If $A = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$ and $B = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix}$,then $\frac{dA}{dx}$ is equal to

The determinant $\left| \begin{array}{ccc} 4 + x^2 & -6 & -2 \\ -6 & 9 + x^2 & 3 \\ -2 & 3 & 1 + x^2 \end{array} \right|$ for $x \neq 0$ is not divisible by:

If $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$, then $f^{\prime}(\pi)$ is equal to

If $a_1, a_2, a_3, \dots, a_n$ form a geometric progression,find the value of the determinant: $\left| \begin{array}{ccc} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{array} \right|$.

The value of the determinant $\left| \begin{array}{ccc} a^2 & a & 1 \\ \cos(nx) & \cos(n+1)x & \cos(n+2)x \\ \sin(nx) & \sin(n+1)x & \sin(n+2)x \end{array} \right|$ is independent of :

Let $f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{vmatrix}$. Then $\lim_{x \to 0} \frac{f'(x)}{x} =$