In deriving the form for the 4-force in special relativity, we begin with
$$\frac{d}{d\tau}p^{\alpha}=m\frac{d}{d\tau}u^{\alpha}=f^\alpha$$
where $\tau$ is the proper time, m is rest mass.
Since $u^\alpha$ is a 4-vector, we have that $u^\alpha \frac{d}{d\tau}u_\alpha$ is frame invariant. My question is why does $u^\alpha u_\alpha = c^2 \implies u^\alpha \frac{d}{d\tau}u_\alpha = 0$? Is the usual product rule for the derivative $\frac{d}{d\tau}$ applied to the inner product of two 4-vectors valid?
