## IsTim 3 years ago Use the first principles definition to determine the first derivative of sqrt(2x-1)

1. IsTim

This is as far as I got:|dw:1340139301477:dw|

2. mashe

Let's look at the definition: $\lim_{h \rightarrow a} f(x)=[f(x+h) -f(x) ]/h$

3. IsTim

Ok.

4. mashe

Your f(x) = sqrt(2x-1), and a=0. So you want to find the limit of f(x) as h approaches 0

5. IsTim

How do I know a=0?

6. mashe

So you have $\lim_{h \rightarrow 0} \sqrt(2x-1)= \sqrt (2(x+h)-1) - \sqrt (2x-1)/h$ where h is the denominator and everything else is on top, the numerator

7. IsTim

As for your latter instructions. I understand, and have reached past that point. The reaffirmation is appreciated.

8. mashe

Look at the third picture here: http://www.intmath.com/differentiation/3-derivative-first-principles.php and the explanation below it states why a=0

9. IsTim

Ok. I think I understand; The IROC is 0 at a right. Let's continue.

10. IsTim

Hello?

11. myininaya

Where are you at @IsTim ? $\lim_{h \rightarrow 0}\frac{\sqrt{2(x+h)-1}-\sqrt{2x-1}}{h}$ This is where you at right?

12. myininaya

You need to rationalize the numerator. We are trying to manipulate this problem so that when we plug in 0 for h we don't get 0 on the bottom.

13. myininaya

Think there is something wrong ... because you should be able to get that h canceled

14. myininaya

$\lim_{h \rightarrow 0}\frac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h} \cdot \frac{\sqrt{2x+2h-1}+\sqrt{2x-1}}{\sqrt{2x+2h-1}+\sqrt{2x-1}}$

15. myininaya

So that is what you did?

16. IsTim

Yes, that was my previous step before this.

17. myininaya

$\lim_{h \rightarrow 0} \frac{(2x+2h-1)-(2x-1)}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}$ Is this what you got next?

18. IsTim

I then cancelled out 2a and -2a.

19. myininaya

what else cancels?

20. myininaya

on top!

21. IsTim

I forgot to cancel 1 and -1.

22. myininaya

yep! :)

23. myininaya

So that means we have $\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}$ Correct?

24. IsTim

yes.

25. myininaya

Do you know what to do from here?

26. IsTim

Yes. thank you very much.