= Solution
The degree-$n$ <Bernstein basis> is
$$
\boxed{b_i^n(t)=\binom ni t^i(1-t)^{n-i},\qquad 0\leq i\leq n.}
$$
For $0\leq t\leq1$, every basis function is nonnegative, and the <binomial theorem> gives
$$
\sum_{i=0}^n b_i^n(t)=[t+(1-t)]^n=1.
$$
Therefore $P(t)=\sum_iP_i b_i^n(t)$ is a <convex combination> of its <control points>. By the definition of their <convex hull>,
$$
\boxed{P([0,1])\subseteq\operatorname{conv}\{P_0,\ldots,P_n\}.}
$$
This includes the endpoints because $P(0)=P_0$ and $P(1)=P_n$. It depends on the specified parameter interval: outside it the weights need not be nonnegative.
For $n\geq1$, differentiating the basis and using the elementary binomial identities gives
$$
\frac{d}{dt}b_i^n(t)=n[b_{i-1}^{n-1}(t)-b_i^{n-1}(t)],
$$
where out-of-range basis indices are zero. Thus
$$
\begin{aligned}
P'(t)&=n\sum_{i=0}^nP_i[b_{i-1}^{n-1}(t)-b_i^{n-1}(t)]\\
&=n\sum_{i=0}^{n-1}(P_{i+1}-P_i)b_i^{n-1}(t).
\end{aligned}
$$
Hence the <Bézier derivative control polygon> has points $n(P_{i+1}-P_i)$, and
$$
\boxed{P'(t)=\sum_{i=0}^{n-1}n\,\Delta P_i\,b_i^{n-1}(t).}
$$
In particular the endpoint <tangent vectors> are $n(P_1-P_0)$ and $n(P_n-P_{n-1})$. For $n=0$ the curve is constant and its <derivative> is zero.
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