<s>[INST] 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.
 [/INST] 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.
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