Prove that $(a,b)\to \mathbb{R}$ etc How do I prove that for $n\in \mathbb{N},f,g:(a,b)\to\mathbb{R}$ $n$ times differentiable and $c\in \mathbb{R}$
1st case:  $(f+g)^{(n)}=f^{(n)}+g^{(n)}$
2nd case: $(cf)^{(n)}=cf^{(n)}$
3rd case: $(fg)^{(n)}=\sum^n_{k=0}{n\choose k}f^{(k)}g^{(n-k)}$
 A: You can prove all by induction, if you already know that $(f+g)' = f' + g'$, $(cf)' = cf'$ and $(fg)' = f'g + fg'$.
The first one and the second one are very easy ; the proof of the third is equivalent to the proof of the binomial theorem.
A: $\forall n \ge 1, f^{n} = (f^{n-1})'$ and $f^{(0)}=f$

Let $P_n:"(f+g)^{(n)} = f^{(n)}+g^{(n)}"$
$\cfrac{(f+g)(t+h)-(f+g)(t)}{h} = \cfrac{f(t+h)+g(t+h)-f(t)-f(h)}{h} = \cfrac{f(t+h)-f(t)}{h}+\cfrac{g(t+h)-g(t)}{h} \underset{h\to0}{\longrightarrow}f'(t)+g'(t)$
So $(f+g)'$ exists and $(f+g)'=f'+g'$ so $P_1$ is true.
Suppose $P_n$ true for $n\ge 1$
$(f+g)^{(n+1)} = ((f+g)^{(n)})' \underset{P_n}{=} (f^{(n)}+g^{(n)})' \underset{P_1}{=} (f^{(n)})' + (g^{(n)})' = f^{(n+1)}+g^{(n+1)}$ so $P_{n+1}$ is true.
Since $P_1$ is true and $P_n \Rightarrow P_{n+1}$, $\forall n \ge 1,P_n$ is true.

Let $Q_n:"(cf)^{(n)}=cf^{(n)}"$
$\cfrac{(cf)(t+h)-(cf)(t)}{h}=\cfrac{cf(t+h)-cf(t)}{h}= c\cfrac{f(t+h)-f(t)}{h}\underset{h\to0}{\longrightarrow}cf'(t)$
So $(cf)'$ exists and $(cf)'=cf'$ so $Q_1$ is true.
Suppose $Q_n$ is true for $n\ge 1$
$(cf)^{(n+1)}=((cf)^{(n)})'\underset{P_n}{=}(cf^{(n)})'\underset{P_1}{=}c(f^{(n)})'=cf^{(n+1)}$ so $Q_{n+1}$ is true.
Since $Q_1$ is true and $Q_n \Rightarrow Q_{n+1}$, $\forall n \ge 1,Q_n$ is true.

Let $R_n:"(fg)^{(n)}=\sum\limits^n_{k=0}{n\choose k}f^{(k)}g^{(n-k)}"$
$\cfrac{(fg)(t+h)-(fg)(t)}{h}=\cfrac{f(t+h)g(t+h)-f(t)g(t)}{h} = \cfrac{f(t+h)g(t+h)+f(t+h)g(t)-f(t+h)g(t)-f(t)g(t)}{h} = \cfrac{f(t+h)(g(t+h)-g(t))+g(t)(f(t+h)-f(t))}{h}=f(t+h)\cfrac{g(t+h)-g(t)}{h}+g(t)\cfrac{f(t+h)-f(t)}{h}\underset{h\to0}{\longrightarrow} f(t) g'(t) + g(t)f'(t)$
So $(fg)'$ exists and $(fg)'=f'g+fg'={1\choose 0}f^{(0)}g^{(1-0)}+{1\choose 1}f^{(1)}g^{(1-1)}=\sum\limits^1_{k=0}{1\choose k}f^{(k)}g^{(n-k)}$ so $R_1$ is true.
Suppose $R_n$ true for $n\ge 1$
$(fg)^{(n+1)}=((fg)^{(n)})'\underset{R_n}{=}(\sum\limits^n_{k=0}{n\choose k}f^{(k)}g^{(n-k)})'\underset{P_k}{=}\sum\limits^n_{k=0}({n\choose k}f^{(k)}g^{(n-k)})'\underset{Q_k}{=}\sum\limits^n_{k=0}{n\choose k}(f^{(k)}g^{(n-k)})'\underset{R_1}{=}\sum\limits^n_{k=0}{n\choose k}\left(f^{(k+1)}g^{(n-k)}+f^{(k)}g^{(n-k+1)}\right)=\sum\limits^n_{k=0}{n\choose k}f^{(k+1)}g^{(n-k)}+\sum\limits^n_{k=0}{n\choose k}f^{(k)}g^{(n-k+1)}=\sum\limits^{n+1}_{k=1}{n\choose k-1}f^{(k)}g^{(n-k+1)}+\sum\limits^n_{k=0}{n\choose k}f^{(k)}g^{(n-k+1)}={n\choose 0}f^{(0)}g^{(n-0+1)}+\sum\limits^n_{k=1}{n\choose k-1}f^{(k)}g^{(n-k+1)}+\sum\limits^{n}_{k=1}{n\choose k}f^{(k)}g^{(n-k+1)}+{n\choose n+1}f^{(n+1)}g^{(n-n-1+1)}=f^{(0)}g^{(n+1)}+\sum\limits^n_{k=1}({n\choose k-1}+{n\choose k})f^{(k)}g^{(n-k+1)}+f^{(n+1)}g^{(0)}={n+1\choose 0}f^{(0)}g^{(n+1)} + \sum\limits^n_{k=1}{n+1\choose k}f^{(k)}g^{(n-k+1)}+{n+1\choose n+1}f^{(n+1)}g^{(0)}=\sum\limits^{n+1}_{k=0}{n+1\choose k}f^{(k)}g^{(n-k)}$ so $R_{n+1}$ is true.
Since $R_1$ is true and $R_n \Rightarrow R_{n+1}$, $\forall n \ge 1,R_n$ is true.
