## anonymous 3 years ago Solve using first derivative principle

1. anonymous

$\huge y=\sqrt{2x-1}$

2. anonymous

|dw:1371162568007:dw|

3. 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.

4. anonymous

conjugate being : $\huge \frac{ \sqrt{2x+h-1} + \sqrt{2x-1}}{ \sqrt{2x+h-1} + \sqrt{2x-1} }$ ?

5. anonymous

i forgot the "2h"

6. phi

yes. the idea is (a-b)*(a+b) = a^2 - b^2 which will get rid of the square roots up top

7. anonymous

{2x+2h-1}

8. anonymous

thank you @phi :)

9. phi

yw

10. anonymous

Im getting |dw:1371163702503:dw|

11. 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 $\sqrt{2x-1} + \sqrt{2x-1}$

12. anonymous

|dw:1371164005685:dw|

13. anonymous

would that be right

14. phi

yes if we use calculus $\frac{d}{dx} (2x -1)^{\frac{1}{2}} = \frac{1}{2} (2x -1)^{-\frac{1}{2}} 2\\= (2x -1)^{-\frac{1}{2}}= \frac{1}{\sqrt{2x-1} }$

Find more explanations on OpenStudy