Both $A_{N \times N}$ and $B_{N \times N}$ are symmetric Toeplitz matrix, $\sigma_s^2$ is a constant.
In detection theory, the computation of $\det(\sigma_s^2A+B)^{1/N}$ arises in the computation of the GLRT (sometimes referred to as sphericity test). In particular, $A_{N \times N}$, $B_{N \times N}$ and $\sigma_s^2$ represent the covariance of received signal, the noise covariance and the received signal power, respectively.
I need to find how $\det(\sigma_s^2A+B)^{1/N}$ varies as $\sigma_s^2$ varies. An approximated expression or bound could work.
Using the Minkowski determinant inequality i found a bound $$\det(\sigma_s^2A+B)^{1/N}\ge \sigma_s^2\det(A)^{1/N}+\det(B)^{1/N}$$ However this inequality yields a not sufficiently tight approximation.
A tight bound or (better an approximation of the above determinant) can be found, maybe under specific conditions on the matrix A and B?
Here is a full computation for the derivative of $$ f(\epsilon) = \det(B + \epsilon A)^{1/N}. $$ Let $g(\epsilon) = \det(I + \epsilon B^{-1}A)$, so that $f(\epsilon) = \det(B)^{1/N}g(\epsilon)^{1/N}$. By the chain rule, we have $$ f'(\epsilon) = \frac 1N \det(B)^{1/N} g(\epsilon)^{1/N - 1} \cdot g'(\epsilon) \\ = \frac 1N \det(B)^{1/N}\det(I + \epsilon B^{-1}A)^{1/N - 1} \cdot \operatorname{tr}(B^{-1}A). $$ Plugging in $\epsilon = 0$ yields $f'(0) = \frac 1N \det(B)^{1/N}\operatorname{tr}(B^{-1}A)$, which leads to the Taylor expansion $$ f(\sigma_s^2) = \det(B)^{1/N} + \frac 1N \det(B)^{1/N}\operatorname{trace}(B^{-1}A)\sigma_s^2 + O(\sigma_s^4) $$ that I mentioned in a comment on this question.