If Delta (x) = left| begin{array}{ccc} x^n & | vedclass

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(B) We are given the determinant $\Delta (x) = \left| \begin{array}{ccc} x^n & \sin x & \cos x \ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \ a

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(B) We are given the determinant $\Delta (x) = \left| \begin{array}{ccc} x^n & \sin x & \cos x \ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \ a & a^2 & a^3 \end{array} ight|$.To find the $n^{th}$ derivative of the determinant,we differentiate the rows (or columns) one by one. Since only the first row contains $x$,we differentiate the first row $n$ times.$\frac{d^n}{dx^n}[\Delta (x)] = \left| \begin{array}{ccc} \frac{d^n}{dx^n}(x^n) & \frac{d^n}{dx^n}(\sin x) & \frac{d^n}{dx^n}(\cos x) \ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \ a & a^2 & a^3 \end{array} ight|$.We know that $\frac{d^n}{dx^n}(x^n) = n!$,$\frac{d^n}{dx^n}(\sin x) = \sin(x + \frac{n\pi}{2})$,and $\frac{d^n}{dx^n}(\cos x) = \cos(x + \frac{n\pi}{2})$.Substituting these,we get:$\frac{d^n}{dx^n}[\Delta (x)] = \left| \begin{array}{ccc} n! & \sin(x + \frac{n\pi}{2}) & \cos(x + \frac{n\pi}{2}) \ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \ a & a^2 & a^3 \end{array} ight|$.Now,evaluating at $x = 0$:$[\Delta^n(x)]_{x=0} = \left| \begin{array}{ccc} n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \ a & a^2 & a^3 \end{array} ight|$.Since the first row $(R_1)$ and the second row $(R_2)$ are identical,the value of the determinant is $0$.

If $\Delta (x) = \left| \begin{array}{ccc} x^n & \sin x & \cos x \\ n! & \sin \frac{n\pi}{2} & \cos \frac{n\pi}{2} \\ a & a^2 & a^3 \end{array} \right|$,then the value of $\frac{d^n}{dx^n}[\Delta (x)]$ at $x = 0$ is

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