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bp mlp 6

bp-mlp-6

Feb 20, 20261 min read

δLδl∂bjl​∂L​∂wjkl​∂L​​=∇a​L⊙σ′(zL)=((Wl+1)Tδl+1)⊙σ′(zl)=δjl​=akl−1​δjl​​​(BP1)(BP2)(BP3)(BP4)​

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