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A trigonometric polynomial of degree at most $n$ obeys the displayed derivative bound. Iteration gives $\Vert T_n^{(r)}{\Vert }_{\infty }\le n^r\Vert T_n{\Vert }_{\infty }$. Applied to differences of approximants at dyadic degrees, this inequality underlies the inverse theorem for trigonometric approximation. It differs from the endpoint-weighted Bernstein inequality for algebraic polynomials.

Apply the Bernstein inequality for algebraic polynomials to the central interval $|x|\le \cos (\pi /(2n))$, using $\sin (\pi /(2n))\ge 1/n$. On the remaining regions apply the Chebyshev nodal derivative comparison to $p^{\prime }$. Since $T_n^{\prime }(\cos \theta )=n\sin (n\theta )/\sin \theta$, the identity expressing $\sin (n\theta )/\sin \theta$ as a sum of $n$ unit complex phases gives $|T_n^{\prime }|\le n^2$. The endpoint limit is $T_n^{\prime }(1)=n^2$, so the constant is sharp.

Assume $n\ge 1$ and $\Vert p{\Vert }_{[-1,1]}\le 1$. The given Bernstein inequality for algebraic polynomials shows that $q=p^{\prime }$ satisfies the nodal hypotheses of (a). Consequently, in the two endpoint regions $\cos (\pi /(2n))\le |x|\le 1$, $|p^{\prime }(x)|\le |T_n^{\prime }(x)|$.

For $0<\theta <\pi$, the finite geometric identitybounds its absolute value by $n$. It follows that $|T_n^{\prime }(\cos \theta )|\le n^2$, and the endpoint limit gives $T_n^{\prime }(1)=n^2$. In the central interval, the Bernstein inequality for algebraic polynomials givessince $\sin t\ge 2t/\pi$ on $[0,\pi /2]$. Combining the regions and rescaling proves the Markov inequality for polynomial derivatives:The Chebyshev polynomial itself attains equality, so this constant is sharp. Constant polynomials have zero derivative.