How does this derivation of the proper time derivative of a covariant vector work? Define the operator $\frac{D}{D\tau}$ by its action on an arbitrary contravariant vector $A^\lambda$:
$$\frac{DA^\lambda}{D\tau} = \frac{dA^\lambda}{d\tau} + \Gamma^\lambda_{\mu\nu} \frac{dx^\mu}{d\tau} A^\nu$$
(with the motivation that this allows us to express the geodesic equation in a nice form).
Now, in an attempt to deduce the corresponding action on a covariant vector $\frac{DA_\lambda}{D\tau}$, contract with an arbitrary covariant vector $B_\lambda$:
$$\begin{align}
B_\lambda\frac{DA^\lambda}{D\tau} &= B_\lambda\frac{dA^\lambda}{d\tau} + B_\lambda\Gamma^\lambda_{\mu\nu} \frac{dx^\mu}{d\tau} A^\nu \\
&= \frac{d(A^\lambda B_\lambda)}{d\tau} - A^\lambda\frac{dB_\lambda}{d\tau} + B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau} A^\lambda \\
&= \frac{d(A^\lambda B_\lambda)}{d\tau} - A^\lambda\left(\frac{dB_\lambda}{d\tau} - B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau}\right) \\
\end{align}$$
The following is the bit which I don't understand - we claim that the term in brackets must be the derivative of this covariant vector, i.e.
$$
\frac{DB_\lambda}{D\tau} = \frac{dB_\lambda}{d\tau} - B_\nu\Gamma^\nu_{\mu\lambda} \frac{dx^\mu}{d\tau}
$$
I agree that it is definitely a vector, but I don't see how we can make the leap to saying that it is certainly the form that this operator takes when acting on a covariant vector. Is there something I'm missing?
I considered that it might have something to do with the fact that if you substitute this definition and rearrange you obtain
$$ 
B_\lambda\frac{DA^\lambda}{D\tau} + A^\lambda\frac{DB_\lambda}{D\tau} = \frac{d(A^\lambda B_\lambda)}{d\tau}
$$
but I can't see exactly where this leads.
 A: We can start with the following reasonable assumptions: (1) the action of the operator $\frac{D}{D\tau}$ reduces to that of $\frac{d}{d\tau}$ on scalar functions i.e. $$\frac{D\phi}{D\tau}=\frac{d\phi}{d\tau}$$ for any scalar function $\phi$ , and (2) the operator $\frac{D}{D\tau}$ satisfies the Leibniz product rule.
These assumptions fix the action of $\frac{D}{D\tau}$ on a covector ($B_{\lambda}$) as follows.
From assumption (1), keeping in mind that $A^{\lambda}B_{\lambda}$ is a scalar, we have:
$$\frac{D(A^{\lambda}B_{\lambda})}{D\tau} = \frac{d(A^{\lambda}B_{\lambda})}{d\tau}$$
From assumption (2), we apply the Leibniz product rule on the left side (and the Leibniz product rule on the right as with any derivative):
$$
B_{\lambda} \frac{DA^{\lambda}}{D\tau}+A^{\lambda} \frac{DB_{\lambda}}{D\tau} = B_{\lambda}\frac{dA^{\lambda}}{d\tau} + A^{\lambda}\frac{dB_{\lambda}}{d\tau} 
$$
Apply the definition of the action of $\frac{D}{D\tau}$ on the vector $A^{\lambda}$:
$$
B_{\lambda} \left(\frac{dA^{\lambda}}{d\tau} + \Gamma_{\mu \nu}^{\lambda} \frac{dx^{\mu}}{d\tau}A^{\nu} \right)+A^{\lambda} \frac{DB_{\lambda}}{D\tau} = B_{\lambda}\frac{dA^{\lambda}}{d\tau} + A^{\lambda}\frac{dB_{\lambda}}{d\tau} 
$$
$$
B_{\lambda} \frac{dA^{\lambda}}{d\tau} + \Gamma_{\mu \nu}^{\lambda} \frac{dx^{\mu}}{d\tau}B_{\lambda}A^{\nu}  +A^{\lambda} \frac{DB_{\lambda}}{D\tau} = B_{\lambda}\frac{dA^{\lambda}}{d\tau} + A^{\lambda}\frac{dB_{\lambda}}{d\tau}
$$
Rearrange:
$$
A^{\lambda} \frac{DB_{\lambda}}{D\tau} = A^{\lambda}\frac{dB_{\lambda}}{d\tau} - \Gamma_{\mu \nu}^{\lambda} \frac{dx^{\mu}}{d\tau}B_{\lambda} A^{\nu}
$$
Rename the indices on the last term on the right:
$$
A^{\lambda} \frac{DB_{\lambda}}{D\tau} = A^{\lambda}\frac{dB_{\lambda}}{d\tau} - \Gamma_{\mu \lambda}^{\nu} \frac{dx^{\mu}}{d\tau} B_{\nu} A^{\lambda}
$$
As this equation must be true for any vector $A^{\lambda}$, we can remove it from all the terms and conclude with the action of $\frac{D}{D\tau}$ on $B_{\lambda}$:
$$
\frac{DB_{\lambda}}{D\tau} = \frac{dB_{\lambda}}{d\tau} - \Gamma_{\mu \lambda}^{\nu} \frac{dx^{\mu}}{d\tau} B_{\nu}
$$
A: This essentially follows from that $\frac{D}{D\tau}=\nabla_{\dot{x}}$ and from how a connection $\nabla$ acts on tensors.
