Before you proceed, please click
**here** to review the concept and
formulae of higher derivatives of a Bézier curve.

The control points of the *k*-th derivative curve of a
Bézier curve of degree *n* are computed with the following diagram,
were, point **D**^{0}_{i} is the control point
**P**_{i} of the given curve, and points
**D**^{k}_{0},
**D**^{k}_{1}, ...,
**D**^{k}_{n-k} are the control points
of the *k*-th derivative curve. Note that the south-east (*resp.*,
north-east) bound arrows are associated with a multiplier of -1
(*resp.*, 1).

How do we compute **D**^{k}_{i},
0 <= *i* <= *n-k*, from the given control points?
From an observation discussed on
**de Casteljau's algorithm**
page, we know that only control points
**P**_{i},
**P**_{i+1}, ...,
**P**_{i+k} will contribute to the computation of
**D**^{k}_{i}.
With the same technique used to prove **
the correctness of de Casteljau's algorithm**, we have the
following diagram:

The contribution of **P**_{i+j} to the computation of
**D**^{k}_{i} can only propagate to the
intermediate points in the parallelogram as shown. This parallelogram has
*j* "columns" and *k-j* "rows." Consequently, along each path from
**P**_{i+j} to **D**^{k}_{i}
the coefficient is (-1)^{k-j} because there
*k-j* (-1)'s and *j* 1's. Because there are
*C*(*k*,*j*) such paths, the total contribution of
**P**_{i+j} to the computation of
**D**^{k}_{i} is
(-1)^{k-j}*C*(*k*,*j*). Therefore, we have