Find the tangent of a point on a cubic bezier curve

Question

For a cubic Bézier curve, with the usual four points a, b, c and d,

for a given value t,

how to most elegantly find the tangent at

Answer 1

The tangent of a curve is simply its derivative. The parametric equation that Michal uses:

P(t) = (1 - t)^3 * P0 + 3t(1-t)^2 * P1 + 3t^2 (1-t) * P2 + t^3 * P3

should have a derivative of

dP(t) / dt =  -3(1-t)^2 * P0 + 3(1-t)^2 * P1 - 6t(1-t) * P1 - 3t^2 * P2 + 6t(1-t) * P2 + 3t^2 * P3 

Which, by the way, appears to be wrong in your earlier question. I believe you're using the slope for a quadratic Bezier curve there, not cubic.

From there, it should be trivial to implement a C function that performs this calculation, like Michal has already provided for the curve itself.