Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 17 / c

The gradient is $\nabla f=(2x,z,y)$, so its value at the point is $(2,-3,1)$. Since the specified direction is already a unit vector, the directional derivative is its dot product with that gradient:

For a unit direction $\mathbf{v}$, the Cauchy-Schwarz inequality gives $\nabla f\cdot \mathbf{v}\le |\nabla f|$, with equality in the gradient direction. Its norm here is $\sqrt{4+9+1}=\sqrt{14}$, so the direction of fastest increase isThe maximal directional derivative itself is $\sqrt{14}$.