Units of the derivative divided by the original function

Question

Let's say you have a function of time $V(t)$ which tells you the amount of volume in liters of a tank after $t$ minutes. The units of $V(t)$ are $L$ and the units of $V'(t)$ are $L/min$. But what are the units of $$\dfrac{d}{dt}\left(\dfrac{V'(t)}{V(t)}\right)$$ Would they be $1/min^2$? That doesn't seem right but it's what I'm getting if I attach the units to each expression and divide them. $\frac{(\frac{L}{min})}{L}=\frac{1}{min}$. However I know $$\dfrac{V'(t)}{V(t)}=\dfrac{d}{dt}ln(V(t))$$ So wouldn't that actually make the units $L/min^2$ but on a logarithmic scale?