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I want to compute the derivative of the determinant of a matrix. This seems to be relatively straightforward for the first derivative using e.g., Jacobi's formula.

$$\frac{d}{dt}\det A(t)=\mathrm{tr}\, \left(\mathrm{adj}(A(t)) \frac{dA(t)}{dt} \right)$$

Let us assume we have the following matrix $A(t)$:

$$A(t) = \begin{bmatrix} -8 \cdot t^3 - 3 \cdot t^2 & 2 \cdot t^3 + 4 \cdot t^2 \\ 6 \cdot t^3 + 4 \cdot t^2 & -8 \cdot t^3 \\ \end{bmatrix}$$

Thus, if I have applied Jacobi's formula correctly, the result of the first derivative should be as follows:

$$\frac{d}{dt}\det A(t) = 312 \cdot t^5 - 40 \cdot t^4 - 64 \cdot t^3 $$

In a related post, there was an attempt to use Jacobi's formula for the second derivative. This is an interesting case in itself.

My results for the second derivative are as follows:

$$\frac{d^2}{dt^2}det\ A(t)= 1560 \cdot t^4 - 160 \cdot t^3 - 192 \cdot t^2$$

I am not overly confident with the matrix calculus and the correct application of Jacobi's formula. Hence, I would be grateful if someone could check if my obtained results are correct? Thanks!

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While using Jacobi's formula may be useful as an exercise, I don't see that in this case it is the simplest method. The determinant can be easily calculated in terms of $t$ as $52t^6-8t^5-16t^4$, giving the second derivative as $1560t^4-160t^3-192t^2$.

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  • $\begingroup$ Thanks for the confirmation and for pointing out that it is not the easiest method. However, in this case I was really interested in using the Jacobi's formula as an exercise. $\endgroup$ – edmond Jan 24 at 19:36

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