= Solution
The remainder in part (b) involves an unspecified $\xi=\xi(x)$. Its existence for every $x$ does not ensure a differentiable selection. Even if such a selection existed and $f$ had an additional <derivative>, the <product rule> and <chain rule> would give
$$
\frac{d}{dx}\left[\frac{f^{(n+1)}(\xi(x))}{(n+1)!}\omega(x)\right]
=\frac{f^{(n+2)}(\xi(x))\xi'(x)\omega(x)+f^{(n+1)}(\xi(x))\omega'(x)}{(n+1)!}.
$$
The first term cannot be discarded or bounded from the given information. Neither $f^{(n+2)}$ nor $\xi'$ is supplied by the original theorem.
For an explicit demonstration, interpolate the <cubic polynomial> $f(x)=x^3$ at zero and one. Then $p_1(x)=x$, and
$$
f(x)-p_1(x)=x^3-x=(x+1)x(x-1).
$$
Differentiating correctly gives
$$
f'(x)-p_1'(x)=x(x-1)+(x+1)(2x-1)=3x^2-1.
$$
Treating the remainder coefficient $x+1$ as a constant would omit $x(x-1)$ and give a wrong answer at every interior point. \b[An existential intermediate point cannot be frozen when differentiating.] Part (i) instead applies <Rolle's theorem> directly to the <polynomial interpolation error>.
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