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markragay

  • 2 years ago

....help in derivatives using delta method

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  1. markragay
    • 2 years ago
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    |dw:1372321232477:dw|

  2. nhorzkie
    • 2 years ago
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    lagyan mo ng pwet

  3. nhorzkie
    • 2 years ago
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    tapos yung mukha mo lagay mo sa tae yun

  4. markragay
    • 2 years ago
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    help @snuggielad

  5. SnuggieLad
    • 2 years ago
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    You know, right now my math brain is fried! I will tag some folks! @ParthKohli @mathslover @mathstudent55 give him a hand please

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

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

  8. markragay
    • 2 years ago
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    @mathstudent55

  9. markragay
    • 2 years ago
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    @Parthkohli I already know the formula please show me the solution and process ...pls..

  10. SnuggieLad
    • 2 years ago
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    Hint...He can only show you the process not the solution

  11. markragay
    • 2 years ago
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    @koikkara

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

  13. markragay
    • 2 years ago
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    @uri

  14. markragay
    • 2 years ago
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    @thomaster

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

  16. ParthKohli
    • 2 years ago
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    Oh

  17. markragay
    • 2 years ago
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    what do you mean by binomial series...please show me the whole process in finding the answer...@Parthkohli

  18. ParthKohli
    • 2 years ago
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    Sorry, yes, you can use a substitution.

  19. markragay
    • 2 years ago
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    please help me

  20. SithsAndGiggles
    • 2 years ago
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    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 \[\lim_{h\to0}\frac{h}{h((x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3})}\\ \lim_{h\to0}\frac{1}{(x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3}}\]

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