Derivative, via definition.

So I'm going over my practice midterms (which all seem to have solutions like this one),

Can anyone help clarify this for me? I understand that you multiply by the reciprocal to get to line two. But after that I'm completely lost, I don't understand how:

$$x^{2} + 1 - [(x + h)^{2} + 1]$$

can become:

$$(x-(x+h))(x+x+h)$$

and so forth, I'm sorry if this is a stupid question the solution doesn't seem to explain it very well.

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They got a bit fancy. The $1$'s cancel and you get a difference of two squares. It would be less mysterious to expand $(x+h)^2$ as $x^2+2xh+h^2$. – André Nicolas Feb 27 '13 at 3:51

This is one of those "verification left to the reader" moments.

If it helps, we can use the intermediate step that $$x^2-1-[(x+h)^2+1]=x^2-(x+h)^2,$$ so the conclusion follows from the fact that $a^2-b^2=(a-b)(a+b).$

Too, they didn't explain how they got from the third line to the fourth, but since $$(x-(x+h))(x+(x+h))=-h(2x+h),$$ that should be fairly straightforward. You'll run into this kind of thing a lot. If you're ever uncertain how they got there, just see if you can get there through some intermediate steps.

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Well, $x^2+1-[(x+h)^2+1]=x^2-(x+h)^2=(x-(x+h))(x+(x+h))$. The last equality comes from the difference of two squares.

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This is easier to see if we work backward:

\begin{align*}(x-(x+h))(x+x+h) & = (x-(x+h))(x+(x+h)) \\ & = x^2 -(x+h)^2 \\ & = x^2 + 1 -(x+h)^2 - 1\\ & = (x^2 + 1) -[(x+h)^2 + 1] \end{align*}

And, as others have pointed out, the trick on the first line here is to see a difference of two squares. This is fairly simple working backward like this, but you might want to manually expand all of the terms yourself, and then simplify and factor again.

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