f(x) to f'(x) help

**snypeshow** · October 28th 2010, 08:22 PM

hello !

this is an excerpt of the solution to a question i had on a test.. I don't understand how you get from the third line (where they use FOIL to multiply) to the fourth line. any ideas?

**blastulation** · October 28th 2010, 08:38 PM

Okay so the first obvious thing is that the two middle terms are the same term, except on is positive and one is negative, so they equal 0. The first term is $((x-1+h)^ \frac {1}2 (x-1=h)^ \frac {1}2$ When multiplying exponentials you add them together, so the exponent becomes 1. The same thing happens to the last term, so you arrive at the simplified answer.

**snypeshow** · October 28th 2010, 08:57 PM

Thank you very much! i didn't consider the exponents!

**TheCoffeeMachine** · October 29th 2010, 04:16 AM

Here again comes my magical way of doing it.

You have $\displaystyle f'(x) = \lim_{x\to{0}}\frac{\sqrt{x-1+h}-\sqrt{x-1}}{h}$

Put $\sqrt{x-1+h} = a[/Math] and <span class=$ [LaTeX ERROR: Convert failed] " alt="\sqrt{x-1+h} = a[/Math] and [LaTeX ERROR: Convert failed] " />[LaTeX ERROR: Convert failed] $and [Math]h = a^2-b^2$ . So:

$\displaystyle f'(x) = \lim_{a\to{b}}\frac{a-b}{a^2-b^2} = \lim_{a\to{b}}\frac{a-b}{(a-b)(a+b)} = \lim_{a\to{b}}\frac{1}{a+b} = \frac{1}{2b} = \frac{1}{2\sqrt{x-1}}$ .

I don't know why it's not taught in schools -- it's far easier, insightful, and less messy.

**Scurmicurv** · October 29th 2010, 04:31 AM

Make sure to understand and memorize the squaring and the conjugate rules and much will be a lot clearer and easier!

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f(x) to f'(x) help