Find the derivative of cube root of x using delta x

To find the derivative of \( f(x) = \sqrt[3]{x} \) using the definition of the derivative, we apply the limit definition:

\[
f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}
\]

Here, \( f(x) = \sqrt[3]{x} \), so:

\[
f'(x) = \lim_{\Delta x \to 0} \frac{\sqrt[3]{x + \Delta x} - \sqrt[3]{x}}{\Delta x}
\]

To simplify this expression, we use the identity for the difference of cubes:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

Setting \( a = \sqrt[3]{x + \Delta x} \) and \( b = \sqrt[3]{x} \), the expression becomes:

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

So, we have:

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

This gives us:

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

Now, substituting into the original limit:

\[
f'(x) = \lim_{\Delta x \to 0} \frac{1}{\left(\sqrt[3]{x + \Delta x}\right)^2 + \sqrt[3]{x + \Delta x} \cdot \sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2}
\]

As \( \Delta x \) approaches 0, \( \sqrt[3]{x + \Delta x} \) approaches \( \sqrt[3]{x} \), so:

\[
f'(x) = \frac{1}{3

Learn how to find the derivative of the cube root of x using the delta x method. This problem involves applying the limit definition of the derivative and simplifying using the difference of cubes identity. Suitable for advanced high school or college-level calculus students.

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