Cube root of x by the Delta Method

\(f\left(x\right)=\sqrt[3]{x}\)

\(\lim _{x\rightarrow 0}\ \frac{\sqrt[3]{x+\Delta x}-\sqrt[3]{x}}{\Delta x}\)

Recall:

\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

Rearrange this formula:

\(\frac{a^3−b^3}{(a^2+ab+b^2)}=a−b\)

Set a as:

\(a=\sqrt[3]{x+\Delta x}\)

Set b as:

\(b=\sqrt[3]{x}\)

Substitute these new a and b values into the rearranged formula:

\(\frac{\frac{x+\Delta x-x}{\left(\sqrt[3]{x+\Delta x}\right)^2+\left(\sqrt[3]{x+\Delta x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2}}{\Delta x}\)

Simplify:

\(\frac{\Delta x}{\left[\left(\sqrt[3]{x+\Delta x}\right)^2+\left(\sqrt[3]{x+\Delta x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2\right]\Delta x}\)

Simplify more:

\(\frac{1}{\left[\left(\sqrt[3]{x+\Delta x}\right)^2+\left(\sqrt[3]{x+\Delta x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2\right]}\)

Now take the bloody limit:

\(=\frac{1}{\left(\sqrt[3]{x}\right)^2+\left(\sqrt[3]{x}\cdot \sqrt[3]{x}\right)+\left(\sqrt[3]{x}\right)^2}\)

\(=\frac{1}{3\left(\sqrt[3]{x}\right)^2}\)