Simplyfy the following complex rational expressions 2

**mj.alawami** · February 21st 2009, 07:19 AM

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

Please write the answer in steps thank you

**skeeter** · February 21st 2009, 07:29 AM

Originally Posted by mj.alawami

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

Please write the answer in steps thank you

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

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

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

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

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

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

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

**ADARSH** · February 21st 2009, 07:29 AM

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

$ (x+h)^2 = x^2 +h^2 +2hx $

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

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

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

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

**earboth** · February 21st 2009, 07:31 AM

Originally Posted by mj.alawami

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

Please write the answer in steps thank you

$\dfrac{\frac1{(x+h)^2}-\frac1{x^2}}{h}$

The denominator is $x^2(x+h)^2$ :

$\dfrac{\frac1{(x+h)^2}-\frac1{x^2}}{h} = \dfrac{\frac{x^2}{x^2 (x+h)^2}-\frac{(x+h)^2}{x^2(x+h)^2}}{h} = \dfrac{x^2-(x+h)^2}{h\cdot x^2(x+h)^2}$

Expand the bracket in the numerator and collect like terms:

$\dfrac{x^2-(x+h)^2}{h\cdot x^2(x+h)^2} = \dfrac{-2hx-h^2}{h\cdot x^2(x+h)^2}$

Factor h out in the numerator and cancel:

$\dfrac{\frac1{(x+h)^2}-\frac1{x^2}}{h} = \dfrac{-2x-h}{x^2(x+h)^2}$

It seems to me as if you were asked to calculate the drivation of $f(x)=\dfrac1x^2$

If so:

$f'(x)=\lim_{h\to 0}\left( \dfrac{-2x-h}{x^2(x+h)^2}\right) = \dfrac{-2x}{x^4} = \dfrac{-2}{x^3}$

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