# Directional Devirative question

The easiest way to do this is to use the fact that the gradient vector always points in the direction of fastest increase. That is, $\nabla f= (ay^2+ 3cx^2z^2)\vec{i}+ (2axy+ bz)\vec{i}+ (by+ 2czx^3)\vec{k}$, evaluated at x= 1, y= 2, z= -1, is just $64\vec{k}$.