= Solution
Write $B_j^n(t)=\binom njt^j(1-t)^{n-j}$ for the <Bernstein basis>, and set $B_j^n=0$ when $j$ is outside its index range. Differentiating and using $j\binom nj=n\binom{n-1}{j-1}$ and $(n-j)\binom nj=n\binom{n-1}j$ gives
$$
\frac{d}{dt}B_j^n(t)=nB_{j-1}^{n-1}(t)-nB_j^{n-1}(t).
$$
Hence, after shifting the first sum by one index,
$$
\begin{aligned}
P'(t)&=n\sum_{j=0}^nP_jB_{j-1}^{n-1}(t)-n\sum_{j=0}^nP_jB_j^{n-1}(t)\\
&=n\sum_{j=0}^{n-1}(P_{j+1}-P_j)B_j^{n-1}(t).
\end{aligned}
$$
Thus \b[the <derivative> is a degree-at-most-$(n-1)$ <Bézier curve> whose <control points> are $n(P_{j+1}-P_j)$]. They are vectors rather than necessarily positions on the original <Bézier curve>. This is the <Bézier derivative control polygon>.
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