# Solving Bezier cubic derivative for t

Getting the derivative of the cubic Bezier curve:

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

Produces the following:

$$P'(t)=3(-P_0-2P_1)+6t(P_0+P_1+P_2)+3t^2(P_3-P_0)$$

Assuming P'(t)=0, is it possible to solve for t?

Let’s assume that your points are 2D, say $$P_0 = (x_0,y_0)$$, $$P_1= (x_1,y_1)$$, and so on.
Then the equation $$P’(t)=0$$ is actually two equations that need to be satisfied: \begin{align} 3(−x_0−2x_1)+6𝑡(x_0+x_1+x_2)+3𝑡^2(x_3−x_0) &=0 \\ 3(−y_0−2y_1)+6𝑡(y_0+y_1+y_2)+3𝑡^2(y_3−y_0) &=0 \\ \end{align} Each of these is a quadratic, which you can solve using the high-school formula, to get values of $$t$$. Then you can check if there’s a value of $$t$$ that satisfies both equations.
For a general cubic, it’s highly unlikely that you will find a point where $$P’(t)=0$$. The equation of $$P’$$ represents a parabola, and $$P’(t)=0$$ means that the parabola passes through the origin, which probably won’t happen, in general.
If your points are 3D, then three equations will have to be satisfied simultaneously to get $$P’(t)=0$$, so this is even less likely to happen.
Maybe you’re thinking that $$P’(t)=0$$ corresponds to places where the curve tangent is horizontal. That’s not true. The tangent is horizontal where $$y’(t)=0$$, and you can find such places by solving the second of the two equations shown above: $$3(−y_0−2y_1)+6𝑡(y_0+y_1+y_2)+3𝑡^2(y_3−y_0) =0$$ This equation won’t always have (real) solutions, but it often will.