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Mathematics: The Question of Evaluating Expression

The question arose about mathgroup, and while I was looking at it, I noticed this thing, and I can’t understand why, I thought that some expert here knows.

When executing Dt [x [1]]

it gives zero, because during the evaluation of x [1] the last value remains equal to 1, as can be seen from the TracePrint below. And therefore, '1' is what is visible on Dt, and therefore Dt [1] is 0.

Therefore, Dt [x [1]] is equal to zero

In[86]:= TracePrint[ Dt[x[1] ]]

During evaluation of In[86]:=  Dt[x[1]]
During evaluation of In[86]:=   Dt
During evaluation of In[86]:=   x[1]
During evaluation of In[86]:=    x
During evaluation of In[86]:=    1
During evaluation of In[86]:=  0

Out[86]= 0

This made sense to me until I dialed x [1] and returned x [1]

In [84]: = x [1] Out [84] = x [1]

But x [1], returning x [1], also makes sense to me, since x [1] doesn't matter, so it should return invaluable.

, , , x [1] "1" , 1?

In[87]:= Evaluate[ x[1] ]
Out[87]= x[1]

+3
1

x[1]

1 - undefined. , , Dt 1 , x , 1 - , 0. ,

In[1]:= Dt[x[y]]

Out[1]= Dt[y] Derivative[1][x][y]


In[2]:= Dt[x[1],x[1]]

Out[2]= 1

1, x[1], . ,

In[3]:= Dt[x[1]^2, x[1]]

Out[3]= 2 x[1]

, , , , ( - x), x[1] Dt, . , , , Dt .

+5

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