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

markragay Group Title

....help in derivatives using delta method

  • one year ago
  • one year ago

  • This Question is Closed
  1. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1372321232477:dw|

    • one year ago
  2. nhorzkie Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    lagyan mo ng pwet

    • one year ago
  3. nhorzkie Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    tapos yung mukha mo lagay mo sa tae yun

    • one year ago
  4. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    help @snuggielad

    • one year ago
  5. SnuggieLad Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    You know, right now my math brain is fried! I will tag some folks! @ParthKohli @mathslover @mathstudent55 give him a hand please

    • one year ago
  6. SithsAndGiggles Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[f(x)=\sqrt[3]{x}\\ f'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\] Is this the "delta" method you're referring to?

    • one year ago
  7. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The definition of a derivative is\[f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\]Now suppose that \(f(x) = \sqrt[3]{x}\)

    • one year ago
  8. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @mathstudent55

    • one year ago
  9. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Parthkohli I already know the formula please show me the solution and process ...pls..

    • one year ago
  10. SnuggieLad Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Hint...He can only show you the process not the solution

    • one year ago
  11. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @koikkara

    • one year ago
  12. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\lim_{h \to 0} \dfrac{(x + h)^{1/3} - x^{1/3}}{h}\]I think you can use a binomial series to expand \((x + h)^{1/3}\)

    • one year ago
  13. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @uri

    • one year ago
  14. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @thomaster

    • one year ago
  15. SithsAndGiggles Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @ParthKohli, I've seen this exact problem done with a substitution, but I forget the details. Something like \(u^3=x\), I think.

    • one year ago
  16. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh

    • one year ago
  17. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    what do you mean by binomial series...please show me the whole process in finding the answer...@Parthkohli

    • one year ago
  18. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry, yes, you can use a substitution.

    • one year ago
  19. markragay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    please help me

    • one year ago
  20. SithsAndGiggles Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I believe it goes something like this: Substitute \[u^3=x~\iff~u=\sqrt[3]x\\ t^3=x+h~\iff~t=\sqrt[3]{x+h}\] Notice that \[\begin{align*}\color{green}{t^3-u^3}&=(t-u)(t^2+tu+u^2)\\ &=\color{red}{\left(\sqrt[3]{x+h}-\sqrt[3]x\right)}\color{blue}{\left((\sqrt[3]{x+h})^2-(\sqrt[3]{x+h})(\sqrt[3]x)+(\sqrt[3]x)^2\right)} \end{align*}\] The red part is what we have in the numerator in the limit. So in the limit, you have to multiply the numerator and denominator by the blue part: \[\lim_{h\to0}\frac{\sqrt[3]{x+h}-\sqrt[3]x}{h}\cdot\frac{(\sqrt[3]{x+h})^2-(\sqrt[3]{x+h})(\sqrt[3]x)+(\sqrt[3]x)^2}{(\sqrt[3]{x+h})^2-(\sqrt[3]{x+h})(\sqrt[3]x)+(\sqrt[3]x)^2}\] I'm going to rewrite the radicals as exponents: \[\lim_{h\to0}\frac{(x+h)^{1/3}-x^{1/3}}{h}\cdot\frac{(x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3}}{(x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3}}\] Using the fact (green part above) that \(t^3-u^3=(x+h)-x=h\), the limit is reduced to \[\lim_{h\to0}\frac{h}{h((x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3})}\\ \lim_{h\to0}\frac{1}{(x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3}}\]

    • one year 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.