burhan101
Solve using first derivative principle
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burhan101
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\[\huge y=\sqrt{2x-1}\]
burhan101
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|dw:1371162568007:dw|
phi
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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.
burhan101
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conjugate being : \[\huge \frac{ \sqrt{2x+h-1} + \sqrt{2x-1}}{ \sqrt{2x+h-1} + \sqrt{2x-1} }\] ?
burhan101
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i forgot the "2h"
phi
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yes. the idea is (a-b)*(a+b) = a^2 - b^2
which will get rid of the square roots up top
burhan101
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{2x+2h-1}
burhan101
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thank you @phi :)
phi
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yw
burhan101
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Im getting |dw:1371163702503:dw|
phi
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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
\[ \sqrt{2x-1} + \sqrt{2x-1} \]
burhan101
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|dw:1371164005685:dw|
burhan101
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would that be right
phi
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yes
if we use calculus
\[ \frac{d}{dx} (2x -1)^{\frac{1}{2}} = \frac{1}{2} (2x -1)^{-\frac{1}{2}} 2\\= (2x -1)^{-\frac{1}{2}}= \frac{1}{\sqrt{2x-1} }\]