= Solution
Assume $n\ge1$ and $\|p\|_{[-1,1]}\le1$. The given <Bernstein inequality for algebraic polynomials> shows that $q=p'$ satisfies the nodal hypotheses of (a). Consequently, in the two endpoint regions $\cos(\pi/(2n))\le|x|\le1$, $|p'(x)|\le|T_n'(x)|$.
For $0<\theta<\pi$, the finite geometric identity
$$
\frac{\sin(n\theta)}{\sin\theta}=\sum_{r=0}^{n-1}e^{i(n-1-2r)\theta}
$$
bounds its absolute value by $n$. It follows that $|T_n'(\cos\theta)|\le n^2$, and the endpoint limit gives $T_n'(1)=n^2$. In the central interval, the <Bernstein inequality for algebraic polynomials> gives
$$
|p'(x)|\le\frac{n}{\sqrt{1-x^2}}\le\frac{n}{\sin(\pi/(2n))}\le n^2,
$$
since $\sin t\ge2t/\pi$ on $[0,\pi/2]$. Combining the regions and rescaling proves the <Markov inequality for polynomial derivatives>:
$$
\boxed{\|p'\|_{[-1,1]}\le n^2\|p\|_{[-1,1]},\qquad T_n'(1)=n^2.}
$$
The <Chebyshev polynomial> itself attains equality, so this constant is sharp. Constant <polynomials> have zero <derivative>.
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