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 March 29th, 2009, 04:15 AM #1 Newbie   Joined: Mar 2009 Posts: 20 Thanks: 0 Quick derivative question For finding the derivative of $f(x)= \frac{x-2}{x+2}$ would it be $\frac{x-2+h}{x+2+h}$ or $\frac{x-2}{x+2} + h$?
 March 29th, 2009, 04:51 AM #2 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Quick derivative question Neither. What do you know about the definition of the derivative and limits? I removed my math symbol program form this computer, but the definition is in any text.
 March 29th, 2009, 05:03 AM #3 Newbie   Joined: Mar 2009 Posts: 20 Thanks: 0 Re: Quick derivative question Oops, left out the rest of the formula. Would it be $\frac{\frac{x-2+h}{x+2+h} - \frac{x-2}{x+2}}{h}$ or $\frac{\frac{x-2}{x+2} + h - \frac{x-2}{x+2}}{h}$ ?
March 29th, 2009, 05:59 AM   #4
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Quote:
 Originally Posted by funsize999 For finding the derivative of $f(x)= \frac{x-2}{x+2}$
From your later post, I will guess that you are needing to "find the derivative from the definition", and that this definition is as follows:

[color=white]. . . . .[/color]$\begin{array}{c}\mbox{limit}\\h\rightarrow 0\end{array}\; \frac{f(x\,+\,h)\, -\, f(x)}{h}$

If so, then start with what you learned back in algebra about evaluating functions, and work in smaller steps.

[color=white]. . . . .[/color]$f(x)\,=\, \frac{x\, -\, 2}{x\, +\, 2}$

[color=white]. . . . .[/color]$f(x\, +\, h)\,=\, \frac{(x\, +\, h)\, -\, 2}{(x\, +\, h)\, +\, 2}$

Then use what you learned back in algebra about subtracting rational expressions to simplify the numerator of the limit expression first:

[color=white]. . . . .[/color]$f(x\, +\, h)\, -\, f(x)\,=\, \frac{(x\, +\, h)\, -\, 2}{(x\, +\, h)\, +\, 2}\,-\,\frac{x\, -\, 2}{x\, +\, 2}$

[color=white]. . . . .[/color]$=\, \frac{(x\, +\, h\, -\, 2)(x\, +\, 2)\, -\, (x\, -\, 2)(x\, +\, h\, +\, 2)}{(x\, +\, h\, +\, 2)(x\, +\, 2)}$

[color=white]. . . . .[/color]$=\, \frac{(x^2\, +\, xh\, +\, 2h\, -\, 4)\, -\, (x^2\, +\, xh\, -\, 2h\, -\, 4)}{(x\, +\, h\, +\, 2)(x\, +\, 2)}$

[color=white]. . . . .[/color]$=\, \frac{4h}{(x\, +\, h\, +\, 2)(x\, +\, 2)}$

Now plug this into the derivative limit expression:

[color=white]. . . . .[/color]$\begin{array}{c}\mbox{limit}\\h\rightarrow 0\end{array}\; \frac{\left(\frac{4h}{(x\, +\, h\, +\, 2)(x\, +\, 2)}\right)}{h}\,=\, \begin{array}{c}\mbox{limit}\\h\rightarrow 0\end{array}\; \frac{4}{(x\, +\, h\, +\, 2)(x\, +\, 2)}$

Then apply the limit as h tends toward zero.

 March 29th, 2009, 07:21 AM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Quick derivative question Limit as h approaches 0, not infinity. You can also write $\frac{x-2}{x+2} \mbox{ as } \frac{(x+2)-4}{x+2}= 1-\frac{4}{x+2}$.
 March 29th, 2009, 07:41 AM #6 Newbie   Joined: Mar 2009 Posts: 7 Thanks: 0 Re: Quick derivative question This is the formulae to derivate a rational function good luck
 March 29th, 2009, 08:53 PM #7 Newbie   Joined: Mar 2009 Posts: 20 Thanks: 0 Re: Quick derivative question Thanks for the help guys, I think I've got it now If a question asked to find the rate of change in a graph (the graph being a parabola) would I use the derivative?

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