Tidal force formula My book explains what happens if we measure the difference in gravitational force by displacing a test mass by $\Delta r\ll r$. They give the following formula:
$$
\Delta F=\frac{2GMm\Delta r}{r^3}.
$$
Now the formule for gravity is
$$
F=-G\frac{Mm}{r^2}
$$
So we get
$$
\Delta F=-G\frac{Mm}{r^2}+G\frac{Mm}{(r+\Delta r)^2}.
$$
This yields to
$$
\Delta F=GMm\left(\frac{1}{(r+\Delta r)^2}-\frac{1}{r^2}\right)=-GMm\left(\frac{\Delta r^2+2\Delta r\cdot r}{r^2(r+\Delta r)^2} \right).
$$
Am I allowed to remove $\Delta r^2$ because it is small? And am I allowed to turn the numerator to $r^4$ just like that? Or can I do this more rigorously?
 A: You are close to the answer, but the way you are dealing with the expression is not the easiest to work with them. Let's start from the beginning:
$$
F(r + \Delta r) = - G Mm \frac{1}{(r+\Delta r)^2} = - G \frac{Mm}{r^2} \frac{1}{(1+\Delta r/r)^2} $$
up to now it is just algebra. Now we expand the denominator using $(1+x)^2 \approx 1 + 2x $:
$$ F(r + \Delta r) \approx - G \frac{Mm}{r^2} \frac{1}{ 1 + 2 \Delta r/r} $$
and now we use $1/(1+y) \approx 1- y$
$$ F(r + \Delta r) \approx - G \frac{Mm}{r^2} (1-2 \Delta r/r) = F(r) + 2G \frac{Mm}{r^3} \Delta r.
$$
Hence
$$
\Delta F = 2G \frac{Mm}{r^3} \Delta r
$$
All these expansions work for small $x$ and $y$; they are proportional to  $\Delta r /r$, so this is the case since $\Delta r \ll r$.
More formally this also follows from
$$
\Delta F(r) = F(r + \Delta r) - F(r) = \frac{F(r + \Delta r) - F(r)}{ \Delta r} \Delta r \approx \frac{dF}{dr} \Delta r
$$
provided $\Delta r \ll r$, that is exactly the condition given.
A: Forget about adding the two fractions, rather take $\frac{1}{r^2}$ outside the bracket and expand $(1+\frac{\Delta r}{r})^{-2}$ using the binomial expansion.
