# Simplifying a complex rational

• November 14th 2010, 06:03 AM
Foxlion
Simplifying a complex rational
I am having a rather frustrating time simplifying this expression...

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

my first step as instructed by my pre-calculus textbook is to multiply [x^2(x+h)^2] to both the numerator and denominator as a result I see:

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

I think my treatment of fractions is the problem because my next action is to eliminate (x+h)^2 from the num. and denom. of the left expression and x ^2 from the num. and denom. of the right expression leaving me with:

[x^2-(x+h)^2]/[hx^2(x+h)^2]

this is my dead end and I know that the only reason I am here is because I cannot see past the error in my mind, please help me.
• November 14th 2010, 07:26 AM
Foxlion
Do you not multiply the h in the denominator by x^2(x+h)^2 as well?
• November 14th 2010, 07:30 AM
Unknown008
Wait, I did a mistake, sorry...

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

Now you multiply by x^2(x+h)^2 on both numerator and denominator.

$\dfrac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h} = \dfrac{x^2 - (x+h)^2}{hx^2(x+h)^2}$
• November 14th 2010, 09:31 AM
Foxlion
yes and that is where I see no further...I am confused as to how to get to the answer in the book which is stated to be:

-{[2x+h]/[x^2(x+h)^2]} with restrictions such that x and h {-=} 0 and x{-=} -h

the majority of my confusion lies in the origin of the number 2...
• November 14th 2010, 09:40 AM
Unknown008
Right. Expand the numerator. What do you get?
• November 14th 2010, 09:42 AM
skeeter
$\displaystyle \frac{1}{h}\left[\frac{1}{(x+h)^2} - \frac{1}{x^2}\right]
$

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

... note that $(x+h)^2 = x^2 + 2xh + h^2$

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

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

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

$\displaystyle -\frac{2x+h}{x^2(x+h)^2}$
• November 14th 2010, 11:09 AM
Foxlion
Thank you, I apologize for my ignorance of mathematical expression.
• November 14th 2010, 07:54 PM
Unknown008
It's okay, we're here to learn (Smile)