## Thomas' Calculus 13th Edition

$r(x,y,z)=-(x^2+y^2+z^2)^{-\frac{3}{2}}(xi+yj+zk)$
The question asks to find the gradient field of the given function. $f(x,y,z)=(x^2+y^2+z^2)^{-\frac{1}{2}}$. This is done by finding the gradient of the function. $f_x=-\frac{1}{2}(x^2+y^2+z^2)^{-\frac{3}{2}} \cdot 2x$ Due to symmetry of the function in all directions, simply replace the last term $2x$ with $2y$ and $2z$ for $f_y$ and $f_z$ respectively. Thus, the final gradient field is given by $r(x,y,z)=-(x^2+y^2+z^2)^{-\frac{3}{2}}(xi+yj+zk)$