$\displaystyle 1\over(x+h)^2$ - $\displaystyle 1\over x^2$

Not very good at that LaTeX sort of stuff but here is what im stuck with and have to simplify it.

Its also all over h but no idea how to get that there :/

Anyways any suggestions?

Results 1 to 6 of 6

- Sep 26th 2009, 12:18 PM #1

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## Bleh need some help :(

$\displaystyle 1\over(x+h)^2$ - $\displaystyle 1\over x^2$

Not very good at that LaTeX sort of stuff but here is what im stuck with and have to simplify it.

Its also all over h but no idea how to get that there :/

Anyways any suggestions?

- Sep 26th 2009, 12:27 PM #2

- Sep 26th 2009, 12:28 PM #3
If it's all over h I would multiply that by $\displaystyle \frac{1}{h}$:

$\displaystyle \frac{1}{h}(\frac{1}{(x+h)^2} - \frac{1}{x^2})$

----------------------------

Ignoring that h for now:

$\displaystyle \frac{1}{x^2+2hx+h^2} - \frac{1}{x^2}$

Cross multiply:

$\displaystyle \frac{2x^2+2hx+h^2}{x^2(x^2+2hx+h^2)}$

I admit it's very messy though. What did you need to find?

- Sep 26th 2009, 12:39 PM #4

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See i dont understand how you got the top part? I mean i did get the bottom somehow :/

Everything on here is all from my university pre calculus math class, and i dont remember ANYTHING from math 12 so im having a pretty rough time.

Like i dont remember most of the information from the units i've already done

- Sep 26th 2009, 12:44 PM #5
$\displaystyle \frac{1}{h} \left[\frac{1}{(x+h)^2} - \frac{1}{x^2}\right]$

$\displaystyle \frac{1}{h} \left[\frac{x^2}{x^2(x+h)^2} - \frac{(x+h)^2}{x^2(x+h)^2}\right]$

$\displaystyle \frac{1}{h} \left[\frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x+h)^2}\right]$

$\displaystyle \frac{1}{h} \left[\frac{-2xh - h^2}{x^2(x+h)^2}\right]$

$\displaystyle \frac{1}{h} \left[\frac{-h(2x + h)}{x^2(x+h)^2}\right]$

$\displaystyle -\frac{2x + h}{x^2(x+h)^2}$

- Sep 26th 2009, 12:48 PM #6

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