## Calculus: Early Transcendentals 8th Edition

As per the definition of a gradient vector $∇f(x,y)=f_{x}i+f_{y}j$ Therefore, $∇f(x,y)=∇(lny)=\frac{∂(lny)}{∂x}i+\frac{∂(lny)}{∂y}j$ $=\frac{1}{y}j$ Thus, $∇f(x,y)=\frac{1}{y}j$ Hence, $∇f(x,y) \ne \frac{1}{y}$