If y = a_0 + a_1x + a_2x^2 + dots + a_nx^n,th | vedclass

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(C) Given the polynomial function $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$. To find the $n^{th}$ derivative $y_n$,we differentiate $y$ with respect

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$n!a_n$

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(C) Given the polynomial function $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$. To find the $n^{th}$ derivative $y_n$,we differentiate $y$ with respect to $x$ repeatedly $n$ times. The first derivative is $y_1 = \frac{dy}{dx} = a_1 + 2a_2x + 3a_3x^2 + \dots + na_nx^{n-1}$. The second derivative is $y_2 = \frac{d^2y}{dx^2} = 2 	imes 1 a_2 + 3 	imes 2 a_3x + \dots + n(n-1)a_nx^{n-2}$. Continuing this process,the $k^{th}$ derivative of $x^n$ is given by $\frac{d^k}{dx^k}(x^n) = \frac{n!}{(n-k)!}x^{n-k}$. For $k=n$,the $n^{th}$ derivative of $x^n$ is $\frac{n!}{(n-n)!}x^{n-n} = n!$. All terms with powers of $x$ less than $n$ will become zero after $n$ differentiations. Therefore,$y_n = \frac{d^n}{dx^n}(a_nx^n) = a_n 	imes n!$.

If $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$,then find the $n^{th}$ derivative $y_n$.

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