Trouble finding the derivative of a fraction using four step process

[cole03k]

I am trying to find the derivative of this problem using the four step process but keep getting stuck when it comes to the third step of f(x+h) - f(x). I do not know what to do once I reach that step. Am I canceling terms out incorrectly? How should I deal with a fraction over a fraction? Any help would be really appreciated.

Work:

 

[topsquark]

Note: Don't take short cuts with the fractions. [math]\dfrac{1}{x - 6 - h} - \dfrac{1}{x - 6}[/math] is not [math]\dfrac{1}{(x - 6 - h) - (x - 6)}[/math]. No matter what your shorthand is you still have to get a common denominator and subtract the two fractions.

[math]f(x) = \dfrac{1}{x - 6}[/math]

[math]f'(x) = \lim_{h \to 0} \dfrac{ \dfrac{1}{x - 6 + h} - \dfrac{1}{x - 6} }{h} [/math]

[math]f'(x) = \lim_{h \to 0} \dfrac{ \dfrac{(x - 6) - (x - 6 + h)}{(x - 6)(x - 6 + h)} }{h}[/math]

[math]f'(x) = \lim_{h \ to 0} \dfrac{ \dfrac{-h}{(x - 6)(x - 6 + h)} }{h}[/math]

[math]f'(x) = \lim_{h \to 0} \dfrac{-1}{(x - 6)(x - 6 + h)}[/math]

[math]f'(x) = - \dfrac{1}{(x - 6)^2}[/math]

-Dan

 

[HallsofIvy]

The problem appears to be that you have never actually learned arithmetic!
First, $-\frac{a}{b}$ is NOT equal to $\frac{-a}{-b}$.
Second, $\frac{1}{a}- \frac{1}{b}$ is NOT equal to $\frac{1}{a- b}$..

Learn arithmetic and algebra before you attempt to learn Calculus!