Week 3: Why dZ^[1] = W^[2]T dZ^[2] * g^[1]'(Z^[1])

Well, notice the dimensions on dZ^{[2]} and W^{[2]T}: they are n^{[2]} x m and n^{[1]} x n^{[2]} respectively, right? So that matrix multiply (dot product style) in the formula you show is not going to work. But it does work if the formula is as Prof Ng and AbdElRhaman show it.

Taking derivatives is a bit more complicated when you’re working with matrices than scalars. This is beyond the scope of this course (by design), but you can find links to background information on matrix calculus and on these specific derivations on this thread.