Remark

Tag 083E, permanent. Cite it as [Bourbaki, Tag 083E].

Functions of a Real Variable, chapter I, § 1. FIRST DERIVATIVE

If $U(x)$ is a square matrix which is differentiable and invertible at the point $x_0$ , then the derivative of its determinant $\Delta(x) = \det(U(x))$ can be expressed through the derivative of $U(x)$ by the formula

\Delta'(x_0) = \Delta(x_0) \cdot \operatorname{Tr}(U'(x_0) U^{-1}(x_0)).

Indeed, let us put

U(x_0 + h) = U(x_0) + h V

; then, by definition,

V

tends to

U'(x_0)

when

h

tends to 0. One can write

\Delta(x_0 + h) = \Delta(x_0) \cdot \det(I + h V U^{-1}(x_0)).

Now

\det(I + h X) = 1 + h \operatorname{Tr}(X) + \sum_{k=2}^n \lambda_k h^k

, the

\lambda_k

(

k \geq 2

) being polynomials in the elements of the matrix

X

; since the elements of

V U^{-1}(x_0)

have a limit when

h

tends to 0, we indeed obtain the formula (3).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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