Help please: Compute the derivative function

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February 11th 2008, 11:17 AM
toop

Help please: Compute the derivative function

Here's the problem:

$f(x) = \frac{2}<br /> {{2x - 1}}$

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

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

$= \mathop {\lim }\limits_{h - > 0} \frac{{\frac{{4x - 2 - 4x - 4h + 2}}<br /> {{2x - 1(2x + 2h - 1)}}}}<br /> {h} = \mathop {\lim }\limits_{h - > 0} \frac{{\frac{{ - 4h}}<br /> {{2x - 1(2x + 2h - 1)}}}}<br /> {h}$
February 11th 2008, 11:35 AM
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?.
February 11th 2008, 03:41 PM
toop

Would I just set $h = 0$ which leaves me with

$<br /> -4 / (2x-1)^2$
February 11th 2008, 05:55 PM
mr fantastic

Quote:

Originally Posted by toop

Would I just set $h = 0$ which leaves me with

$<br /> -4 / (2x-1)^2$

Yes.

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