A community for students.
Here's the question you clicked on:
 0 viewing
IsTim
 2 years ago
Use the first principles definition to determine the first derivative of sqrt(2x1)
IsTim
 2 years ago
Use the first principles definition to determine the first derivative of sqrt(2x1)

This Question is Closed

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0This is as far as I got:dw:1340139301477:dw

mashe
 2 years ago
Best ResponseYou've already chosen the best response.0Let's look at the definition: \[\lim_{h \rightarrow a} f(x)=[f(x+h) f(x) ]/h\]

mashe
 2 years ago
Best ResponseYou've already chosen the best response.0Your f(x) = sqrt(2x1), and a=0. So you want to find the limit of f(x) as h approaches 0

mashe
 2 years ago
Best ResponseYou've already chosen the best response.0So you have \[\lim_{h \rightarrow 0} \sqrt(2x1)= \sqrt (2(x+h)1)  \sqrt (2x1)/h\] where h is the denominator and everything else is on top, the numerator

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0As for your latter instructions. I understand, and have reached past that point. The reaffirmation is appreciated.

mashe
 2 years ago
Best ResponseYou've already chosen the best response.0Look at the third picture here: http://www.intmath.com/differentiation/3derivativefirstprinciples.php and the explanation below it states why a=0

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0Ok. I think I understand; The IROC is 0 at a right. Let's continue.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3Where are you at @IsTim ? \[\lim_{h \rightarrow 0}\frac{\sqrt{2(x+h)1}\sqrt{2x1}}{h}\] This is where you at right?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3You 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.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3Think there is something wrong ... because you should be able to get that h canceled

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3\[\lim_{h \rightarrow 0}\frac{\sqrt{2x+2h1}\sqrt{2x1}}{h} \cdot \frac{\sqrt{2x+2h1}+\sqrt{2x1}}{\sqrt{2x+2h1}+\sqrt{2x1}}\]

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3So that is what you did?

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, that was my previous step before this.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3\[\lim_{h \rightarrow 0} \frac{(2x+2h1)(2x1)}{h(\sqrt{2x+2h1}+\sqrt{2x1})}\] Is this what you got next?

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0I then cancelled out 2a and 2a.

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0I forgot to cancel 1 and 1.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3So that means we have \[\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h1}+\sqrt{2x1})}\] Correct?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.3Do you know what to do from here?

IsTim
 2 years ago
Best ResponseYou've already chosen the best response.0Yes. thank you very much.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.