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

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

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

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

SithsAndGiggles
 one year 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 \[\lim_{h\to0}\frac{h}{h((x+h)^{2/3}(x+h)^{1/3}x^{1/3}+x^{2/3})}\\ \lim_{h\to0}\frac{1}{(x+h)^{2/3}(x+h)^{1/3}x^{1/3}+x^{2/3}}\]
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