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

1. markragay

|dw:1372321232477:dw|

2. nhorzkie

lagyan mo ng pwet

3. nhorzkie

tapos yung mukha mo lagay mo sa tae yun

4. markragay

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

6. SithsAndGiggles

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

7. ParthKohli

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

@mathstudent55

9. markragay

@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

@koikkara

12. ParthKohli

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

13. markragay

@uri

14. markragay

@thomaster

15. SithsAndGiggles

@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

Oh

17. markragay

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

18. ParthKohli

Sorry, yes, you can use a substitution.

19. markragay

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