Let $(M, g)$ be a Riemannian manifold, equipped with the Riemannian connection. Let $f:I\to M$ be the constant curve, mapping all elements of $I$ to a point $p$ on the manifold. Let $V$ be a vector field along $f$. Thus $V$ can be thought of as a smooth map $V:I\to T_pM$.
Question. What is the covariant derivative of $V$ along $f$?
By definition, the covariant derivative $\frac{DV}{dt}(t_0)=\nabla_{f'(t_0)}\tilde V$, where $\tilde V$ is any extension of $V$ in a neihborhood of $f'(t_0)$. But $f'(t_0)=0$ since $f$ is constant. Thus $\nabla_{f'(t_0)}\tilde V=0$, and thus the required covariant derivative is also $0$.
This is not the right answer, for Problem 6 of Chapter 2 of Do Carmo's Riemannian Geometry asks to prove that the covariant derivative is same as the derivative of the smooth map $V:I\to T_pM$, which might not be $0$.
I am unable to locate my mistake.
Note that $V$ is not induced by a vector field on $M$. Hence we can not apply proposition 2.2 c)
If $V=\sum_iV_i(t)E_i$ where $E_i$ is a coordinate vector field then $$\frac{DV}{dt}=\sum_i \{V_i'(t) E_i +V_i(t) \nabla_{f'(t)} E_i \} =\frac{dV}{dt} $$