (1 - \tanh^2(x))$$ (b) Concave up on $(-\infty, 0)$, concave down on $(0, \infty)$. (c) Concavity affects how gradients change; near zero, gradients are stronger, but for large $|x|$, gradients vanish due to $g'(x) \to 0$, impacting learning in neural networks.