On: Calculus Made Easy (1910)

[mettamage]

I notice that I stumble over math over small but important details. I understand the big ideas, but then at chapter 4 in the book it says:

y+dy = (x+dx)^-2

is equal to

x^−2 * (1 + dx/x)^−2

[1]

To me (not that strong at math) this isn't apparent at all.

I have a couple of options here:

1. Spend a couple of hours fiddling around and trying to figure out the answer.

2. Hopefully find some app.

3. Ask a friend.

Regarding the options: I don't have a friend and I don't have an app. If you wouldn't know how to solve this, then what other strategies for understanding this are there?

[1] The LaTeX version:

y+dy &= (x+dx)^{-2} \\ &= x^{-2} \left(1 + \frac{dx}{x}\right)^{-2}

[fesoliveira → mettamage]

From your other post, you might have been a bit tired due to sleep deprivation, so don't be too hard on yourself for not getting it ;)

As for the expression, you just have to expand it:

  y+dy = (x+dx)^-2 
       = 1 / (x+dx)^2 
       = 1 / (x^2 + 2xdx + d^2x) 
       = 1 / [x^2 * (1 + 2dx/x + d^2x / x^2)] 
       = 1 / [x^2 * (1 + dx/x)^2]
       = [x^2 * (1 + dx/x)^2]^-1
       = x^-2 * (1 + dx/x)^-2

It looks like a bit of a jump at first, but he just skipped the expansion of the expression. When I see this kind of thing, it helps me to just mess around with both start and end to see if I can find a way to get from one to the other.

[gspetr → mettamage]

4. Try to look for a different (one you might understand better) explanation of the same concept in different sources.

These can be youtube videos, other books, math.stackexchange.com, math forums, etc.

[mettamage → gspetr]

Haha I'm now looking at Khan Academy.

I went to https://tutorme.com/ and went on a free trial.