Let f(x) = left| begin{array}{ccc} x^3 & sin | vedclass

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(D) Given $f(x) = \left| \begin{array}{ccc} x^3 & \sin x & \cos x \ 6 & -1 & 0 \ p & p^2 & p^3 \end{array} ight|$.Since the derivative of a determ

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Independent of $p$

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(D) Given $f(x) = \left| \begin{array}{ccc} x^3 & \sin x & \cos x \ 6 & -1 & 0 \ p & p^2 & p^3 \end{array} ight|$.Since the derivative of a determinant is the sum of determinants obtained by differentiating one row at a time,and the second and third rows are constants,the third derivative $f'''(x)$ is given by differentiating the first row three times:$f'''(x) = \left| \begin{array}{ccc} \frac{d^3}{dx^3}(x^3) & \frac{d^3}{dx^3}(\sin x) & \frac{d^3}{dx^3}(\cos x) \ 6 & -1 & 0 \ p & p^2 & p^3 \end{array} ight|$Calculating the derivatives:$\frac{d^3}{dx^3}(x^3) = 6$$\frac{d^3}{dx^3}(\sin x) = -\cos x$$\frac{d^3}{dx^3}(\cos x) = \sin x$Thus,$f'''(x) = \left| \begin{array}{ccc} 6 & -\cos x & \sin x \ 6 & -1 & 0 \ p & p^2 & p^3 \end{array} ight|$.At $x = 0$:$f'''(0) = \left| \begin{array}{ccc} 6 & -\cos(0) & \sin(0) \ 6 & -1 & 0 \ p & p^2 & p^3 \end{array} ight| = \left| \begin{array}{ccc} 6 & -1 & 0 \ 6 & -1 & 0 \ p & p^2 & p^3 \end{array} ight|$.Since the first two rows are identical,the value of the determinant is $0$.Therefore,the result is $0$,which is independent of $p$.

Let $f(x) = \left| \begin{array}{ccc} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{array} \right|$,where $p$ is a constant. Then $\frac{d^3}{dx^3} \{f(x)\}$ at $x = 0$ is

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