Higher Order Derivatives Formula 2: Let (*junk*) be a Bezier curve.
The r-th derivative of the curve is given by (*junk*) Click here for
an explanation of notation and the idea of the proof.

Directional derivatives When we discussed derivatives for tensor
product patches, we considered partials because they are easily computed
for those surfaces. The situation is different for triangular patches;
the appropriate derivative here is the directional derivative. Let
u1 and u2 be two points in the domain. Their difference d = u2 - u1
defines a vector. The directional derivative of a surface at x(u)
with respect to d is given by A geometric interpretation: in the domain,
draw the straight line through u that is parallel to d. This straight
line will be mapped to a curve on the patch. Its tangent vector at
x(u) is the desired directional derivative. The partials of a Bezier
patch are not hard to compute; we have, for the u-partial: Working
out similar expressions for the v- and w-partials, we have found our
directional derivative: A closer look at the terms in square brackets
brings to mind the de Castlejau algorithm, and we may write the above
equation as Thus a directional derivative is obtained by performing
one step of the de Castlejau algorithm with respect to the direction
vector d, and n-1 more with respect to the position u. Such configurations
can be succintly expressed in blossom form:

We may continue taking derivatives, arriving at The r-th directional
derivative at b(u) is therefore found by performing r steps of the
de Castlejau algorithm with respect to d, and then by performing n-r
more steps with respect to u. Of course it is irrelevant in which
order we take these steps.

Mixed directional derivatives In the same way, we may compute mixed
directional derivatives: if d1 and d2 are two vectors in the domain,
then their mixed directional derivatives are This blossom result may
also be expressed in terms of Bernstein polynomials. Taking n-r steps
of the de Castlejau algorithm with respect to u, and then r more with
respect to d, gives Or, we might have taken r steps with respect to
d first, and then n-r ones with respect to u. This gives