Differentiability of the remainder in Taylor's theorem Suppose we have a function that's differentiable $m$ times over $[a,b]$, we have $a< \alpha < x < b$ and $n < m$.  Then
$$
f(x) = \sum_{i = 0}^{n-1} \frac{f^{(i)}(\alpha)}{i!}(x - \alpha)^i + \frac{f^{(n)}(t)}{n!}(x - \alpha)^n 
$$ where $t\in(\alpha, x)$. Now, we can regard $t$ as a function of $x$ (if there are multiple valid $t$ values associated with each $x$ we can impose a condition of $t$ being the max or sup or something else as needed). So suppose we have $t(x)$.  Then under what conditions on $f$ is $t(x)$ and thus $f^{(n)}(t)$ differentiable?  Of course the LHS of the above expression is differentiable as a function of $x$ but I am not sure if that is sufficient to show that $f^{(n)}(t)$ is differentiable... Or it is really that simple?
 A: As Ted Shifrin said, the composition  $f^{(n)}(t(x))$ is differentiable for $x\ne \alpha$  by virtue of 
$$f^{(n)}(t(x)) = \frac{n!}{(x - \alpha)^n} \left\{ f(x) - \sum_{i = 0}^{n-1} \frac{f^{(i)}(\alpha)}{i!}(x - \alpha)^i \right\} \tag1$$
The point $x=\alpha$ takes some more work. Let $R_{n-1}$ denote the content of curly braces in (1). By the integral form of remainder, 
$$
R_{n-1}(x) = \int_\alpha^x \frac{f^{(n)}(s)}{n!}  (x-s)^{n-1}\,ds
$$
Change the variable of integration by $s=\alpha+ u (x-\alpha)$, so that $ds=(x-\alpha)\,du$ and
$$\begin{split}
R_{n-1}(x) & = \int_0^1 \frac{f^{(n)}(\alpha+ u (x-\alpha))}{(n-1)!}  (x-\alpha -u(x-\alpha))^{n-1}\,du \\
& = (x-\alpha)^{n}\int_0^1 \frac{f^{(n)}(\alpha+ u (x-\alpha))}{(n-1)!}  (1 -u)^n \,du
\end{split}$$
Thus, 
$$
f^{(n)}(t(x)) =  n \int_0^1  {f^{(n)}(\alpha+ u (x-\alpha))} \,  (1 -u)^{n-1} \,du \tag2
$$
which is valid for $x=\alpha$ too.
The right hand side of (2) is as smooth  with respect to $x$  as the function $f^{(n)}$. That is, you have $m-n$ derivatives. 

As for $t(x)$ itself... you can try to invert $f^{(n)}$, with success depending on it being invertible and having nonzero derivative.
