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markragay Group TitleBest ResponseYou've already chosen the best response.0
dw:1372321232477:dw
 one year ago

nhorzkie Group TitleBest ResponseYou've already chosen the best response.0
lagyan mo ng pwet
 one year ago

nhorzkie Group TitleBest ResponseYou've already chosen the best response.0
tapos yung mukha mo lagay mo sa tae yun
 one year ago

markragay Group TitleBest ResponseYou've already chosen the best response.0
help @snuggielad
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.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

SithsAndGiggles Group TitleBest 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?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.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

markragay Group TitleBest ResponseYou've already chosen the best response.0
@mathstudent55
 one year ago

markragay Group TitleBest ResponseYou've already chosen the best response.0
@Parthkohli I already know the formula please show me the solution and process ...pls..
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.0
Hint...He can only show you the process not the solution
 one year ago

markragay Group TitleBest ResponseYou've already chosen the best response.0
@koikkara
 one year ago

ParthKohli Group TitleBest 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 \((x + h)^{1/3}\)
 one year ago

markragay Group TitleBest ResponseYou've already chosen the best response.0
@thomaster
 one year ago

SithsAndGiggles Group TitleBest 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.
 one year ago

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Sorry, yes, you can use a substitution.
 one year ago

markragay Group TitleBest ResponseYou've already chosen the best response.0
please help me
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.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 \[\begin{align*}\color{green}{t^3u^3}&=(tu)(t^2+tu+u^2)\\ &=\color{red}{\left(\sqrt[3]{x+h}\sqrt[3]x\right)}\color{blue}{\left((\sqrt[3]{x+h})^2(\sqrt[3]{x+h})(\sqrt[3]x)+(\sqrt[3]x)^2\right)} \end{align*}\] 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 \[\lim_{h\to0}\frac{h}{h((x+h)^{2/3}(x+h)^{1/3}x^{1/3}+x^{2/3})}\\ \lim_{h\to0}\frac{1}{(x+h)^{2/3}(x+h)^{1/3}x^{1/3}+x^{2/3}}\]
 one year ago
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