A community for students.
Here's the question you clicked on:
 0 viewing
markragay
 2 years ago
....help in derivatives using delta method
markragay
 2 years ago
....help in derivatives using delta method

This Question is Closed

markragay
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1372321232477:dw

nhorzkie
 2 years ago
Best ResponseYou've already chosen the best response.0tapos yung mukha mo lagay mo sa tae yun

SnuggieLad
 2 years ago
Best ResponseYou've already chosen the best response.0You know, right now my math brain is fried! I will tag some folks! @ParthKohli @mathslover @mathstudent55 give him a hand please

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.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?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1The definition of a derivative is\[f'(x) = \lim_{h \to 0} \dfrac{f(x + h)  f(x)}{h}\]Now suppose that \(f(x) = \sqrt[3]{x}\)

markragay
 2 years ago
Best ResponseYou've already chosen the best response.0@Parthkohli I already know the formula please show me the solution and process ...pls..

SnuggieLad
 2 years ago
Best ResponseYou've already chosen the best response.0Hint...He can only show you the process not the solution

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1\[\lim_{h \to 0} \dfrac{(x + h)^{1/3}  x^{1/3}}{h}\]I think you can use a binomial series to expand \((x + h)^{1/3}\)

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.1@ParthKohli, I've seen this exact problem done with a substitution, but I forget the details. Something like \(u^3=x\), I think.

markragay
 2 years ago
Best ResponseYou've already chosen the best response.0what do you mean by binomial series...please show me the whole process in finding the answer...@Parthkohli

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry, yes, you can use a substitution.

SithsAndGiggles
 2 years ago
Best ResponseYou've already chosen the best response.1I 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 \[\begin{align*}\color{green}{t^3u^3}&=(tu)(t^2+tu+u^2)\\ &=\color{red}{\left(\sqrt[3]{x+h}\sqrt[3]x\right)}\color{blue}{\left((\sqrt[3]{x+h})^2(\sqrt[3]{x+h})(\sqrt[3]x)+(\sqrt[3]x)^2\right)} \end{align*}\] 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 \(t^3u^3=(x+h)x=h\), the limit is reduced to \[\lim_{h\to0}\frac{h}{h((x+h)^{2/3}(x+h)^{1/3}x^{1/3}+x^{2/3})}\\ \lim_{h\to0}\frac{1}{(x+h)^{2/3}(x+h)^{1/3}x^{1/3}+x^{2/3}}\]
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.