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In Calculus Made Easy, on pg. 56, the author performs the following algebra and I can't seem to figure out how they did it.

$$ \begin{align} y + dy & = (x + dx)^{-2} \\[5pt] & = x^{-2} \Bigg(1 + \frac{dx}{x}\Bigg)^{-2} \end{align} $$

It's been a minute since I've been in a calculus class, let alone an algebra class. I'm able to do the follwing, but can't seem to figure out how the 2 was eliminated as well as how the second factor is raised by -2 instead of -1.

$$ \begin{align} y + dy & = (x + dx)^{-2} \tag 1 \\[5pt] & = \frac{1}{x^2 + 2 x \cdot dx + dx^2} \tag 2 \\[5pt] & = \frac{1}{x^2 + 2 x \cdot dx} \tag 3 \\[5pt] & = \frac{1}{x^2 \Biggl(1 + \frac{2 \cdot dx}{x}\Biggr)} \tag 4 \\[5pt] & = x^{-2} \Biggl(1 + \frac{2 \cdot dx}{x}\Biggr)^{-1} \tag 5 \end{align} $$

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  • $\begingroup$ I don't understand your confusion. What $2$ was eliminated? And what factor was raised to the $-2$ and why should it have be raise to $1$? $\endgroup$
    – fleablood
    Mar 4, 2021 at 6:38
  • $\begingroup$ @fleablood the top equation is the author's. The equation with the labels is mine. My step 5 doesn't match the conclusion of the top equation. $\endgroup$ Mar 4, 2021 at 6:42

3 Answers 3

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We have $$(x+dx) = \left[x\left( 1+\frac{dx}{x}\right)\right]$$

Hence \begin{align} (x+dx)^{-2} &= \left[x\left( 1+\frac{dx}{x}\right)\right]^{-2} \\ &=x^{-2}\left( 1+\frac{dx}{x}\right)^{-2} \end{align}

Also notice that

$$\left( 1+\epsilon\right)^2 \approx 1+2\epsilon$$

if $\epsilon$ is small enough, it is the first order approximation.

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  • $\begingroup$ This is why I dropped out after multivariable calculus, I can't remember basic algebra rules and completely forgot this "power of a product" rule for exponents. Immediately clicked when I saw this. @Siong Thye Goh, thank you very much. $\endgroup$ Mar 4, 2021 at 6:51
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The first computation is exact. Your computation is correct up to infinitesimals of order, higher than $dx$. Up to such infinitesimals, both computations are correct.

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There's nothing wrong.

$(1 + \frac {dx}{x})^2 =(1 + \frac {2dx}x + \frac {dx^2}{x^2}) = (1 + \frac {2dx}{x})^{1}$

So either expression is correct.

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