## Calculus: Early Transcendentals 8th Edition

$D_{k}f(x,y,z)=∇ f.k$ $=[f_{x}(x,y,z),f_{y}(x,y,z),f_{z}(x,y,z)].(0,0,1)$ $=f_{z}(x,y,z)$ Hence, the statement is true.