# Thread: function --- derivative (help!)

1. ## function --- derivative (help!)

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

2. Originally Posted by gomes

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..

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

now using the definition

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

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

simplify this and get your proof

3. Originally Posted by gomes

but instead of f(x)=x^-2, let f(x) = x^(1/2 )
$\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

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

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

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

finish it.

4. Thanks everyone

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

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

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

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

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!

5. Sorry, one more, im stuck

6. Originally Posted by gomes
Sorry, one more, im stuck

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

7. 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
Thanks! i understand it, but what about for sin2x?

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

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

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

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

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

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

$= 2cos(2x)$