Biomedical Engineering Reference
In-Depth Information
and obtain
⊡
⊤
M
L
M
L
∂
∂
ʱ
1
2
1
2
⊣
1
ʻ
j
ʱ
A
j
,
⊦
log
(ʻ
j
ʱ
)
−
j
=
1
=
1
j
=
1
=
⊡
⊣
j
=
1
ʻ
j
A
j
,
1
⊤
⊦
0
...
0
j
=
1
ʻ
j
A
j
,
2
...
0
0
M
2
ʱ
−
1
1
2
=
−
.
.
.
0
0
0
...
j
=
1
ʻ
j
A
j
,
L
0
0
1
2
(
ʱ
−
1
A
T
=
M
−
diag
[
A
ʛ
]
),
(5.77)
where diag
indicates a diagonal matrix whose diagonal entries are equal to those
of a matrix in the brackets.
4
[·]
The relationship
⊡
⊤
M
M
M
M
1
ʻ
j
[
ʨ
−
1
⊣
1
ʻ
j
A
j
,
⊦
=
A
j
,
+
A
j
,
+
[
ʨ
−
1
E
A
1
ʻ
j
1
ʻ
j
]
,
=
1
ʻ
j
M
]
,
j
=
j
=
j
=
j
=
also holds. (Note that Eq. (
5.168
) shows the general case of computing
E
A
[
A
i
,
k
A
j
,
]
.)
Thus, we finally obtain
⊡
⊡
⊤
⊤
M
L
M
L
∂
∂
ʱ
F
[
ʱ
,
ʛ
]=
∂
∂
ʱ
1
2
1
2
⊣
⊣
⊦
⊦
1
ʻ
j
ʱ
A
j
,
E
A
log
(ʻ
j
ʱ
)
−
j
=
1
=
1
j
=
1
=
⊡
⊤
j
=
1
ʻ
j
A
j
,
1
...
0
0
j
=
1
ʻ
j
⊣
⊦
A
j
,
2
...
0
0
1
2
M
1
2
ʱ
−
1
=
−
.
.
.
0
0
0
...
j
=
1
ʻ
j
A
j
,
L
0
0
⊡
⊣
⊤
⊦
[
ʨ
−
1
]
1
,
1
0
...
0
[
ʨ
−
1
0
]
2
,
2
...
0
1
2
M
−
.
.
.
0
0
0
...
[
ʨ
−
1
]
L
,
L
0
0
1
2
(
[
¯
A
T
ʛ
¯
A
ʱ
−
1
[
ʨ
−
1
=
M
−
diag
]−
M
diag
]
).
(5.78)
4
Here computing the derivative of a scalar
X
with a diagonal matrix
A
is equal to creating a diagonal
matrix whose
(
j
,
j
)
th diagonal element is equal to
∂
X
/∂
A
j
,
j
where
A
j
,
j
is the
(
j
,
j
)
th diagonal
element of
A
.
