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Consider a function $A: \mathbb{R} \rightarrow \mathbb{R}^{n \times n}, t \rightarrow A(t)$. I want to compute the derivative of the determinant

$$\frac{d}{dt} \det(A(t)) \; .$$

Suppose $A(t)$ is invertible at some $t = t_0$, then a closed form solution exists (cf. e.g. The matrix cookbook, p.7),

$$\frac{d}{dt} \det(A(t)) \vert_{t=t_0} = \det(A(t_0)) \;\mathrm{tr}\left(A(t_0)^{-1} \frac{d A(t)}{dt}\vert_{t=t_0}\right) \;.$$

Now, what happens if $A(t_0)$ is not invertible? Is there still a closed form solution?

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    $\begingroup$ Yes, use the adjugate. Take a look at Jacobi's formula. $\endgroup$ Commented Mar 21, 2017 at 13:58
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    $\begingroup$ In other words, just move the determinant factor on the right side into the parentheses. $\endgroup$
    – Deane Yang
    Commented Mar 21, 2017 at 15:40

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