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Finding the derivative is a critical step in solving various calculus problems, including calculating arc length. Derivatives tell us the rate of change or the slope of a function at any given point. Here, our function is given as \( y = \frac{3}{2} x^{2/3} \). Calculating its derivative using established rules helps in subsequent steps of determining arc length.
To compute the derivative, use the power rule. This rule states that if \( y = ax^n \), then the derivative \( y' \) is given by \( y' = n \, ax^{n-1} \). Thus, for \( y = \frac{3}{2} x^{2/3} \):

• Apply the power rule: The exponent \( 2/3 \) comes down in front as a multiplier.
• Subtract one from the exponent to adjust \( x ^ {2/3} \) to \( x^{-1/3} \).

This gives \( y' = 2x^{-1/3} \). Calculating this derivative is the foundation for determining other properties, such as the slope and subsequent formula adjustments that relate to arc length.

The power rule simplifies the differentiation of expressions where variables are raised to a power. It’s essential for quickly finding derivatives, especially useful in this exercise.
With the power rule, if a function is in the form of \( ax^n \), the derivative can be calculated straightforwardly:

• Bring the exponent \( n \) as a coefficient in front of the term.
• Reduce the exponent by one to adjust the power of \( x \).

For the function \( y = \frac{3}{2} x^{2/3} \), applying the power rule efficiently transforms it into \( 2x^{-1/3} \). This step is crucial as it sets the stage for further calculus operations, like integrating or finding arc lengths. Without understanding the power rule, differentiation would be much more complex.

Integral calculus is the tool used to "aggregate" infinitesimally small data over an interval, allowing us to find areas, volumes, and, in this context, arc lengths. Once a derivative is computed, we modify it to find the arc length along a curve.
Using integral calculus, the key step involves integrating the function \( \sqrt{1 + (y')^2} \) over the interval [1, 8].

• This expression accounts for both the slope and distance below the curved path.
• The integrand \( \sqrt{1 + 4x^{-2/3}} \) reflects the squaring of the calculated derivative and adding 1 to fit the arc length formula.

Integrating this complex function isn't always straightforward and may require further algebraic manipulation or numerical methods, particularly when the antiderivative is not easily obtained. The process results in an approximate solution which, in this case, gives an arc length of around 8.77 units.

Function graphs help visualize mathematical concepts by providing a visual representation of functions and their behaviors. For the function \( y = \frac{3}{2} x^{2/3} \), a graph would show us how the function behaves across the domain, offering insights that pure algebra might miss.

• The graph allows us to see the curve from \( x = 1 \) to \( x = 8 \) and understand where the function increases or decreases.
• It provides a context for the arc length calculation as we seek the distance along this curve within the specified interval.

Function graphs are tools for comprehension that aid in verifying and illustrating the solutions obtained from calculus. They showcase the application of concepts such as derivatives and integrals in a way that complements the algebraic side of problem-solving.