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

Hello!

Now, I have to count derivative from :

$\displaystyle (\frac{x}{x-1}$

it would of course not be a problem IF the result would not be:

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

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:

$\displaystyle = \frac{x+h}{x+h-1}-\frac{x}{x-1}$

and got of course:

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

can someone help?

Did I make a mistake?

Many thanks!

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:

$\displaystyle = \frac{x+h}{x+h-1}-\frac{x}{x-1}$

and got of course:

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

You forgot that it is:

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

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:

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?

Re: derivative of x/x-1-including h!!!

Quote:

Originally Posted by

**Boo** Hello!

Now, I have to count derivative from :

$\displaystyle (\frac{x}{x-1}$

it would of course not be a problem IF the result would not be:

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

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:

$\displaystyle = \frac{x+h}{x+h-1}-\frac{x}{x-1}$

and got of course:

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

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 $\displaystyle \displaystyle \begin{align*} f(x) = \frac{x}{x - 1} \equiv 1 + \frac{1}{x - 1} \end{align*}$.

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