I know that if $\int(\frac{1}{x}dx) =ln(x) $, then $\frac{1}{d(ln(x))}=\frac{x}{dx}$
My first instinct is to just replace every differential operator with a nabla, but how do I work through this to determine if this rule- or any for that matter -is still true in multiple dimensions, i.e., that $\frac{1}{\nabla(ln(x))}=\frac{x}{\nabla(x)}$ where x is now a vector rather than a scalar?