Abel's identity states that if $X(t)$ and $A(t)$ are $n\times n$ matrix-valued functions such that $X'(t)=A(t)X(t)$, then $\frac{d}{dt}(\det X(t)) = \mathrm{tr}\,A(t) \cdot \det X(t)$.
The question is whether there's a nice high-brow basis-independent way to see this. Given that you can state the problem without ever referring to matrix entries, I want prove it without ever referring to matrix entries. I'd expect the identity $\det e^A = e^{\mathrm{tr}(A)}$ to play a central role in the proof, but I have yet to come up with such an argument.
You can prove this without too much trouble in a bare-hands way, but I don't see how to turn this into a basis-free proof. Suppose $X_i(t)$ is the matrix you get from $X(t)$ by taking the derivative of every entry in the $i$-th row. Then $\frac{d}{dt}(\det X(t)) = \sum_{i=1}^n \det(X_i(t))$. Using the relation $X'(t)=A(t)X(t)$ you can express the derivative of the $i$-th row of $X'(t)$ in terms of the entries of $A$ and $X$. When you do this, you find that $\det X_i(t)$ is simply $\det X(t)$ multiplied by the $(i,i)$-th entry of $A(t)$.