Derivative of a determinant whose entries are functions

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.

The determinant is like a generalized product of vectors (in fact, it is related to the outer product). So considering the rows as factors in this generalized product, this formula reflects the product rule of differentiation.

If $D(a,b,c)$ is generally a function of vectors that is linear in each argument, and you apply it to vector functions in one variable, then

\begin{align} &D(a(t+h),b(t+h),c(t+h))-D(a(t),b(t),c(t))\\[0.5em] &=\ D(a(t+h),b(t+h),c(t+h))-D(a(t),b(t+h),c(t+h))\\ &\ +D(a(t),b(t+h),c(t+h))-D(a(t),b(t),c(t+h))\\ &\ +D(a(t),b(t),c(t+h))-D(a(t),b(t),c(t))\\[0.5em] &=\ D([a(t+h)-a(t)],b(t+h),c(t+h)) +D(a(t),[b(t+h)-b(t)],c(t+h)) +D(a(t),b(t),[c(t+h)-c(t)]) \end{align}

and from that the claimed generalized product rule can be obtained.

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.