What is the cut-off point of $\frac{\mathrm{d}F}{\mathrm{d}r}$ beyond which no tidal locking will occur? I am a beginner, please pardon my non-use of proper astronomical terms. I would appreciate proper edits to my questions.

I have learned that when the $\frac{\mathrm{d}F}{\mathrm{d}r}$ is a very big negative number, that would mean that the far side will have much less gravitational tug than the near side of a satellite. The differential gravity or $\frac{\mathrm{d}F}{\mathrm{d}r}$ will be as follows

$$\frac{\mathrm{d}F}{\mathrm{d}r} = \frac{-2 GMm}{r^3}$$

If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Earth-Moon system we get $-1.308 \times 10^{-5} \ \mathrm{M_o^2G/AU^3} $

I was surprised to know that Ganymede, being almost 3 times as far away from Jupiter as our Moon is from Earth, still is tidally locked.

If we compute $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Jupiter-Ganymede system we get $-1.565 \times 10^{-6} \ \mathrm{M_o^2G/AU^3}$

I then wondered what is the $\frac{\mathrm{d}F}{\mathrm{d}r}$ of Sun-Earth that our Earth is not tidally locked to it and I found $-2.422 \times 10^{-8} \ \mathrm{M_o^2G/AU^3}$ 

Which is a very small negative number so the difference in tug is not significant.

So what is the minimum $\frac{\mathrm{d}F}{\mathrm{d}r}$ (or I guess "maximum") value or a boundary value at which the object can be tidally locked and beyond that, the tidal locking won't occur?
I can't comment here because I don't have enough reputation, but this is very related to your problem.

https://en.wikipedia.org/wiki/Tidal_locking#Timescale

Tidal locking will always occur albeit potentially not in a Hubble time. In order for there to be no tidal locking, you would need:

$$\frac{\mathrm{d}F}{\mathrm{d}r} = 0$$

Which any extended object will not have. Problems like this always involve timescales and not definite yes/no answers. I hope this helped.