## markragay Group Title ....help in derivatives using delta method one year ago one year ago

1. markragay Group Title

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2. nhorzkie Group Title

lagyan mo ng pwet

3. nhorzkie Group Title

tapos yung mukha mo lagay mo sa tae yun

4. markragay Group Title

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

6. SithsAndGiggles Group Title

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

7. ParthKohli Group Title

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 Group Title

@mathstudent55

9. markragay Group Title

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

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

11. markragay Group Title

@koikkara

12. ParthKohli Group Title

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

13. markragay Group Title

@uri

14. markragay Group Title

@thomaster

15. SithsAndGiggles Group Title

@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 Group Title

Oh

17. markragay Group Title

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

18. ParthKohli Group Title

Sorry, yes, you can use a substitution.

19. markragay Group Title

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 $\lim_{h\to0}\frac{h}{h((x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3})}\\ \lim_{h\to0}\frac{1}{(x+h)^{2/3}-(x+h)^{1/3}x^{1/3}+x^{2/3}}$