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Posts: 23920
6 months ago
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
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