G gomes May 2010 39 0 May 22, 2010 #1 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

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

harish21 Feb 2010 1,036 386 Dirty South May 22, 2010 #2 gomes said: 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 Click to expand... 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 Reactions: gomes

gomes said: 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 Click to expand... 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

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 22, 2010 #3 gomes said: but instead of f(x)=x^-2, let f(x) = x^(1/2 ) Click to expand... \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \) \(\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \) \(\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\) finish it. Reactions: gomes

gomes said: but instead of f(x)=x^-2, let f(x) = x^(1/2 ) Click to expand... \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \) \(\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \) \(\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\) finish it.

G gomes May 2010 39 0 May 22, 2010 #4 Thanks everyone skeeter said: \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \) \(\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \) \(\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\) finish it. Click to expand... 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!

Thanks everyone skeeter said: \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) \(\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} \) \(\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \) \(\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\) finish it. Click to expand... 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!

AllanCuz Apr 2010 384 153 Canada May 22, 2010 #6 gomes said: Sorry, one more, im stuck Click to expand... I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Resources/sincosfirstprinciples.pdf If you have any questions we can answer them here (Hi)

gomes said: Sorry, one more, im stuck Click to expand... I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Resources/sincosfirstprinciples.pdf If you have any questions we can answer them here (Hi)

G gomes May 2010 39 0 May 23, 2010 #7 AllanCuz said: I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Resources/sincosfirstprinciples.pdf If you have any questions we can answer them here (Hi) Click to expand... Thanks! i understand it, but what about for sin2x?

AllanCuz said: I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Resources/sincosfirstprinciples.pdf If you have any questions we can answer them here (Hi) Click to expand... Thanks! i understand it, but what about for sin2x?

harish21 Feb 2010 1,036 386 Dirty South May 23, 2010 #8 gomes said: Thanks! i understand it, but what about for sin2x? Click to expand... \(\displaystyle \lim_{h \to 0} \frac{sin(2x+2h)-sin(2x)}{h}\) \(\displaystyle = \lim_{h \to 0} \frac{2}{2} \times \frac{sin(2x+2h)-sin(2x)}{h} \) \(\displaystyle = \lim_{h \to 0} \frac{2(sin(2x+2h)-sin(2x))}{2h}\) \(\displaystyle = \lim_{h \to 0} \frac{2(sin(X+H)-sin(X))}{H}\) where X=2x and H=2h \(\displaystyle = 2cos(X)\) ; which you proved before doing this. \(\displaystyle = 2cos(2x)\)

gomes said: Thanks! i understand it, but what about for sin2x? Click to expand... \(\displaystyle \lim_{h \to 0} \frac{sin(2x+2h)-sin(2x)}{h}\) \(\displaystyle = \lim_{h \to 0} \frac{2}{2} \times \frac{sin(2x+2h)-sin(2x)}{h} \) \(\displaystyle = \lim_{h \to 0} \frac{2(sin(2x+2h)-sin(2x))}{2h}\) \(\displaystyle = \lim_{h \to 0} \frac{2(sin(X+H)-sin(X))}{H}\) where X=2x and H=2h \(\displaystyle = 2cos(X)\) ; which you proved before doing this. \(\displaystyle = 2cos(2x)\)