Derivative of a taylor polynomial with respect to the center Consider a taylor polynomial of degree $n$ and centered at $a$ :$$T_n(x,a) = \sum\limits_{k=0}^n\dfrac{f^{(k)}(a)}{k!}(x-a)^k$$
Differentiating $T_n$ with respect to $a$ gives a nice result :
$$\frac{\partial}{\partial a} T_n(x,a) = \dfrac{f^{(n+1)}(a)}{n!}(x-a)^n$$
I've been trying to understand this result. It seems to suggest that if we change the center $a$ by a small bit, the change in the value of taylor polynomial depends only on the $(x-a)^n$ term. All other preceding terms vanished. I couldn't wrap my head around why $\frac{\partial}{\partial a}$ doesn't depend on the lesser order terms. I'm pretty sure there exists some intuitive explanation for this. Any help?
(I got above result using telescoping. Kindly let me know if you want to see the work...)
 A: Let's state that the Taylor series has only terms up to $f^{(n)}(a)$, i.e. all further derivatives $f^{(m)}(a) =  0$ for $m >n$. Then careful telescoping gives
$$\frac{\partial}{\partial a} T_n(x,a) \\
= \sum\limits_{k=0}^{n-1}\dfrac{f^{(k+1)}(a)}{k!}(x-a)^k - \sum\limits_{k=1}^n\dfrac{f^{(k)}(a)}{(k-1)!}(x-a)^{k-1} \\
=\sum\limits_{k=0}^{n-1}\dfrac{f^{(k+1)}(a)}{k!}(x-a)^k - \sum\limits_{k=0}^{n-1}\dfrac{f^{(k+1)}(a)}{k!}(x-a)^k \\
= 0
$$
If infinitely many nonzero derivatives exist, then telescoping also gives $\frac{\partial}{\partial a} T_n(x,a)  = 0$ as the trick with the "last term" doesn't exist.
In summary, $\frac{\partial}{\partial a} T_n(x,a)  = 0$ which is interpreted: it doesn't matter where you expand, the function is always the same. As it should be, since Taylor series can be performed about any point as long as the function is continuous and $n$-times (where all further derivatives are zero, or $n$ be $\infty$) differentiable in the considered range.
A: 
I couldn't wrap my head around why $\frac{\partial}{\partial a}$ doesn't depend on the lesser order terms.

For any polynomial $p$ of degree $\le n$ we have
$$
 T_n(f+p; x, a) = T_n(f; x, a) + T_n(p; x, a) = T_n(f; x, a) + p(x)
$$
because the Taylor polynomial of $p$ is $p$ itself. It follows that
$$
\frac{\partial}{\partial a} T_n(f+p; x,a) = \frac{\partial}{\partial a} T_n(f; x,a)
$$
which means that $\frac{\partial}{\partial a} T_n(f; x,a)$ does not change if $f$ is modified in way that changes only $f, f', \ldots f^{(n)}$.
Another interpretation is that $T(\cdot, a)$ is independent of $a$ if and only if $f^{(n+1)} \equiv 0$, i.e. if $f$ is a polynomial of degree $\le n$.
