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.
 A: Consider the expression $$A=(1+a)^{-2}=\frac 1{(1+a)^2}=\frac 1 {1+2a+a^2}$$ and, hoping you already know it, perform the long division. Limiting to first terms, you will arrive to $$A=1-2 a+3 a^2-4 a^3+5 a^4+\cdots$$ Now, replace in this last expression $a$ by $\frac{dx}x$; this will lead to $$A=1 - 2\left(\frac{dx}{x}\right) +3\left(\frac{dx}{x}\right)^2 - 4\left(\frac{dx}{x}\right)^3 + 5\left(\frac{dx}{x}\right)^4+\cdots$$ 
Another way, would be to consider  $$B=(1+a)^{-1}=\frac 1{1+a}$$ and perform the long division again. This will give $$B=1-a+a^2-a^3+a^4-a^5+\cdots$$ But $$\frac{dB}{da}=-\frac 1{(1+a)^2}=-A$$ which makes $$-A=\frac{d}{da}\Big(1-a+a^2-a^3+a^4-a^5+\cdots \Big)=-1+2 a-3 a^2+4 a^3-5 a^4+\cdots$$ Multiplying both sides by $-1$ leads to the previous result.
A: 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.
A: It can't be rigorously explained at high school level, because it uses a notion called "limited development" which can be seen as an application or extension of limits. 
But... Some "less" rigorous explanation can be given in the following form, with high school level. 
First, you must know the formula (high school level!) 
(1-q^n)/(1-q)=1+q+q^2+q^3+ ... + q^(n-1) (0) 
This formula is true whenever |q|<1. 
If n tends towards infinity then it becomes 
1/(1-q)=1+q+q^2+q^3+ ... + q^n + ... (1) 
Please note that in (1) the second member is a sum with an infinite numbers of terms (it's called a series) and that's why we put "..." at the end.
Then, there's something interesting that happens when q is "very small" (here, "very small" means "much smaller than 1" ; it may means that |q|<0.01 for example)
If you take q=-2*t-t^2 the formula (1) becomes
1/(1+2t+t^2) = 1 -2t-t^2 + (-2t-t^2)^2 + (-2t-t^2)^3 + ... (2)
If you simplify the second member in (2) and factor the first member it gives you
1/(1+t)^2 = 1 -2t + 3t^2 -4*t^3 +5t^4 + ... (3)
It's almost what you asked for. Just say that t=dx/x and here you are. 
Please note that typically, dx can be defined as an "infinitesimal", that means "very small". It's deeply related to the notations of derivation f'(x)=df/dx and of integration [integral of f(x)dx]
