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burhan101Best ResponseYou've already chosen the best response.1
\[\huge y=\sqrt{2x1}\]
 10 months ago

burhan101Best ResponseYou've already chosen the best response.1
dw:1371162568007:dw
 10 months ago

phiBest ResponseYou've already chosen the best response.1
try multiplying top and bottom by the conjugate ( sq rt + sq rt) so you get a difference of squares up top that can be simplified.
 10 months ago

burhan101Best ResponseYou've already chosen the best response.1
conjugate being : \[\huge \frac{ \sqrt{2x+h1} + \sqrt{2x1}}{ \sqrt{2x+h1} + \sqrt{2x1} }\] ?
 10 months ago

phiBest ResponseYou've already chosen the best response.1
yes. the idea is (ab)*(a+b) = a^2  b^2 which will get rid of the square roots up top
 10 months ago

burhan101Best ResponseYou've already chosen the best response.1
Im getting dw:1371163702503:dw
 10 months ago

phiBest ResponseYou've already chosen the best response.1
yes, but what you want to do is \[\lim_{h \rightarrow 0}\frac{f(x+h)  f(x)}{h} \] the next step is let h>0 so the bottom becomes \[ \sqrt{2x1} + \sqrt{2x1} \]
 10 months ago

burhan101Best ResponseYou've already chosen the best response.1
dw:1371164005685:dw
 10 months ago

burhan101Best ResponseYou've already chosen the best response.1
would that be right
 10 months ago

phiBest ResponseYou've already chosen the best response.1
yes if we use calculus \[ \frac{d}{dx} (2x 1)^{\frac{1}{2}} = \frac{1}{2} (2x 1)^{\frac{1}{2}} 2\\= (2x 1)^{\frac{1}{2}}= \frac{1}{\sqrt{2x1} }\]
 10 months ago
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