Vector Derivatives of a Log Expression

Formulas to address a question posted in https://mathematica.stackexchange.com/questions/262063/third-fourth-derivative-of-cross-entropy-loss/262193

Setup


Results

• Original expression:
In[]:=
xent
Out[]=
Log
n
∑
i
z
i

-
n
∑
i
q
i
z
i
• First derivative:
In[]:=
D[xent,z[i]]
Out[]=
n
∑
K[1]
z
K[1]

δ
i,K[1]
n
∑
i
z
i

-
n
∑
K[1]
δ
i,K[1]
q
K[1]
In[]:=
%//DeltaSimplify
Out[]=
-
q
i
+
z
i

n
∑
i
z
i

In[]:=
%//pSimplify
Out[]=
p
i
-
q
i
• Second derivative:
In[]:=
D[xent,z[i],z[j]]
Out[]=
-
n
∑
K[1]
0-

n
∑
i
z
i

δ
i,j

n
∑
K[1]
z
K[1]

δ
i,K[1]
2

n
∑
i
z
i


+
n
∑
K[1]
z
K[1]

δ
i,K[1]
δ
j,K[1]
n
∑
i
z
i

In[]:=
%//DeltaSimplify
Out[]=
-
z
i
+
z
j

2

n
∑
i
z
i


+
z
i

δ
i,j
n
∑
i
z
i

In[]:=
%//pSimplify
Out[]=
δ
i,j
p
i
-
p
i
p
j
• Third derivative:
In[]:=
D[xent,z[i],z[j],z[k]]
Out[]=
-
n
∑
K[1]
0+
2
n
∑
i
z
i

δ
i,j

n
∑
i
z
i

δ
i,k

n
∑
K[1]
z
K[1]

δ
i,K[1]
3

n
∑
i
z
i


-

n
∑
i
z
i

δ
i,j
δ
i,k

n
∑
K[1]
z
K[1]

δ
i,K[1]
2

n
∑
i
z
i


-

n
∑
i
z
i

δ
i,k

n
∑
K[1]
z
K[1]

δ
i,K[1]
δ
j,K[1]
2

n
∑
i
z
i


-

n
∑
i
z
i

δ
i,j

n
∑
K[1]
z
K[1]

δ
i,K[1]
δ
k,K[1]
2

n
∑
i
z
i


+
n
∑
K[1]
z
K[1]

δ
i,K[1]
δ
j,K[1]
δ
k,K[1]
n
∑
i
z
i

In[]:=
%//DeltaSimplify
Out[]=
2
z
i
+
z
j
+
z
k

3

n
∑
i
z
i


-
z
i
+
z
k

δ
i,j
2

n
∑
i
z
i


-
z
i
+
z
j

δ
i,k
2

n
∑
i
z
i


-
z
i
+
z
j

δ
j,k
2

n
∑
i
z
i


+
z
i

δ
i,j
δ
i,k
n
∑
i
z
i

In[]:=
%//pSimplify
Out[]=
δ
i,j
δ
i,k
p
i
-
δ
i,k
p
i
p
j
-
δ
j,k
p
i
p
j
-
δ
i,j
p
i
p
k
+2
p
i
p
j
p
k
Obviously the result is highly symmetric (as it must be due to the symmetry of partial derivatives).
• Fourth derivative:
In[]:=
D[xent,z[i],z[j],z[k],z[l]]//DeltaSimplify//pSimplify
Out[]=
δ
i,j
δ
i,k
δ
i,l
p
i
-
δ
i,k
δ
i,l
p
i
p
j
-
δ
i,l
δ
j,k
p
i
p
j
-
δ
i,k
δ
j,l
p
i
p
j
-
δ
j,k
δ
j,l
p
i
p
j
-
δ
i,j
δ
i,l
p
i
p
k
-
δ
i,j
δ
k,l
p
i
p
k
+2
δ
i,l
p
i
p
j
p
k
+2
δ
j,l
p
i
p
j
p
k
+2
δ
k,l
p
i
p
j
p
k
-
δ
i,j
δ
i,k
p
i
p
l
+2
δ
i,k
p
i
p
j
p
l
+2
δ
j,k
p
i
p
j
p
l
+2
δ
i,j
p
i
p
k
p
l
-6
p
i
p
j
p
k
p
l
• Fifth derivative:
In[]:=
D[xent,z[i],z[j],z[k],z[l],z[m]]//DeltaSimplify//pSimplify
Out[]=
δ
i,j
δ
i,k
δ
i,l
δ
i,m
p
i
-
δ
i,k
δ
i,l
δ
i,m
p
i
p
j
-
δ
i,l
δ
i,m
δ
j,k
p
i
p
j
-
δ
i,k
δ
i,m
δ
j,l
p
i
p
j
-
δ
i,m
δ
j,k
δ
j,l
p
i
p
j
-
δ
i,k
δ
i,l
δ
j,m
p
i
p
j
-
δ
i,l
δ
j,k
δ
j,m
p
i
p
j
-
δ
i,k
δ
j,l
δ
j,m
p
i
p
j
-
δ
j,k
δ
j,l
δ
j,m
p
i
p
j
-
δ
i,j
δ
i,l
δ
i,m
p
i
p
k
-
δ
i,j
δ
i,m
δ
k,l
p
i
p
k
-
δ
i,j
δ
i,l
δ
k,m
p
i
p
k
-
δ
i,j
δ
k,l
δ
k,m
p
i
p
k
+2
δ
i,l
δ
i,m
p
i
p
j
p
k
+2
δ
i,m
δ
j,l
p
i
p
j
p
k
+2
δ
i,l
δ
j,m
p
i
p
j
p
k
+2
δ
j,l
δ
j,m
p
i
p
j
p
k
+2
δ
i,m
δ
k,l
p
i
p
j
p
k
+2
δ
j,m
δ
k,l
p
i
p
j
p
k
+2
δ
i,l
δ
k,m
p
i
p
j
p
k
+2
δ
j,l
δ
k,m
p
i
p
j
p
k
+2
δ
k,l
δ
k,m
p
i
p
j
p
k
-
δ
i,j
δ
i,k
δ
i,m
p
i
p
l
-
δ
i,j
δ
i,k
δ
l,m
p
i
p
l
+2
δ
i,k
δ
i,m
p
i
p
j
p
l
+2
δ
i,m
δ
j,k
p
i
p
j
p
l
+2
δ
i,k
δ
j,m
p
i
p
j
p
l
+2
δ
j,k
δ
j,m
p
i
p
j
p
l
+2
δ
i,k
δ
l,m
p
i
p
j
p
l
+2
δ
j,k
δ
l,m
p
i
p
j
p
l
+2
δ
i,j
δ
i,m
p
i
p
k
p
l
+2
δ
i,j
δ
k,m
p
i
p
k
p
l
+2
δ
i,j
δ
l,m
p
i
p
k
p
l
-6
δ
i,m
p
i
p
j
p
k
p
l
-6
δ
j,m
p
i
p
j
p
k
p
l
-6
δ
k,m
p
i
p
j
p
k
p
l
-6
δ
l,m
p
i
p
j
p
k
p
l
-
δ
i,j
δ
i,k
δ
i,l
p
i
p
m
+2
δ
i,k
δ
i,l
p
i
p
j
p
m
+2
δ
i,l
δ
j,k
p
i
p
j
p
m
+2
δ
i,k
δ
j,l
p
i
p
j
p
m
+2
δ
j,k
δ
j,l
p
i
p
j
p
m
+2
δ
i,j
δ
i,l
p
i
p
k
p
m
+2
δ
i,j
δ
k,l
p
i
p
k
p
m
-6
δ
i,l
p
i
p
j
p
k
p
m
-6
δ
j,l
p
i
p
j
p
k
p
m
-6
δ
k,l
p
i
p
j
p
k
p
m
+2
δ
i,j
δ
i,k
p
i
p
l
p
m
-6
δ
i,k
p
i
p
j
p
l
p
m
-6
δ
j,k
p
i
p
j
p
l
p
m
-6
δ
i,j
p
i
p
k
p
l
p
m
+24
p
i
p
j
p
k
p
l
p
m
• Sixth derivative:
In[]:=
D[xent,z[i],z[j],z[k],z[l],z[m],z[n]]//DeltaSimplify//pSimplify
Out[]=