# Help please: Compute the derivative function

• Feb 11th 2008, 12:17 PM
toop
Help please: Compute the derivative function
Here's the problem:

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

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

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

$= \mathop {\lim }\limits_{h - > 0} \frac{{\frac{{4x - 2 - 4x - 4h + 2}}
{{2x - 1(2x + 2h - 1)}}}}
{h} = \mathop {\lim }\limits_{h - > 0} \frac{{\frac{{ - 4h}}
{{2x - 1(2x + 2h - 1)}}}}
{h}$
• Feb 11th 2008, 12:35 PM
galactus
These things are mostly about the algebra. As you can well see.

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

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

Now, factor:

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

Now, see it?.
• Feb 11th 2008, 04:41 PM
toop
Would I just set $h = 0$ which leaves me with

$
-4 / (2x-1)^2$
• Feb 11th 2008, 06:55 PM
mr fantastic
Quote:

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

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

Yes.