$\frac{d^2}{dx^2}\left\{det\begin{bmatrix}x^3&x\\2&e^x\end{bmatrix}\right\}$ equals

$xe^x (x^2+6x+6)$

The correct answer is Option (3) → $xe^x (x^2+6x+6)$

Given expression:

$\frac{d^{2}}{dx^{2}}\left\{\det\begin{bmatrix}x^{3}&x\\2&e^{x}\end{bmatrix}\right\}$

First, compute determinant:

$\det = x^{3}e^{x} - 2x = e^{x}x^{3} - 2x$

First derivative:

$\frac{d}{dx}(x^{3}e^{x} - 2x) = e^{x}(x^{3} + 3x^{2}) - 2$

Second derivative:

$\frac{d^{2}}{dx^{2}} = \frac{d}{dx}[e^{x}(x^{3} + 3x^{2}) - 2]$

$= e^{x}(x^{3} + 3x^{2}) + e^{x}(3x^{2} + 6x)$

$= e^{x}(x^{3} + 6x^{2} + 6x)$

$= x e^{x}(x^{2} + 6x + 6)$

$x e^{x}(x^{2} + 6x + 6)$