function --- derivative (help!)

May 2010
39
0

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 :)
 
Feb 2010
1,036
386
Dirty South

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
 
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skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas

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.
 
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May 2010
39
0
Thanks everyone :)

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

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

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

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

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!

 
Feb 2010
1,036
386
Dirty South
Thanks! i understand it, but what about for sin2x?

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

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

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

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

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

\(\displaystyle = 2cos(2x)\)