Covariant derivative. Is this solution correct?

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$\textbf{Problem:}$ If $(u_1,u_2)$ are coordinates on a surface S, define the covariant derivatives $DV/\partial{u_i}$ of a vector field V in the $u_i$-direction, in terms of ordinary derivatives in $R^3$. Show that $$ \frac{\partial}{\partial{u_i}} \langle V, W \rangle = \left\langle \frac{DV}{\partial{u_i}}, W \right\rangle + \left\langle V, \frac{DW}{\partial{u_i}} \right\rangle $$

To show the equality, I wrote:

$$ \left\langle \frac{DV}{\partial{u_i}}, W \right\rangle =\left\langle\frac{\partial{V}}{\partial{u_i}} - \left(\frac{\partial{V}}{\partial{u_i}} \cdot N\right)N, W\right\rangle = \left\langle\frac{\partial{V}}{\partial{u_i}}, W\right\rangle - \left(\frac{\partial{V}}{\partial{u_i}} \cdot N\right) \langle N, W \rangle $$ and the last term is zero since $W$ is in the tangent plane to which the normal is perpendicular. Similarly, we have $$ \left\langle V, \frac{DW}{\partial{u_i}} \right\rangle = \left\langle V, \frac{\partial{W}}{\partial{u_i}}\right\rangle $$ It follows that $$ \frac{\partial}{\partial{u_i}} \langle V, W \rangle = \left\langle\frac{\partial{V}}{\partial{u_i}}, W\right\rangle + \left\langle V, \frac{\partial{W}}{\partial{u_i}}\right\rangle = \left\langle \frac{DV}{\partial{u_i}}, W \right\rangle + \left\langle V, \frac{DW}{\partial{u_i}} \right\rangle $$