http://img15.imageshack.us/img15/705/123gm.jpg

but instead of f(x)=x^-2, let f(x) = x^(1/2 )

So, I did

[ (x+h)^1/2 - (x)^1/2 ]/ h

But im not sure how to simplify it. What would be my next step? Thanks alot :)

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- May 22nd 2010, 12:01 PMgomesfunction --- derivative (help!)
http://img15.imageshack.us/img15/705/123gm.jpg

but instead of f(x)=x^-2, let f(x) = x^(1/2 )

So, I did

[ (x+h)^1/2 - (x)^1/2 ]/ h

But im not sure how to simplify it. What would be my next step? Thanks alot :) - May 22nd 2010, 12:26 PMharish21
- May 22nd 2010, 01:49 PMskeeter
$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}

$

$\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}

$

$\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

finish it. - May 22nd 2010, 02:37 PMgomes
Thanks everyone :)

Thanks, I'm having trouble with sin(2x), i think my double angle formulae, i might be doing something wrong/missing something. Any help? Most appreciated!

http://img199.imageshack.us/img199/7870/qwertymj.jpg - May 22nd 2010, 02:47 PMgomes
Sorry, one more, im stuck :(

http://img38.imageshack.us/img38/4512/qweryn.jpg - May 22nd 2010, 07:07 PMAllanCuz
I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Reso...principles.pdf

If you have any questions we can answer them here (Hi) - May 23rd 2010, 10:03 AMgomes
Thanks! i understand it, but what about for sin2x?

http://img199.imageshack.us/img199/7870/qwertymj.jpg - May 23rd 2010, 12:47 PMharish21
$\displaystyle \lim_{h \to 0} \frac{sin(2x+2h)-sin(2x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{2}{2} \times \frac{sin(2x+2h)-sin(2x)}{h} $

$\displaystyle = \lim_{h \to 0} \frac{2(sin(2x+2h)-sin(2x))}{2h}$

$\displaystyle = \lim_{h \to 0} \frac{2(sin(X+H)-sin(X))}{H}$ where X=2x and H=2h

$\displaystyle = 2cos(X)$ ; which you proved before doing this.

$\displaystyle = 2cos(2x)$