Derivative of a determinant whose entries are functions I do not understand a remark in Adams' Calculus (page 628 $7^{th}$ edition). This remark is about the derivative of a determinant whose entries are functions as quoted below.

Since every term in the expansion of a determinant of any order is a product involving one element from each row, the general product rule implies that the derivative of an $n\times n$ determinant whose elements are functions will be the sum of $n$ such $n\times n$ determinants, each with the elements of one of the rows differentiated. For the $3\times 3$ case we have
$$\frac{d}{dt}\begin{vmatrix}
  a_{11}(t) & a_{12}(t) & a_{13}(t) \\
  a_{21}(t) & a_{22}(t) & a_{23}(t) \\
  a_{31}(t) & a_{32}(t) & a_{33}(t)
  \end{vmatrix}=\begin{vmatrix}
  a'_{11}(t) & a'_{12}(t) & a'_{13}(t) \\
  a_{21}(t) & a_{22}(t) & a_{23}(t) \\
  a_{31}(t) & a_{32}(t) & a_{33}(t)
  \end{vmatrix}+\begin{vmatrix}
  a_{11}(t) & a_{12}(t) & a_{13}(t) \\
  a'_{21}(t) & a'_{22}(t) & a'_{23}(t) \\
  a_{31}(t) & a_{32}(t) & a_{33}(t)
  \end{vmatrix}+\begin{vmatrix}
  a_{11}(t) & a_{12}(t) & a_{13}(t) \\
  a_{21}(t) & a_{22}(t) & a_{23}(t) \\
  a'_{31}(t) & a'_{32}(t) & a'_{33}(t)
  \end{vmatrix}.$$


It is not difficult to check this equality by simply expanding both sides. However, the remark sounds like using some clever trick to get this result. Can anyone explain it to me, please? Thank you!

That remarks has said most of what it needs to explain.However, I think a more precise explaination for the example  is necessary.Hence, I'll cite one.

i) $a_{11}(t).a_{23}(t).a_{32}(t) $ is abitrary term in expansion of left determinant
ii) $ (a_{11}(t).a_{23}(t).a_{32}(t))' = a'_{11}(t)a_{23}(t).a_{32}(t)+a_{11}(t)a'_{23}(t).a_{32}(t)+a_{11}(t)a_{23}(t).a'_{32}(t)$
iii) $a'_{11}(t)a_{23}(t).a_{32}(t)$ is a term in expansion of the first determinant on the left of equation. This is a determinant in which the first row is differentiated
$a_{11}(t)a'_{23}(t).a_{32}(t),a_{11}(t)a_{23}(t).a'_{32}(t)$  are similar.
