The $n$th-derivative of $q(x) = x^4 - 8x^3 - 4x^2 + 3x - 2$, where $n \le 4$

Question

Some factors will be $\frac{4!}{(4-n)!}\cdot a_nx^n, \frac{3}{(3-n)!} \cdot a_{n-1}x^{n-1}, \ldots, \frac{1!}{(1-n)!} \cdot a_0x^0$, but the lowest degree one will always become zero in the next derivative and I don't know how to express that in a formula. I know $f^{(4)}(x)=4!\cdot a_nx^0$ and all the following derivatives are $0$ since $f^{(4)}(x)$ is a constant but how do I get rid of the lower degree terms in $n$th-derivative notation?

Answer 1

I think it can be done if we do not use the factorial notation.

Let's consider the simplest case $y = x^k$. By writing out the first few derivatives, we easily see that $$y^{(n)} = k(k-1)\cdots(k-n+1) x^{k-n} = \bigg(\prod_{i=1}^n(k+1-i)\bigg)x^{k-n}.$$ There is no problem when the number of times we take the derivative is less than or equal to the degree $k$, that is, no problem when $n \leq k$. But there is also no problem if $n > k$ because somewhere in the product $\prod_{i=1}^n(k+1-i)$, there will be a $0$ term, so that $y^{(n)} = 0$.

So applying the above to the function $q(x)$ in the title, we have $$ q^{(n)}(x) = \bigg(\prod_{i=1}^n(5-i)\bigg) x^{4-n} - 8\bigg(\prod_{i=1}^n(4-i)\bigg) x^{3-n} -4 \bigg(\prod_{i=1}^n(3-i)\bigg) x^{2-n} + 3 \bigg(\prod_{i=1}^n(2-i)\bigg) x^{1-n} -2 \bigg(\prod_{i=1}^n(1-i)\bigg) $$ for any $n \geq 1$.