# derivative of x/x-1-including h!!!

• February 15th 2013, 05:04 AM
Boo
derivative of x/x-1-including h!!!
Hello!
Now, I have to count derivative from :

$(\frac{x}{x-1}$
it would of course not be a problem IF the result would not be:
$-\frac{1}{(x-1)(x-1+h)}$
So, obviously it is the whole procedure including the proof of the rule for the derivative of the rational function and that is what I am asking about.

(Of course I am avare h is almost 0, but I am interested to the whoel procedure how didi they get it!)
I tried like this:
$= \frac{x+h}{x+h-1}-\frac{x}{x-1}$
and got of course:

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

can someone help?
Did I make a mistake?
Many thanks!
• February 15th 2013, 05:23 AM
Plato
Re: derivative of x/x-1-including h!!!
Quote:

Originally Posted by Boo
Hello!
Now, I have to count derivative from :
I tried like this:
$= \frac{x+h}{x+h-1}-\frac{x}{x-1}$
and got of course:

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

You forgot that it is:
$=\frac{ \frac{x+h}{x+h-1}-\frac{x}{x-1}}{h}=\frac{-1}{(x+h-1)(x-1)}$
• February 15th 2013, 06:13 AM
Boo
Re: derivative of x/x-1-including h!!!
O !
Thanks, Plato!!!
So, we obvioulsy get the rule for the derivative of the rational function out of it...
so:
• February 15th 2013, 06:36 AM
HallsofIvy
Re: derivative of x/x-1-including h!!!
Quote:

Originally Posted by Boo
O !
Thanks, Plato!!!
So, we obvioulsy get the rule for the derivative of the rational function out of it...
so:

WHAT? What is the limit as h goes to 0?
• February 15th 2013, 06:41 AM
Prove It
Re: derivative of x/x-1-including h!!!
Quote:

Originally Posted by Boo
Hello!
Now, I have to count derivative from :

$(\frac{x}{x-1}$
it would of course not be a problem IF the result would not be:
$-\frac{1}{(x-1)(x-1+h)}$
So, obviously it is the whole procedure including the proof of the rule for the derivative of the rational function and that is what I am asking about.

(Of course I am avare h is almost 0, but I am interested to the whoel procedure how didi they get it!)
I tried like this:
$= \frac{x+h}{x+h-1}-\frac{x}{x-1}$
and got of course:

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

can someone help?
Did I make a mistake?
Many thanks!

A first principles approach for this problem may be easier if you can see that \displaystyle \begin{align*} f(x) = \frac{x}{x - 1} \equiv 1 + \frac{1}{x - 1} \end{align*}.

\displaystyle \begin{align*} f'(x) &= \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0}\frac{\left( 1 + \frac{1}{x + h - 1} \right) - \left( 1 + \frac{1}{x - 1} \right) }{ h } \\ &= \lim_{h \to 0}\frac{\frac{1}{x + h - 1} - \frac{1}{x - 1}}{h} \\ &= \lim_{h \to 0}\frac{ \frac{x - 1}{\left(x - 1 \right) \left( x + h - 1 \right) } - \frac{x + h - 1}{\left( x - 1 \right) \left( x + h - 1 \right) } }{h} \\ &= \lim_{h \to 0}\frac{ -h}{h \left( x - 1 \right) \left( x + h - 1 \right) } \\ &= \lim_{h \to 0}\frac{-1}{\left( x - 1 \right) \left( x + h - 1 \right) } \\ &= -\frac{1}{ \left( x - 1 \right) \left( x + 0 - 1 \right) } \\ &= -\frac{1}{ \left( x - 1 \right) ^2 } \end{align*}