Prove that if $f$ is divisible by $h^2$ then h divides gcd(f,f')

Question

Let $F$ be a field. For $f = a_0 + a_1x + ... + a_nx^n \in F[x]$, we consider the formal derivative: $f' = a_1 + 2a_2x + ... + n(a_n)^{n-1}$

Prove that if $f$ is divisible by $h^2$ then h divides gcd(f,f')

So, what I have achieved so far is the following:

Suppose $h^2|f$. Then, $f = qh^2 and f'=2hh'q + h^2q$.

Now, since $f = qh^2 and f'=h(2h'q + hq)$, we see that $h|f'$.

I'm not really sure where to go from here to show that $h|gcd(f,f')$

Answer 1

''f′=2hh′q+h2q''.

Nope, its $f'=2hh'q + h^2q'$.

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You are right, $h$ divides $f$ and $h$ divides $f'$. Thus by the definition of the gcd, $h$ divides the gcd of $f,f'$.