Showing that $\nabla (\alpha f) = \alpha \nabla f$ for constant $\alpha$

Question

I want show that del of alpha times a vector function for is equal to alpha times del of fun using. Alphar is a constant hence it should be factories out after finding partial derivetives,but how do I show this by applying the rules of differentiation.

Answer 1

\begin{eqnarray} \frac{\partial (\alpha f)}{\partial x_{i}} &=& f \cdot \frac{\partial \alpha}{\partial x_{i}}+ \alpha \frac{\partial f}{\partial x_{i}} \\ &=& 0+ \alpha \frac{\partial f}{\partial x_{i}}\\ &=& \alpha \frac{\partial f}{\partial x_{i}} \end{eqnarray}