1. ## Help please: Compute the derivative function

Here's the problem:

$\displaystyle f(x) = \frac{2} {{2x - 1}}$

Here's what I did so far...I have no idea if I'm even on the right track..

$\displaystyle f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{2} {{2(x + h) - 1}} - \frac{2} {{2x - 1}}}} {h}$

$\displaystyle = \mathop {\lim }\limits_{h - > 0} \frac{{\frac{{4x - 2 - 4x - 4h + 2}} {{2x - 1(2x + 2h - 1)}}}} {h} = \mathop {\lim }\limits_{h - > 0} \frac{{\frac{{ - 4h}} {{2x - 1(2x + 2h - 1)}}}} {h}$

2. These things are mostly about the algebra. As you can well see.

$\displaystyle \lim_{h\rightarrow{h}}\frac{\frac{2}{2(x+h)-1}-\frac{2}{2x-1}}{h}$

$\displaystyle =\lim_{h\rightarrow{h}}\left[\frac{2}{h(2x+2h-1)}-\frac{2}{h(2x-1)}\right]$

Now, factor:

$\displaystyle =\lim_{h\rightarrow{h}}\frac{-4}{(2x-1)(2x+2h-1)}$

Now, see it?.

3. Would I just set $\displaystyle h = 0$ which leaves me with

$\displaystyle -4 / (2x-1)^2$

4. Originally Posted by toop
Would I just set $\displaystyle h = 0$ which leaves me with

$\displaystyle -4 / (2x-1)^2$
Yes.