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Delta method calculus for ³√x²
bio_man
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8 months ago
Delta method calculus for ³√x²
Set:
\(f\left(x\right)=\sqrt[3]{x^2}\)
Apply into limit expression, do not take the limit yet:
\(\lim _{x\rightarrow 0}\ \frac{\sqrt[3]{\left(x+\Delta x\right)^2}-\sqrt[3]{x^2}}{\Delta x}\)
Set a and b:
\(a=\sqrt[3]{\left(x+\Delta x\right)^2}\)
Solve for a
3
:
\(a^3=\left(x+\Delta x\right)^2\)
Next, we look at b:
\(b=\sqrt[3]{x^2}\)
Solve for b
3
:
\(b^3=x^2\)
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:
\(\frac{2}{3\sqrt[3]{x}}\)
Source
First principles, Calculus
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