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Understanding the derivative of a function is pivotal for various applications in calculus, including finding the length of a curve. The derivative represents the rate at which a function is changing at any given point and it is a fundamental tool for identifying the slope of a function.
For a function represented as \(y = f(x)\), the derivative, denoted as \(f'(x)\), provides us with a new function that tells us how fast \(y\) changes with respect to \(x\). In the context of our exercise, the derivative of \(f(x)=x^{1/3} + x^{2/3}\) reveals how the slope of the curve changes as we move from one point to another along the curve.
When calculating the length of a curve between two points, we need to find the integral of the square root of \(1\) plus the derivative squared. The derivative informs us about the steepness and direction of the curve at each infinitesimal segment along our interval, which is critical in assessing the true length of the curve.

One of the most routinely used rules in calculus for finding derivatives is the power rule for differentiation. This rule simplifies the process of taking derivatives of functions that are polynomial-like, where the variable \(x\) is raised to a power, like in \(x^n\).
According to the power rule, the derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\). If we apply this to our function \(f(x) = x^{1/3} + x^{2/3}\), we find the derivatives individually for each term. The first term's derivative is \((1/3)x^{1/3 - 1} = (1/3)x^{-2/3}\) and the second term's is \((2/3)x^{2/3 - 1} = (2/3)x^{-1/3}\).
In our exercise, applying the power rule efficiently provides us with function's rate of change, which is crucial for calculating the length of the curve. It's through this fundamental rule that we can handle even more complex functions, to analyze the nature of curves and other varied phenomena.

The definite integral is a central concept in calculus that quantifies the accumulation of quantities, such as areas under curves. In our exercise, the definite integral is used to calculate the actual length of a curve over a closed interval, which in this case is from \(x = 0\) to \(x = 2\).
For the formula \(l = \int_a^b \sqrt{1 + (f'(x))^2} dx\), the definite integral aims to sum up the lengths of infinitely small line segments along the curve, providing the total length. Each infinitesimal segment's length is approximated by the square root of \(1\) plus the square of the derivative, which accounts for both the vertical and horizontal changes along the curve.
The challenge with our specific function is that its square, when inside the square root, makes the integral non-elementary, meaning it cannot be evaluated using the basic rules of integration. This is where special mathematical functions or numerical methods come into play to find an approximation of the integral, thus giving us an estimate of the length of the curve.