Sigmoid derivative from the activation

Background

The lesson uses sigmoid everywhere partly because its derivative has a tidy form that reuses the value you already computed:

$$\sigma'(z) = \sigma(z)\bigl(1 - \sigma(z)\bigr) = a\,(1 - a)$$

where $a = \sigma(z)$ is the activation from the forward pass. So once you've cached $a$, you get the local gradient for free — no need to recompute the exponential.

Problem statement

Implement sigmoid_prime(a) that, given the sigmoid activation a (not the pre-activation $z$), returns $\sigma'(z) = a(1-a)$ elementwise.

Input

• a — array-like (or scalar) of sigmoid activations, each in $(0, 1)$.

Output

Returns an np.ndarray (same shape as a) of local gradients $a(1-a)$.

Examples

Example 1 — steepest at 0.5

Input:  a = 0.5
Output: 0.25

Explanation: $0.5\times0.5 = 0.25$, the maximum slope of the sigmoid (at $z=0$).

Example 2 — saturated activations have tiny gradient

Input:  a = [0.5, 0.99, 0.01]
Output: [0.25   0.0099 0.0099]

Explanation: near 0 or 1 the slope collapses — the vanishing-gradient regime.

Constraints

• Use the cached form a * (1 - a) — do not take an exponential.
• Return an np.ndarray.

Notes

• This is the factor the backward pass multiplies by at every sigmoid; it's largest at $a=0.5$ and near zero when the unit saturates.