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!

The author is probably referring to the fact that the determinant is given by: 
$$
\sum_{i,j,k=1}^n\varepsilon_{ijk}a_{1i}a_{2j}a_{3k}
$$
where $\varepsilon_{ijk}$ is $1$ if $(ijk)$ is an even permutation of $(123)$, $-1$ if $(ijk)$ is an odd permutation of $(123)$ and $0$ if two or more of $i,j,k$ are equal.  Differentiating this expression immediately gives: 
$$
\sum_{i,j,k=1}^n\varepsilon_{ijk}(a_{1i}'a_{2j}a_{3k}+a_{1i}a_{2j}'a_{3k}+a_{1i}a_{2j}a_{3k}')
$$
which is easily seen to be the sum of the three determinants given, using the same formula again.  
