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10 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 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:$$\frac{2}{3\sqrt[3]{x}}$$ Source  First principles, Calculus Read 183 times
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