Thanks to the answer here:

how to calculate the value of "t" for the highest point in a quadratic bezier curve?

I know that the derivative of the quadratic bézier curve of

$P(t) = P_0(1−t)^2 + P_12t(1−t) + P_2t^2$


$P′(t) = −2(P_0 − P_1) + 2t_∗(P_0 − P_1) + 2t_∗(P_2 − P_1)$

But I'm afraid I don't understand how that's derived.

I know Wikipedia lists the equation for a bézier derivative as

$B′(t) = n \sum_{i=0}^{n-1} b_{i, n-1} (t)(P_{i+1} - P_i)$

But I frankly understand that even less. Just that it's series notation?

So then the derivative of the linear curve

$P(t) = (1-t)P_0 + tP_1$

would be?

$P′(t) = P_1 − P_0$

(Thanks to bubba)

And what would be the derivative of the cubic curve?

$P(t) = P_0(1-t)^3 + P_13t(1-t)^2 + P_23t^2(1-t) + P_3t^3$

  • 1
    $\begingroup$ The formatting makes the question hard to read (I fixed some of it), and I suspect that there are several typos. $\endgroup$
    – bubba
    Oct 1, 2022 at 7:13
  • 1
    $\begingroup$ And your link to the other answer is broken. $\endgroup$
    – bubba
    Oct 1, 2022 at 7:14
  • $\begingroup$ The derivatives are just calculated the way you would in Calculus 101. You can pretend the points are just numbers. So the answer for the linear case is $P’(t)= P_1 - P_0$. $\endgroup$
    – bubba
    Oct 1, 2022 at 7:19
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – bubba
    Oct 1, 2022 at 7:23
  • $\begingroup$ @bubba: The link to another Question (and its answer) was not so much broken as not proper MarkDown. So I fixed it up. $\endgroup$
    – hardmath
    Oct 5, 2022 at 16:43

1 Answer 1


You just treat the points as numerical constants, and differentiate the way they teach you in calculus classes. So, for the quadratic curve, we have $$ P(t) = P_0\,(1-t)^2 + P_1\,2t(1-t) + P_2\,t^2 $$ We expand the “$t$” terms to get $$ P(t) = P_0\,(1-2t+t^2)+P_1\,(2t - 2t^2) + P_2\,t^2 $$ Then we differentiate, which gives \begin{align*} P’(t) &= P_0\,(-2+2t)+P_1\,(2 - 4t) + P_2\,(2t). \\ &=2\big\{ P_0\,(t-1)+P_1\,(1-2t) + P_2\,(t) \big\} \\ &=2\big\{ P_0\,(t-1)+P_1\,(1-t) - P_1\,(t) + P_2\,(t) \big\} \\ &= 2\big\{ (P_1 - P_0)\,(1-t) + (P_2 - P_1)\,(t) \big\} \end{align*} This is slightly different from the formula you quoted, but it’s nicer because it shows that the derivative of a quadratic (degree 2) Bézier curve is actually a linear (degree 1) Bézier curve.

By similar reasoning, you can show that the derivative of the cubic curve is $$ 3\big\{ (1-t)^2(P_1 - P_0) + 2t(1-t)(P_2 - P_1) + t^2 (P_3- P_2)\big\} $$ So again we can see that the derivative of a cubic Bézier curve is a quadratic Bézier curve.

  • $\begingroup$ I'm afraid I don't follow how you got from step 2 to step 3 in your differentiation of the quadratic. Can you add the intermediate steps? $\endgroup$
    – Darkreaper
    Oct 12, 2022 at 1:20
  • $\begingroup$ I added a step. The key is that $1-2t = (1-t)-t$. $\endgroup$
    – bubba
    Oct 12, 2022 at 23:26

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