Find the n -th derivative of y =

Find the $n$-th derivative of $y = a_{0}x^{n}+a_{1}x^{\,n-1}+\cdots+a_{n-1}x+a_{n}$.

Step 1: The $n$-th derivative of a single power

From Problem 287, for any integer $m\ge0$: \[\frac{d^{n}}{dx^{n}}x^{m} = \begin{cases} \dfrac{m!}{\bigl(m-n\bigr)!}\,x^{\,m-n}, & m\ge n,\\[6pt] 0, & m\lt n. \end{cases} \tag{1}\]

The second line is the decisive one: differentiating a power more times than its degree annihilates it.

Step 2: Apply term by term

Differentiation is linear, so each term may be treated separately. The term $a_{k}x^{\,n-k}$ has degree $n-k$.

The leading term $\bigl(k = 0\bigr)$ has degree exactly $n$, so by the first line of (1):

\[ \frac{d^{n}}{dx^{n}}\bigl(a_{0}x^{n}\bigr) = a_{0}\cdot n!. \]

Every other term $\bigl(k\ge1\bigr)$ has degree $n-k\le n-1\lt n$, so by the second line of (1) it vanishes.

Step 3: Sum

\[ y^{(n)} = a_{0}\,n!+0+\cdots+0 = a_{0}\,n!. \]

Step 4: Verification

\[y = 3x^{2}+5x-7:\quad y' = 6x+5,\ y'' = 6 = 3\cdot2! \;\;\checkmark\]

\[y^{(n+1)} = \bigl(a_{0}n!\bigr)' = 0, \text{ as required of a degree-}n\text{ polynomial} \;\;\checkmark\]

\[\text{Problem 287 is the case } a_{0} = 1,\ a_{1} = \cdots = a_{n} = 0 \;\;\checkmark\]