Question

What is the solution of:

Answer 1

I write the scalar product like this:
\[a \cdot x = \sum\limits_{j = 1}^3 {{a_j}{x_j}} = {a_j}{x_j}\]
so, we get:
\[\begin{gathered}
{\nabla _i}\cos \left( {a \cdot x} \right) = \frac{\partial }{{\partial {x_i}}}\cos \left( {{a_j}{x_j}} \right) = - \sin \left( {a \cdot x} \right){a_j}{\delta _{ij}} = \hfill \\
\hfill \\
= - \sin \left( {a \cdot x} \right){a_i} \hfill \\
\end{gathered} \]
where \( \delta_{ij} \) is the Kronecker's symbol
therefore:
\[\nabla \cos \left( {\vec a \cdot \vec x} \right) = - - \sin \left( {\vec a \cdot \vec x} \right)\vec a\]

Answer 2

Of course, where there are repeated indices, we have an implicit summation (Einstein notation)