Question

....help in derivatives using delta method

Answer 1

lagyan mo ng pwet

Answer 2

tapos yung mukha mo lagay mo sa tae yun

Answer 3

help @snuggielad

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

@mathstudent55

Answer 8

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

Answer 9

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

Answer 10

@koikkara

Answer 11

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

Answer 12

@uri

Answer 13

@thomaster

Answer 14

@ParthKohli, I've seen this exact problem done with a substitution, but I forget the details.

Something like \(u^3=x\), I think.

Answer 15

Oh

Answer 16

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

Answer 17

Sorry, yes, you can use a substitution.

Answer 18

please help me

Answer 19

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