# function --- derivative (help!)

• May 22nd 2010, 12:01 PM
gomes
function --- 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 PM
harish21
Quote:

Originally Posted by gomes
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 :)

wait..

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

now using the definition

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

$\displaystyle \lim_{h \to 0} \frac{{\frac{1}{(x+h)^2}}-{\frac{1}{x^2}}}{h}$

simplify this and get your proof
• May 22nd 2010, 01:49 PM
skeeter
Quote:

Originally Posted by gomes
http://img15.imageshack.us/img15/705/123gm.jpg
but instead of f(x)=x^-2, let f(x) = x^(1/2 )

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

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

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

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

finish it.
• May 22nd 2010, 02:37 PM
gomes
Thanks everyone :)

Quote:

Originally Posted by skeeter
$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

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

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

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

finish it.

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 PM
gomes
• May 22nd 2010, 07:07 PM
AllanCuz
Quote:

Originally Posted by gomes

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 AM
gomes
Quote:

Originally Posted by AllanCuz
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)

Thanks! i understand it, but what about for sin2x?

http://img199.imageshack.us/img199/7870/qwertymj.jpg
• May 23rd 2010, 12:47 PM
harish21
Quote:

Originally Posted by gomes
Thanks! i understand it, but what about for sin2x?

http://img199.imageshack.us/img199/7870/qwertymj.jpg

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

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

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

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

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

$\displaystyle = 2cos(2x)$