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burhan101 Group TitleBest ResponseYou've already chosen the best response.1
\[\huge y=\sqrt{2x1}\]
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
dw:1371162568007:dw
 one year ago

phi Group TitleBest 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.
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
conjugate being : \[\huge \frac{ \sqrt{2x+h1} + \sqrt{2x1}}{ \sqrt{2x+h1} + \sqrt{2x1} }\] ?
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
i forgot the "2h"
 one year ago

phi Group TitleBest 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
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
{2x+2h1}
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
thank you @phi :)
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
Im getting dw:1371163702503:dw
 one year ago

phi Group TitleBest 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 \[ \sqrt{2x1} + \sqrt{2x1} \]
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
dw:1371164005685:dw
 one year ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.1
would that be right
 one year ago

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