Formal power series, the Chain Rule and the Product Rule. Definitons
Let $$\mathbb{C}[[x]] := \left\{ \sum_{n\geq 0} a_n x^n : a_n \in \mathbb{C} \right\}$$ be the set of formal power series of $x$.
Exercise
i)
If $F_1(x)$ and $F_2(x)$ are power series in $\mathbf{C}[[x]]$, and $F_3(x)=F_1(x)F_2(x)$ is their product, then their formal derivatives satisfy the usual 'derivative of the product' formula
$$
F_3'(x)=F_1(x)F_2'(x)+F_1'(x)F_2(x).
$$
I tried making a proof using just the substitution and product rules, but I can't seem to come to an answer. Thanks in advance.
 A: We want to show that
$$
F_3'(x) = F_1(x)F_2'(x) + F_1(x)F_2'(x)
$$
In other words, we want to show that the coefficients of $x^n$ on either side are the same. However, by the Cauchy Product, the coefficient of $x^n$ on the LHS and on the RHS is only dependent on finitely many terms in the series describing $F_1(x)$ and $F_2(x)$.
So we may assume without loss of generality that $F_1(x), F_2(x)$ and $F_3(x)$ are polynomials, where we know it is true.
A: It’s good practice working with summations.
Let $F_1(x)=\sum_{n\ge 0}a_nx^n$, $F_2(x)=\sum_{n\ge 0}b_nx^n$, and $F_3(x)=F_1(x)F_2(x)=\sum_{n\ge 0}c_nx^n$, where we know that
$$c_n=\sum_{k=0}^na_kb_{n-k}\;.$$ 
Now
$$\begin{align*}
F_1(x)F_2'(x)&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}nb_nx^{n-1}\\
&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}(n+1)b_{n+1}x^n\\
&=\sum_{n\ge 0}\sum_{k=0}^na_k(n-k+1)b_{n-k+1}x^n
\end{align*}$$
and
$$\begin{align*}
F_1'(x)F_2(x)&=\sum_{n\ge 0}na_nx^{n-1}\sum_{n\ge 0}b_nx^n\\
&=\sum_{n\ge 0}(n+1)a_{n+1}x^n\sum_{n\ge 0}b_nx^n\\
&=\sum_{n\ge 0}\sum_{k=0}^n(k+1)a_{k+1}b_{n-k}x^n\;,
\end{align*}$$
so the coefficient of $x^n$ in $F_1(x)F_2'(x)+F_1'(x)F_2(x)$ is
$$\begin{align*}
&\sum_{k=0}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=0}^n(k+1)a_{k+1}b_{n-k}\\\\
&\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=0}^{n-1}(k+1)a_{k+1}b_{n-k}+(n+1)a_{n+1}b_0\\\\
&\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=1}^nka_kb_{n-k+1}+(n+1)a_{n+1}b_0\\\\
&\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n+1)a_kb_{n-k+1}+(n+1)a_{n+1}b_0\\\\
&\quad=(n+1)\sum_{k=0}^{n+1}a_kb_{n+1-k}\\\\
&\quad=(n+1)c_{n+1}\;,
\end{align*}$$
which is of course the coefficient of $x^n$ in $$F_3'(x)=\sum_{n\ge 0}(n+1)c_{n+1}x^n\;.$$
