# Is this partial derivative correct

For

$$a^{2} = \sigma(w^2 a^{1}+b^2) = \sigma(z^2)\\ \\ \text{where} \; z^l =w^l a^{l-1} +b^l$$ If we write the derivate

$${\frac{\partial a^{l} }{\partial w^{l}} = \frac{\partial \sigma (z^{l}) }{\partial w^{l}} = \sigma' (z^{l}) \quad \rightarrow ( {a})} \\$$ Then is the below proper $$\frac{\partial(a^2)}{\partial(a^1)} = \frac{\partial(\sigma(w^2 a^{1}+b^2))}{\partial(a^1)} = w^2.\sigma'(a^1) \; \rightarrow ( {b})\\ \\$$

Assuming that $$a^2$$ and $$a^1$$ are distinct variables, then you are mostly correct. It should be $$\frac{\partial a^2}{\partial a^1} = w^2\sigma'(w^2 a^1 + b^2)$$ rather than $$w^2\sigma'(a^1)$$.