Derivative

**madman1611** · Jul 20th 2007, 09:03 PM

Hey, guys. Can someone please assist me with...

In each case, find the derivative dy/dx.
b.) y=x+1/x-1

I obtained the answer using the quotient rule, but we are required to use f'(x)= f(x+h)-f(x)/h (which I'm having trouble using for this question).

**Jhevon** · Jul 20th 2007, 09:12 PM

Originally Posted by madman1611

Hey, guys. Can someone please assist me with...

In each case, find the derivative dy/dx.
b.) y=x+1/x-1

I obtained the answer using the quotient rule, but we are required to use f'(x)= f(x+h)-f(x)/h (which I'm having trouble using for this question).

This one was actually kind of straight forward

$f(x) = \frac {x + 1}{x - 1}$

$\Rightarrow f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

$= \lim_{h \to 0} \frac { \frac {x + h + 1}{x + h - 1} - \frac {x + 1}{x - 1}}{h}$

$= \lim_{h \to 0} \frac {(x - 1)(x + h + 1) - (x + 1)(x + h - 1)}{h(x + h - 1)(x - 1)}$

$= \lim_{h \to 0} \frac {x^2 + xh + x - x - h - 1 - x^2 - xh + x - x - h + 1}{h(x + h - 1)(x - 1)}$

$= \lim_{h \to 0} \frac {-2h}{h(x + h - 1)(x - 1)}$

$= \lim_{h \to 0} \frac {-2}{(x + h - 1)(x - 1)}$

Now take the limit as $h \to 0$

**Soroban** · Jul 21st 2007, 04:12 AM

Hello, madman!

Find the derivative: . $y\:=\:\frac{x+1}{x-1 }$

Defintion: . $f'(x) \;=\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

This can be done in four steps: .**

. . [1] Find $f(x+h)$ . . . replace $x$ with $x+h$ ... and simplify

. . [2] Subtract $f(x)$ . . . subtract the original function ... and simplify

. . [3] Divide by $h$ . . . and simplify

. . [4] Take the limit as $h\to0$

. . . . ** .Am I the only one that teaches it this way?

Here we go . . .

$[1]\;\;f(x+h) \;=\;\frac{(x+h) +1}{(x+h)-1}$

$[2]\;\;f(x+h)-f(x) \;=\;\frac{x+h+1}{x+h-1} \:- \:\frac{x+1}{x-1}$

. . . . . . . . . . . . . $=\;\frac{x-1}{x-1}\cdot\frac{x+h+1}{x+h-1} \:-\:\frac{x+h-1}{x+h-1}\cdot\frac{x+1}{x+1}$

. . . . . . . . . . . . . $=\;\frac{(x^2 + xh + x - x - h - 1) - (x^2 + x + xh + h - x - 1)}{(x-1)(x+h-1)}$

. . . . . . . . . . . . . $= \;\frac{x^2 + xh + x - x - h - 1 - x^2 - x - xh - h + x + 1}{(x-1)(x+h-1)}$

. . . . . . . . . . . . . $=\;\frac{\text{-}2h}{(x-1)(x+h-1)}$

$[3]\;\;\frac{f(x+h)-f(x)}{h}\;=\;\frac{\text{-}2h}{h(x-1)(x+h-1)} \;=\;\frac{\text{-}2}{(x-1)(x+h-1)}$

$[4]\;\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}\frac{\text{-}2}{(x-1)(x+h-1)} \;=\;\frac{\text{-}2}{(x-1)(x+0-1)}$

Therefore: . $f'(x) \;=\;\frac{\text{-}2}{(x-1)^2}$

**Jhevon** · Jul 21st 2007, 08:28 AM

Originally Posted by Soroban

. . . . ** .Am I the only one that teaches it this way?

Apparently so. When I first did calculus (not in the US) no one had trouble with huge formulas, so evaluating it in pieces is not something any teacher thought of doing. In the US, they might have taught it, but usually (in my experience) professors can't pace themselves here, and eventually get into such a rush that your head starts spinning by the time you leave class and you wonder what just happened.

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