I am working on a question and I think I have it right but I'm not sure. Now, the question reads:
If $f$ is a function on a surface and $X$ is a vector field, expand the covariant derivative: $$\frac{D(fX)}{\partial u_i}$$
Now, a vector field has $3$ components. Thus, we must have the basis $[u_1, u_2, u_3]$. Then, the Covariant Derivative should become (I think):
$$\frac{D(f)}{\partial u_1}\cdot X + f\cdot\frac{D(X)}{\partial u_1}+\frac{D(f)}{\partial u_2}\cdot X+f\cdot\frac{D(X)}{\partial u_2}+\frac{D(f)}{\partial u_3}\cdot X+f\cdot\frac{D(X)}{\partial u_3}$$
Is this way off? I knew I have to apply the Product rule for the Covariant Derivative, but I'm not sure if I'm doing it right.
Apply the point-wise principle. Since $f: \Sigma \to \mathbb{R}$ and $X: \Sigma \to T\Sigma$ then locally we have,
$$X_p = \sum_j \omega_j(p)\ u^j(p)$$
And so by the point-wise principle,
$$(fX)(p) = f(p) X_p = \sum_j f(p) \cdot\omega_j(p) \ u^j(p)$$
Now we take the covariant derivative over the vector field $V = u^i$. I will use the traditional notation for the covariant derivative $\nabla_VW$, as well as $v[f]$ is the derivative of $f$ in the direction of $v$, and so:
\begin{align*} \nabla_{u^i} fX (p) &= \sum_j u^i\left[f(p) \cdot \omega_j(p)\right] \ u^j(p)\\ &=\sum_j \left(f(p) u^i[\omega_j(p)] + u^i[f]\omega_j(p)\right) \ u^j(p) \\&= \sum_j f(p) u^i[\omega_j(p)] \ u^j(p) + \sum_j u^i[f] \omega_j(p) \ u^j(p) \end{align*}
Hence $\nabla_{u^i} fX = f \nabla_{u^i}X + X \nabla_{u^i}f $.