$\mathcal{L}$ is the loss function, $\mathcal{L} = y_i \text{log} \sigma(z) + (1-y_i) \text{log} (1-\sigma(z))$, where $z = \sum_i w_ix_i$, with $w_i$ representing the weights and $x_i$ the features. Typically, it is required to take a derivative of $\mathcal{L}$ with respect to $w_1$ or $w_2$. But I am required to provide derivative with respect to parameters which includes all weights.

Could anyone please recommend a solution or some tutorials? Thanks

  • $\begingroup$ What do you mean by providing derivative wrt parameters which includes all weights'?' $\endgroup$
    – gunes
    May 21, 2020 at 8:08
  • $\begingroup$ What is $\sigma(x)$? The sigmoid function? $\endgroup$
    – Tinu
    May 21, 2020 at 8:18

1 Answer 1


Since $\mathcal{L}(w_1, w_2, ..., w_n)$ is a scalar loss function depending on multiple weights $w_i$ (neglecting features $x$ and labels $y$ for a moment) taking the gradient with respect to all would give you a vector

$\nabla \mathcal{L} = [\frac{\partial \mathcal{L}}{\partial w_1}, ...,\frac{\partial \mathcal{L}}{\partial w_n}]$.

Now you can calculate each individual component with the following formula, using the sigmoid derivative $\sigma'(x) = \sigma(x)(1 - \sigma(x))$ and chain rule:

$\frac{\partial \mathcal{L}}{\partial w_i} = \frac{y_ix_i\sigma(z)(1 - \sigma(z))}{\sigma(z)} - \frac{(1 - y_i)x_i\sigma(z)(1-\sigma(z))}{1 - \sigma(z)}$.


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