TITLE: Proof for the derivative of the determinant of a matrix
QUESTION [28 upvotes]: I was looking for theorems that might be helpful in order for some proofs that I have and I came across the following one:
$$\frac{d}{dt} [\det A(t)]=\det A(t) \cdot \operatorname*{tr}[A^{-1}(t)\cdot \frac{d}{dt} A(t)]$$
where $A(t)$ is a matrix with a variable $t$.
The problem is that I have neither a reliable source for this theorem nor am I able to prove it.
Did anyone come across the aforementioned equation or is able to prove it?

REPLY [31 votes]: Another way to obtain the formula is to first consider the derivative of the determinant at the identity:
$$
 \frac{d}{dt} \det (I + t M) = \operatorname{tr} M.
$$
Next, one has 
$$
\begin{split}
   \frac{d}{dt} \det A (t)
&=\lim_{h \to 0} \frac{\det \bigl(A (t + h)\bigr) - \det A (t)}{h}\\
&=\det A (t) \lim_{h \to 0}  \frac{\det \bigl(A (t)^{-1} A (t + h)\bigr) - 1}{h}\\
&=\det A (t) \operatorname{tr} \Bigl(A (t)^{-1}\frac{d A}{dt} (t) \Bigr).
\end{split}
$$