Title: Delta method calculus for ³√x² Post by: bio_man on Aug 29, 2018 Set:
\(f\left(x\right)=\sqrt[3]{x^2}\) Apply into limit expression, do not take the limit yet: \(\lim _{x\rightarrow 0}\ \frac{\sqrt[3]{\left(x+\Delta x\right)^2}-\sqrt[3]{x^2}}{\Delta x}\) \(a=\sqrt[3]{\left(x+\Delta x\right)^2}\) Solve for a3: \(a^3=\left(x+\Delta x\right)^2\) Next, we look at b: \(b=\sqrt[3]{x^2}\) Solve for b3: \(b^3=x^2\) Recall: \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\) Rearrange this formula: \(\frac{a^3−b^3}{(a^2+ab+b^2)}=a−b\) \(\frac{\left(x+\Delta x\right)^2−x^2}{\left(\sqrt[3]{\left(x+\Delta x\right)^4}\right)+\sqrt[3]{\left(x+\Delta x\right)^2x^2}+\sqrt[3]{x^4}}=a−b\) We expand the numerator: \(\left(x+\Delta x\right)^2−x^2=2\Delta x\cdot x+\Delta x^2\) Common factor it further! \(2\Delta x\cdot x+\Delta x^2=\Delta x\left(2x+\Delta x\right)\) Apply into limit expression, do not take the limit yet: \(\lim _{x\rightarrow 0}\frac{\Delta x\left(2x+\Delta x\right)}{\left[\left(\sqrt[3]{\left(x+\Delta x\right)^4}\right)+\sqrt[3]{\left(x+\Delta x\right)^2x^2}+\sqrt[3]{x^4}\right]\Delta x}\) The Δx cancel each other out: \(\lim _{x\rightarrow 0}\frac{2x+\Delta x}{\left[\left(\sqrt[3]{\left(x+\Delta x\right)^4}\right)+\sqrt[3]{\left(x+\Delta x\right)^2x^2}+\sqrt[3]{x^4}\right]}\) Take the limit now: \(=\frac{2x}{\left(\sqrt[3]{x^4}\right)+\sqrt[3]{x^4}+\sqrt[3]{x^4}}\) \(=\frac{2x}{3\left(\sqrt[3]{x^4}\right)}\) Technically you're done, but you can use the laws of exponents to go further: \(=\frac{2x}{3\left(\sqrt[3]{x^4}\right)}\rightarrow \frac{2}{3}\cdot \frac{x}{x^{\frac{4}{3}}}\rightarrow \ \frac{2}{3}x^{1-\frac{4}{3}}\rightarrow \ \frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x}}\) Final answer: \(\frac{2}{3\sqrt[3]{x}}\) Title: Re: Delta method calculus for ³√x² Post by: duddy on Nov 7, 2019 👍
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