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Trouble finding the derivative of a fraction using four step process

cole03k

New member
May 21, 2020
1
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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:

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topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,103
The Astral plane
Note: Don't take short cuts with the fractions. \(\displaystyle \dfrac{1}{x - 6 - h} - \dfrac{1}{x - 6}\) is not \(\displaystyle \dfrac{1}{(x - 6 - h) - (x - 6)}\). No matter what your shorthand is you still have to get a common denominator and subtract the two fractions.

\(\displaystyle f(x) = \dfrac{1}{x - 6}\)

\(\displaystyle f'(x) = \lim_{h \to 0} \dfrac{ \dfrac{1}{x - 6 + h} - \dfrac{1}{x - 6} }{h} \)

\(\displaystyle f'(x) = \lim_{h \to 0} \dfrac{ \dfrac{(x - 6) - (x - 6 + h)}{(x - 6)(x - 6 + h)} }{h}\)

\(\displaystyle f'(x) = \lim_{h \ to 0} \dfrac{ \dfrac{-h}{(x - 6)(x - 6 + h)} }{h}\)

\(\displaystyle f'(x) = \lim_{h \to 0} \dfrac{-1}{(x - 6)(x - 6 + h)}\)

\(\displaystyle f'(x) = - \dfrac{1}{(x - 6)^2}\)

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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!