Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

burhan101

Solve using first derivative principle

  • 10 months ago
  • 10 months ago

  • This Question is Closed
  1. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\huge y=\sqrt{2x-1}\]

    • 10 months ago
  2. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1371162568007:dw|

    • 10 months ago
  3. phi
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 10 months ago
  4. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 10 months ago
  5. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    i forgot the "2h"

    • 10 months ago
  6. phi
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 10 months ago
  7. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    {2x+2h-1}

    • 10 months ago
  8. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    thank you @phi :)

    • 10 months ago
  9. phi
    Best Response
    You've already chosen the best response.
    Medals 1

    yw

    • 10 months ago
  10. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    Im getting |dw:1371163702503:dw|

    • 10 months ago
  11. phi
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[ \sqrt{2x-1} + \sqrt{2x-1} \]

    • 10 months ago
  12. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1371164005685:dw|

    • 10 months ago
  13. burhan101
    Best Response
    You've already chosen the best response.
    Medals 1

    would that be right

    • 10 months ago
  14. phi
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 10 months ago
    • Attachments:

See more questions >>>

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.