## Thomas' Calculus 13th Edition

$\nabla g=\nabla F=F(proved)$
We have $F=\nabla F$ and $\int_{0,0,0}^{x,y,z} F \cdot \space dr=\int_{0,0,0}^{x,y,z} \nabla \space F \space dr$ $\implies dr=f(x,y,z)-f(0,0,0)$ Since, $\dfrac{\partial g}{\partial x}=\dfrac{\partial f}{\partial x}-0,\dfrac{\partial f}{\partial y}=0$ This implies that $\nabla g=\nabla F=F(proved)$