Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$? Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$?
Source: calculus made easy by S. Thompson.
I have looked up the formula for binomial theorem with negative exponents but it is confusing. The expansion stated in the text is:
$$\left[1-\frac{2\,dx}{x}+\frac{2(2+1)}{1\cdot2}\left(\frac{dx}{x}\right)^2 - \text{etc.}\right] $$
Please explain at a high school level.

First, recall the expansion $\frac{1}{1-x}=1+x+x^2+\cdots$.  If you don't believe this yet, let $z=1+x+x^2+\cdots$, then $(1-x)z=z-xz=(1+x+x^2+\cdots)-(x+x^2+x^3+\cdots)=1$, so $z=\frac{1}{1-x}$.
Next, we can derive an expansion of $\frac{1}{(1-x)^2}$ by squaring this sequence: $\frac{1}{(1-x)^2}=\left(1+x+x^2+\cdots\right)^2=1+2x+3x^2+\cdots+(i+1)x^i+\dots$ (count how many ways there are to get any particular power, for instance $x^2\cdot 1 + x\cdot x + 1\cdot x^2=3x^2$).
With this, let us put $-\frac{dx}{x}$ into the expansion:
$$1 + 2\left(-\frac{dx}{x}\right) + 3\left(-\frac{dx}{x}\right)^2 + \cdots $$
which is
$$1 - 2\frac{dx}{x} + 3\left(\frac{dx}{x}\right)^2 - 4\left(\frac{dx}{x}\right)^2 + \cdots$$
This doesn't explain why factorials show up in the derivation from the book, but it is an equivalent expansion.
