Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Use the first principles definition to determine the first derivative of sqrt(2x-1)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

This is as far as I got:|dw:1340139301477:dw|
Let's look at the definition: \[\lim_{h \rightarrow a} f(x)=[f(x+h) -f(x) ]/h\]
Ok.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Your f(x) = sqrt(2x-1), and a=0. So you want to find the limit of f(x) as h approaches 0
How do I know a=0?
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
As for your latter instructions. I understand, and have reached past that point. The reaffirmation is appreciated.
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
Ok. I think I understand; The IROC is 0 at a right. Let's continue.
Hello?
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?
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.
Think there is something wrong ... because you should be able to get that h canceled
\[\lim_{h \rightarrow 0}\frac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h} \cdot \frac{\sqrt{2x+2h-1}+\sqrt{2x-1}}{\sqrt{2x+2h-1}+\sqrt{2x-1}}\]
So that is what you did?
Yes, that was my previous step before this.
\[\lim_{h \rightarrow 0} \frac{(2x+2h-1)-(2x-1)}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\] Is this what you got next?
I then cancelled out 2a and -2a.
what else cancels?
on top!
I forgot to cancel 1 and -1.
yep! :)
So that means we have \[\lim_{h \rightarrow 0}\frac{2h}{h(\sqrt{2x+2h-1}+\sqrt{2x-1})}\] Correct?
yes.
Do you know what to do from here?
Yes. thank you very much.

Not the answer you are looking for?

Search for more explanations.

Ask your own question