Suppose $$f=(x-a)^r h \in C[x] \text{ where } h(a)\neq 0$$ prove that $$f'=(x-a)^{r-1} h_1 \text{ where } h_1 \in \mathbb{C}[x]$$ where does not vanish at $a$.
Hint: use product rule
the derivative of $$f=c_0 x^n =c_1 x^{n-1} +\dots + c_{n-1}x +c_n \in C[x]$$
the formal derivative is defined by the usual formulas from calculus:
$$ f'=nc_0 x^{n-1} +(n-1 ) c_1 x^{n-2} + \dots + c_{n-1}+0$$ the following rules of differentiation apply
$$ \begin{aligned} (a f) '&=af'&& a\in C \\ (f+g)'&=f'+g' \\ (fg)'&=f'g+fg' \end{aligned}$$
not to sure if $h$ is just a constant I will first assume that it is then try to show when its not.
leting $f=g h $ where $g(x)=(x-a)^r_1$ and $h=h$. Taking the derivatives $g'(x)=(x-a)^{r_1-1} $ and $h'=h'$
assuming $h$ is a constant $h'=0$ so $$\begin{aligned} f'=g'h+g*h'=(x-a)^{r-1}*h+(x-a)^r_1*0=(x-a)^{r-1}*h \end{aligned} $$
h does not vanish at a.
**Assuming h ** is not a constant $$\begin{aligned} f'&=g'h+g*h'=(x-a)^{r-1}*h+(x-a)^r_1*h' \\ &\vdots ( Not sure) \\f'&=(x-a)^{r-1}*h_1 \end{aligned} $$
where $h_1$ does not vanish at a