I would like to find those points where a Bezier curve $\mathbf{C} = [(t(u) , P(u)]^T$ has zero gradient. Following the chain rule, I have tried the following
$$ \frac{dP}{dt} = \left( \frac{dP}{du} \right) \left( \frac{dt}{du} \right)^{-1} $$
The trouble is that setting $\frac{dP}{du}$ to zero imples that $\frac{dt}{du}$, or more specifically $\mathbf{t}(u)$ plays no part in the determining the turning point(s) -- and that I find highly unusual.
Is my thinking right?
The conventional expression for a Bezier curve is $$ \mathbf P(t) = (1-t)^3\mathbf P_0 + 3t(1-t)^2\mathbf P_1 +3t^2(1-t)\mathbf P_2 + t^3\mathbf P_3 $$ Or, writing this out component-wise (assuming a 2D curve): $$ x(t) = (1-t)^3x_0 + 3t(1-t)^2x_1 +3t^2(1-t)x_2 + t^3x_3 \\ y(t) = (1-t)^3y_0 + 3t(1-t)^2y_1 +3t^2(1-t)y_2 + t^3y_3 $$ The gradient $dy/dx$ will be zero when $dy/dt$ is zero. The expression for $x(t)$ is irrelevant in this computation. It sounds like that's what's bothering you. But it should make sense -- the curve's movement in the $x$-direction is not related to its movement in the $y$-direction.