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Finding the derivative of a function is a core concept in calculus. Derivatives represent the rate of change of a function with respect to one of its variables. For instance, when you want the slope of a tangent line at a particular point on a curve, you use the derivative.

In our specific problem, we have the function \( y = \left( x - \frac{1}{x} \right)^3 \). To find its derivative, we need to identify how \( y \) changes as \( x \) changes. We start by substituting \( u = x - \frac{1}{x} \), making \( y = u^3 \).

Next, differentiate \( u \) concerning \( x \), which gives us \( \frac{du}{dx} = 1 + \frac{1}{x^2} \). Similarly, differentiate \( y \) concerning \( u \), resulting in \( \frac{dy}{du} = 3u^2 \). Combining these using the chain rule gives us the complete derivative \( \frac{dy}{dx} = 3u^2 \left( 1 + \frac{1}{x^2} \right) \).

Understanding derivatives will help you describe the slope of the line that just touches a curve at a point -- the tangent line.

The chain rule is a fundamental tool in calculus used to compute the derivative of composite functions. A composite function is one formed by "substituting" one function within another.

For example, if you have \( y = (u)^3 \) where \( u = x - \frac{1}{x} \), then \( y \) is a function of \( u \), and \( u \) is a function of \( x \). The chain rule allows us to compute \( \frac{dy}{dx} \) by multiplying the derivative of \( y \) with respect to \( u \) by the derivative of \( u \) with respect to \( x \).

This gives us the complete derivative: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \left( 1 + \frac{1}{x^2} \right) \]

Without the chain rule, differentiating composite functions would be a cumbersome task. Breaking functions into smaller, manageable parts eases the differentiation process significantly.

Function evaluation is the process of determining the output of a function for a particular input. This concept helps you understand how a function behaves at specific points, which is crucial in calculus, especially when finding tangent lines to curves.

In this exercise, we started with the function \( y = \left( x - \frac{1}{x} \right)^3 \) and a given \( x = 2 \). Evaluating the function at \( x = 2 \) involves substituting this value into the function.

Substituting gives: \[ y = \left( 2 - \frac{1}{2} \right)^3 = \left( \frac{3}{2} \right)^3 \]

Calculating further yields \( y = \frac{27}{8} \). This value represents the point on the curve at \( x = 2 \), which is essential for finding the tangent line. Evaluating functions is a straightforward yet pivotal part of solving calculus problems, especially in practical applications such as calculating slopes or rates of change at a given point.