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bp mlp 6
bp-mlp-6
Feb 20, 2026
1 min read
δ
L
δ
l
∂
b
j
l
∂
L
∂
w
jk
l
∂
L
=
∇
a
L
⊙
σ
′
(
z
L
)
=
(
(
W
l
+
1
)
T
δ
l
+
1
)
⊙
σ
′
(
z
l
)
=
δ
j
l
=
a
k
l
−
1
δ
j
l
(BP1)
(BP2)
(BP3)
(BP4)
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