Gradient of a scalar function f is a vector.
I just read a proof of why gradient is a vector. The proof follows from the fact that Directional derivative is not depended on choice of coordinates.
But the book says $[\partial f/\partial x,2\partial f/\partial y,\partial f/\partial z]$ is NOT a vector.
Could you please explain this.
Given a scalar function $f:\>{\mathbb R}^3\to{\mathbb R}$ the formula $${\bf v}(x,y,z):=\left({\partial f\over\partial x},2{\partial f\over\partial y},{\partial f\over\partial z}\right)\tag{1}$$ defines a bona fide vector field in ${\mathbb R}^3$. But this vector field is not canonically attached to $f$; it has been arbitrarily defined my me through the formula $(1)$. This formula is valid only with respect to the $(x,y,z)$-coordinate system given at the beginning.
Contrasting this the vector (resp., the vector field) $\nabla f$ is canonically attached to $f$ through the formula $$df({\bf x}).{\bf X}=\nabla f({\bf x})\cdot{\bf X}\qquad({\bf X}\in T_{\bf x})\ .$$ This formula does not require any additional data (like the "$2$" in $(1)$), is not confined to a particular coordinate system, and only requires that a scalar product $\cdot$ has been defined in ${\mathbb R}^3$.