$$\mbox{Logistic regression: }\mathbf{z} = \sigma(\mathbf{h}) = \frac{1}{1 + e^{-\mathbf{h}}}$$

$$\mbox{Cross-entropy loss: } J(\mathbf{w}) = -(\mathbf{y} log(\mathbf{z}) + (1 - \mathbf{y})log(1 - \mathbf{z})) $$ $$ \mbox{Use chain rule: } \frac{\partial{J(\mathbf{w})}}{\partial{\mathbf{w}}} = \frac{\partial{J(\mathbf{w})}}{\partial{\mathbf{z}}} \frac{\partial{\mathbf{z}}}{\partial{\mathbf{h}}} \frac{\partial{\mathbf{h}}}{\partial{\mathbf{\mathbf{w}}}}$$

$$\mbox{Gradient descent: } \mathbf{w} = \mathbf{w} - \alpha \frac{\partial{J(\mathbf{w})}}{\partial{\mathbf{w}}} $$

Suppose there's a random variable where (for binary classification), then the Bernoulli probability model will give us:

Its often easier to work with the derivatives when the metric is in terms of log and additionally, the min/max of loglikelihood is the same as the min/max of likelihood. The inherent meaning of a cost or loss function is such that the more it deviates from the 0, the worse the model performs. The negative sign the just preserves that meaning and is easier to interpret. Maximizing the above function will lead to the same result.

Here is the definition of cross-entropy for Bernoulli random variables $\operatorname{Ber}(p),\operatorname{Ber}(q)$, taken from Wikipedia: $$ H(p,q) = p \log \frac{1}{q} + (1-p) \log \frac{1}{1-q}. $$ This is exactly what your first function computes.

The partial derivative of this function with respect to $p$ is $$ \frac{\partial H(p,q)}{\partial p} = \log \frac{1}{q} - \log \frac{1}{1-q} = \log \frac{1-q}{q}. $$ The partial derivative of this function with respect to $q$ is $$ \frac{\partial H(p,q)}{\partial q} = -\frac{p}{q} + \frac{1-p}{1-q} = \frac{(1-p)q-p(1-q)}{q(1-q)} = \frac{q-p}{q(1-q)}. $$ This is exactly what your third function computes.

I'm not sure what your second function computes. Also, there is no reason to expect that the partial derivative with respect to one variable will be the same as the partial derivative with respect to another variable.