Cannot understand calculations (Taylor's Theorem) I have been studying Taylor's Theorem in Bartle's book. However, I cannot understand some simple (?) calculations, and this is really bothering me.
Theorem (as stated in the book):
Let $n \in \mathbb{N},$ let $I:=[a, b],$ and let $f: I \rightarrow \mathbb{R}$ be such that $f$ and its derivatives $f^{\prime}, f^{\prime \prime}, \ldots, f^{(n)}$ are continuous on $I$, and that $f^{(n+1)}$ exists on $(a, b)$. If $x_{0} \in I,$ then for any $x$ in $I$ there exists a point $c$ between $x$ and $x_{0}$ such that:
$$
f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f^{\prime \prime}\left(x_{0}\right)}{2 !}\left(x-x_{0}\right)^{2}
$$
$$
+\cdots+\frac{f^{(n)}\left(x_{0}\right)}{n !}\left(x-x_{0}\right)^{n}+\frac{f^{(n+1)}(c)}{(n+1) !}\left(x-x_{0}\right)^{n+1}
$$
Proof:
Let $x_{0}$ and $x$ be given, and let $J$ denote the closed interval with endpoints $x_{0}$ and $x$. We define the function $F$ on $J$ by:
$$
F(t):=f(x)-f(t)-(x-t) f^{\prime}(t)-\cdots-\frac{(x-t)^{n}}{n !} f^{(n)}(t)
$$
for $t \in J$. Then an easy calculation shows that we have:
$$
F^{\prime}(t)=-\frac{(x-t)^{n}}{n !} f^{(n+1)}(t)
$$
The proof continues here, but the rest is fine, I get it.
What I cannot understand is the following:

Then an easy calculation shows that we have:
$$
F^{\prime}(t)=-\frac{(x-t)^{n}}{n !} f^{(n+1)}(t)
$$

There are a lot of terms that just seem to have vanished from the expression regarding $F(t)$, and I just don't know how that happened. Furthermore, the last term $$
F^{\prime}(t)=-\frac{(x-t)^{n}}{n !} f^{(n+1)}(t)
$$
does not make sense to me.
What I got from the derivative of the last term $-\frac{(x-t)^{n}}{n !} f^{(n)}(t)$ was:
$$
\left((x-t)^{n} \cdot f^{(n)}(t)\right)^{\prime}=n(x-t)^{n-1} \cdot(-1) \cdot f^{(n)}(t)-(x-t)^{n} \cdot f^{(n+1)}(t)
$$
Now, dividing by $n!$ what I get is:
$$
\frac{-n(x-t)^{n-1} \cdot f^{(n)}(t)}{n \cdot(n-1) !}-\frac{(x-t)^{n} \cdot f^{(n+1)}(t)}{n !}
$$
which does not even resemble the last term regarding the $F(t)$ expression.
Can someone help?
 A: You have that $F(t)$ is a constant ($f(x)$) plus$$-f(t)-(x-t)f'(t)-\frac{(x-t)^2}2f''(t)-\cdots-\frac{(x-t)^n}{n!}f^{(n)}(t).$$Forget the constant; if you differentiate it, you get $0$. Then:

*

*if you differentiate $-f(t)$, you get $-f'(t)$;

*if you differentiate $-(x-t)f'(t)$, you get $f'(t)-(x-t)f''(t)$;

*if you differentiate $-\frac{(x-t)^2}{2}f''(t)$, you get $(x-t)f''(t)-\frac{(x-t)^2}2f^{(3)}(t)$;

*$\vdots$

*if you differentiate $-\frac{(x-t)^n}{n!}f^{(n)}(t)(t)$, you get $\frac{(x-t)^{n-1}}{(n-1)!}f^{(n)}(t)-\frac{(x-t)^n}{n!}f^{(n+1)}(t)$.

Now, sum up all of this. Everything gets cancelled, except for $-\frac{(x-t)^n}{n!}f^{(n+1)}(t)$.
A: This is a case where an easy calculation really means a tedious but straightforward calculation that I don't want to write down, and not an obvious and simple calculation. There is loads of cancellation that is occurring.
Examine the first few terms
$$f(x) - f(t) - f'(t)(x - t) - f''(t)(x-t)^2/2.$$
Differentiating with respect to $t$, this becomes
$$ [0] - [f'(t)] - [f''(t)(x-t) - f'(t)] - [f'''(t)(x-t)^2/2 - f''(t)(x-t)].$$
I have written brackets around the derivatives of each term in the previous expression, to better see where each term comes from.
The first term is $0$. Note that the $-f'(t)$ in the second term cancels with the $f'(t)$ in the third term. Similarly, the $f''(t)(x-t)$ cancels with the $f''(t)(x-t)$ in the fourth term.
You could prove that this happens more generally with either an inductive argument or with careful use of precise summation notation. The only terms that don't cancel with other terms are the first (which is $0$ on its own) and the last (which is the term Bartle claims).
A: $\big( f^{i}(x) \frac{(x - t)^i}{i!} \big)' = f^{i+1}(x) \frac{(x - t)^i}{i!} + f^{i}(x) \frac{(x - t)^{i-1}}{(i-1)!}$ by product rule.
Sum over $i$, to see that the series is telescoping.
