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Solve using first derivative principle

Mathematics
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\[\huge y=\sqrt{2x-1}\]
|dw:1371162568007:dw|
  • phi
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.

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Other answers:

conjugate being : \[\huge \frac{ \sqrt{2x+h-1} + \sqrt{2x-1}}{ \sqrt{2x+h-1} + \sqrt{2x-1} }\] ?
i forgot the "2h"
  • phi
yes. the idea is (a-b)*(a+b) = a^2 - b^2 which will get rid of the square roots up top
{2x+2h-1}
thank you @phi :)
  • phi
yw
Im getting |dw:1371163702503:dw|
  • phi
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 \[ \sqrt{2x-1} + \sqrt{2x-1} \]
|dw:1371164005685:dw|
would that be right
  • phi
yes if we use calculus \[ \frac{d}{dx} (2x -1)^{\frac{1}{2}} = \frac{1}{2} (2x -1)^{-\frac{1}{2}} 2\\= (2x -1)^{-\frac{1}{2}}= \frac{1}{\sqrt{2x-1} }\]

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