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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 $$ \frac{\partial(a^2)}{\partial(a^1)} = \frac{\partial(\sigma(w^2 a^{1}+b^2))}{\partial(a^1)} = w^2.\sigma'(a^1) \; \rightarrow ( {b})\\ \\ $$

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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 $w^2\sigma'(a^1)$.

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