How strong are the tides raised by Io on Jupiter relative to the ones raised by the Moon on Earth? There seems to be 2 ways of calculating tidal forces that are contradictory. Either:

*

*By calculating the difference of Io's gravitational acceleration on a point on Jupiter's near side and the gravitational acceleration felt by a point on Jupiter's far side, using the equation:

$$\frac{GM_{io}}{(d-r_{jupiter})^{2}}-\frac{GM_{io}}{(d+r_{jupiter})^{2}}$$
For Io, this is $24.0807\times10^{-6}$ Newtons, while for the Moon it is $2.2009 \times 10^{-6} $ Newtons.
So if you were to calculate tidal forces that way, the tides raised on Jupiter by Io would be $10.94$ times as strong as the tides raised by the Moon on Earth

However there is also another way I found, which is:


*By using the derivative of Newton's law of universal gravitation, using the equation:

$$\frac{-2G\times M_{jupiter}\times{m_{io}}}{d^{3}}$$
For Io, this is $3.015 \times 10^{14}$ Newtons per meter, whereas for the Moon it is $1.031 \times 10^{12}$ Newtons per meter.
So if you were to calculate tidal forces that way, the tides raised on Jupiter by Io would be $292.5$ times as strong as the tides raised by the Moon on Earth

So, which one is it? Are the tides raised on Jupiter by Io 10.94 or 292.5 times as strong as the ones raised by the Moon on Earth? My intuition leads me to believe the first answer is correct since it takes the primary's radius into account but I just want a concise answer.
 A: The formulae are not quite correct. The first one gives an acceleration, not a force, so it should be in $m/s^2$, not Newtons.
The second is a force divided by a distance, which is not Newtons either.
If you replace $M_\text{jupiter}$ with $r_\text{jupiter}$, then it is an acceleration. Up to a factor of 2 it is the acceleration obtained by Taylor expanding the first formula in $r_\text{jupiter}/d$, which is approximately $70,000/400,000 \sim 0.175$, so the approximation is only good to about 20%. The more accurate answer is the first one.
A: In your first part the formula for the tidal acceleration difference
$$a_\text{tide}=
\frac{GM_{io}}{(d-r_{jupiter})^{2}}-\frac{GM_{io}}{(d+r_\text{jupiter})^{2}} \tag{1}$$
is correct. However, because it is an acceleration, the result
needs to be in m/s$^2$ or Newton/kg, but not in Newton.
Plugging in the numbers and units the result is $a_\text{tide}=24\cdot 10^{-6}\text{ N/kg}$
which is the same result as yours (except for the unit).
Because $r_\text{jupiter}\ll d$ in equation (1) you can approximate it by using differential calculus:
$$\begin{align}a_\text{tide}
&=\frac{GM_\text{io}}{(d-r_\text{jupiter})^{2}}-\frac{GM_\text{io}}{(d+r_\text{jupiter})^{2}} \\
&\approx\left.\frac{\partial}{\partial r}\left(\frac{GM_{io}}{r^2}\right)\right|_{r=d} (-2 r_\text{jupiter}) \\
&=\left.-\frac{2GM_\text{io}}{r^3}\right|_{r=d} (-2 r_\text{jupiter}) \\
&=\frac{4GM_\text{io}r_\text{jupiter}}{d^3}
\tag{2}\end{align}$$
Plugging in the numbers and units here you get
the result $a_\text{tide}=22\cdot 10^{-6}\text{ N/kg}$
which is roughly the same as from (1).
Anyway, this tidal acceleration difference, as calculated from (1)
or (2), seems to be a meaningful indicator of the tidal effect.

What is meaning of the expression given in your second part
$$\frac{-2G M_{jupiter}{m_{io}}}{d^{3}}$$
is not clear to me, especially because of its obscure dimension (Newton/m).
