$\newcommand{\Reals}{\mathbf{R}}$In our lecture, when we introduced the Levi-Civita connection, we had as an example the directional derivative of a vector field $X$ in direction of another vectorfield $Y$ in $\Reals^n$ defined by $$ D_XY(p) := \lim_{t \to 0} \frac{Y(p + tX(p))-Y(p)}{t}. $$ We have written down that this definition fullfils the Levi-Civita connection definition, but actually I don't even see why it is a connection. For example why does $D_{f \cdot X}Y = f \cdot D_XY$ for an arbitrary $f \in C^{\infty}(\Reals^n)$ hold?
My ideas: Interpreting p as a tangent vector one could use the $\Reals$-linearity of the vectorfield: \begin{align*} D_{f \cdot X}Y(p) &= \lim_{t \to 0} \frac{Y(p + t(f \cdot X)(p))-Y(p)}{t} \\ &= \lim_{t \to 0} \frac{Y(p + tf(p) \cdot X(p))-Y(p)}{t} \\ &= \lim_{t \to 0} \frac{Y(p) + tf(p)Y(X(p))-Y(p)}{t} \\ &= f(p) \cdot \lim_{t \to 0} \frac{tY(X(p))}{t} \\ &= f(p) \cdot \lim_{t \to 0} \frac{Y(p + tX(p)) - Y(p)}{t} \\ &= f(p) D_X Y. \end{align*} Is that correct? Thank you a lot!
The point is that $f(p)$ is just a number, and the derivative depends only on $f(p)$, not on values of $f$ at other points. If $f(p) = 0$, there's nothing to prove. Otherwise, $t f(p) \to 0$ when $t \to 0$, and \begin{align*} D_{f\cdot X} Y(p) &= \lim_{t \to 0} \frac{Y\bigl(p + t(f \cdot X)(p)\bigr) - Y(p)}{t} \\ &= \lim_{t \to 0} \frac{Y\bigl(p + tf(p)X(p)\bigr) - Y(p)}{t} \\ &= f(p)\lim_{t \to 0} \frac{Y\bigl(p + tf(p)X(p)\bigr) - Y(p)}{tf(p)} \\ &= f(p) \lim_{t \to 0} \frac{Y\bigl(p + tX(p)\bigr) - Y(p)}{t} \\ &= f(p) D_{X} Y. \end{align*}