\(\displaystyle\frac{d^2}{dx^2}\left\{\det\begin{bmatrix}x^3&x\\2&e^x\end{bmatrix}\right\}\) equals
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(xe^x (x^2+6x+6)\)
The determinant of the 2 × 2 matrix is
\(\displaystyle \det\begin{bmatrix}x^3&x\\2&e^x\end{bmatrix} = x^3e^x - 2x\)
By the product rule:
\(\displaystyle \frac{d}{dx}\left(x^3e^x - 2x\right) = 3x^2e^x + x^3e^x - 2\)
\(= e^x(x^3 + 3x^2) - 2\)
Differentiating again (the constant −2 drops out):
\(\displaystyle \frac{d^2}{dx^2}\left(x^3e^x - 2x\right) = e^x(x^3 + 3x^2) + e^x(3x^2 + 6x)\)
\(= e^x(x^3 + 6x^2 + 6x)\)
\(= xe^x(x^2 + 6x + 6)\)