TL;DR

It’s easy to calculate derivatives of Bézier curves (they’re other Bézier curves).

In Bézier curves we got a look at a great resource for learning about these curves: A Primer on Bézier Curves.

It includes hints on how to calculate the bounding box (parallel to the coordinate axes), leveraging the derivatives. Good! Although… I found it a bit too dispersive to follow.

Main take-aways:

• the derivative of an $n$-degree Bézier curve is an $n-1$-degree Bézier curve;
• it’s easy to find the control point from the $n$ to the $n-1$ degree derivative;
• if you need the derivatives to find the zero-crossing for the maxima and minima, there’s a very quick way to do it with matrices.

The first bullet is explained thoroughly in the linked article, so I will not bother you here.

The other two bullets are somehow complementary. Either you need the derivative per-se, or most probably you need it to find the zeroes for the maxima and minima within the boundary for the parameter $t$ (which is why, in my opinion, there’s no need to bother with the second derivative).

The second bullet can be summarized as follows (again, see the article for the details). The $i$-th control point $Q_i$ of the derivative can be found from the control points of the starting $n$-degree Bézier curve as follow:

\[Q_i = n \cdot (P_{i+1} - P_i)\]

for $i$ that spans from $1$ to $n$ (because the derivative has one degree less, hence one control point less).

This can be expressed in matrix form:

\[\hat{Q} = D_n \cdot \hat{P}\]

where $D_n$ is a $n \times (n+1)$ matrix that expresses the difference between two consecutive items, diagonally:

\[D_n = n \cdot \begin{bmatrix} -1 & 1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & -1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \vdots & -1 & 1 & 0 \\ 0 & 0 & 0 & \vdots & 0 & -1 & 1 \end{bmatrix}\]

This can be pre-combined with the matrix for the $n-1$-degree Bézier curve, which yields:

\[M_{D,n} = M_{n-1} \cdot D_n\]

and this matrix can be used to find the parameters for the parametric components in the X and Y axes *by multiplying it directy times the control points of the original Bézier curve:

\[M_{D,n} \cdot \hat{P}\]

Hence the whole thing boils down to:

• pre-calculate $M_{D,n}$ for the values of $n$ we are interested into (usually 2 and 3), once and for all;
• multiply it times the control points matrix $\hat{P}$, yielding a $n \times 2$ matrix whose columns are the two derivatives for the X and Y axes
• zero them and find the candidate extremes for the bounding box.

For quadratic Bézier curves this means using the following matrix:

\[M_{D,2} = M_1 \cdot D_2 = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \cdot 2 \cdot \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \\ M_{D,2} = 2 \cdot \begin{bmatrix} -1 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\]

For cubic Bézier curves we have instead:

\[M_{D,3} = M_2 \cdot D_3 = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 2 & 0 \\ 1 & -2 & 1 \end{bmatrix} \cdot 3 \cdot \begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \end{bmatrix} \\ M_{D,3} = 3 \cdot \begin{bmatrix} -1 & 1 & 0 & 0 \\ 2 & -4 & 2 & 0 \\ -1 & 3 & -3 & 1 \\ \end{bmatrix}\]