Throughout, suppose we're working over a field of characteristic $0$.
Let $f(x) = \displaystyle \sum_{k=0}^n a_kx^k$ be a polynomial of degree $n$, so that $a_n \ne 0$. Then its derivative $f'(x) = \displaystyle \sum_{k=0}^{n-1} (k+1)a_{k+1}x^{k}$ has leading coefficient $na_n$, which is nonzero if $n>0$. So if $n>0$ then this has degree $n-1$. Apply some sort of induction argument and you'll see that this implies that a polynomial of degree $n$ (over a field of characteristic $0$) has exactly $n$ nonzero derivatives, and that its $(n+1)^{\text{th}}$ derivative is $0$.
So if you're working over $\mathbb{C}$ then a consequence of the fundamental theorem of algebra is that $\#\text{nonzero derivatives} = \deg f = \#\text{roots}$
This is true over any algebraically closed field of characteristic $0$. In fact, it's true for a polynomial of degree $n$ over any algebraically closed field of characteristic $0$ or $p>n$.