Bezier curve, change of parametrized variable From the NURBS book (pg 48 B-spline basis functions), given that we have a Bezier curve $C_1(u)$, the following condition holds:
$\frac{1}{u_1 - u_0} C'_1(v = 1)  = C'_1 (u_1)$ for $u_0<u<u_1$...(1)
Where:
$v = \frac{u - u_0}{u_1 - u_0}$ for $0<v<1$
Given that we know the derivative of the endpoint of the Bezier curve of degree n is (which does not depend on the parametric variable but purely on the control points):
$C'_1(v=1) = n(P_n - P_{n-1})$
How does one obtain the expression (1)? I tried the following:
$C'_1(v=1) = C'_1(\frac{u_1-u_0}{u_1-u_0})=n(P_n - P_{n-1})$
$\implies C_1'(u_1-u_0)=n(u_1-u_0)(P_n-P_{n-1})$
$\implies C_1'(u_1)=u_0 +n(u_1-u_0)(P_n-P_{n-1})$
$\implies C_1'(u_1)=u_0+(u_1-u_0)C_1'(v=1)$
 A: The derivative at the end-point of a Bezier curve does depend on the parameterization (not just on the control points).
Typically, we use a parameter $v\in [0,1]$ to describe a Bezier curve in isolation. Then, if the control points are $\mathbf{P}_0, \ldots, \mathbf{P}_m$, the curve is
$$
\mathbf{C}(v) = \sum_{i=0}^m \phi^m_i(v)\mathbf{P}_i
\quad\quad (0 \le v \le 1)
$$
where $\phi^m_i$ is the i-th Bernstein polynomial of degree $m$. Then the derivative at the end-point is:
$$
\frac{d\mathbf{C}}{dv}(v=1) = m(\mathbf{P}_m - \mathbf{P}_{m-1})
$$
But the Bezier segments that form a b-spline curve are not parameterized over the interval $[0,1]$, they use a more general parameter interval, say $u \in [u_0,u_1]$. Then, if we set
$$
v = \frac{u-u_0}{u_1 - u_0}
$$
the equation of the Bezier curve is:
$$
\mathbf{C}(u) = \sum_{i=0}^m \phi^m_i\left(\frac{u-u_0}{u_1-u_0}  \right)\mathbf{P}_i
\quad\quad (u_0 \le u \le u_1)
$$
Then, by the chain rule for differentiation
$$
\frac{d\mathbf{C}}{du}(u=u_1) = 
\frac{d\mathbf{C}}{dv}(v=1) \frac{dv}{du} =
\frac{m(\mathbf{P}_m - \mathbf{P}_{m-1})}{u_1-u_0}
$$
