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Calculating a derivative can be thought of as finding the rate at which a function changes at any given point. For the function \( y = x^3 \), the derivative is found by using the power rule. The power rule is a basic derivative rule that states if you have a function \( f(x) = x^n \), then its derivative \( f'(x) \) is \( nx^{n-1} \).

For our function, \( y = x^3 \), applying the power rule gives us:

• The derivative \( f'(x) = 3x^2 \).

This means the rate of change of \( y \) with respect to \( x \) is given by this formula. When calculated at a specific point, such as \( x = 2 \), you simply substitute the value of \( x \) into the derivative:

• Substituting \( x = 2 \) into \( f'(x) \) gives: \( f'(2) = 3(2)^2 = 12 \).

Thus, when \( x = 2 \), the function \( y = x^3 \) changes 12 units of \( y \) for each unit of \( x \).

Once you know the rate at which the function changes, you can calculate how much the function value changes when you make a small change to \( x \). This is done by finding \( \Delta y \), the actual change in \( y \).

In this case, \( \Delta x \), or the change in \( x \), is 0.01, moving from \( x = 2 \) to \( x = 2.01 \). The change in \( y \) can be calculated by evaluating the function at these points:

• First, calculate the function at the new \( x \) value: \( f(2.01) = (2.01)^3 = 8.120601 \).
• Then, calculate the function at the original \( x \) value: \( f(2) = (2)^3 = 8 \).

The change in the function's value, \( \Delta y \), is:

• \( \Delta y = f(2.01) - f(2) = 8.120601 - 8 = 0.120601 \).

When rounded to four decimal places, \( \Delta y \) is 0.1206.

Derivatives also offer us a powerful tool known as linear approximation. Instead of precisely calculating \( \Delta y \), we can approximate it by using the derivative and \( \Delta x \). This approximation is represented by \( f'(x) \Delta x \).

In this case, here’s how the approximation works:

• We previously found the derivative at \( x = 2 \), which is 12.
• Multiply this by the change in \( x \), which is 0.01: \( f'(2) \Delta x = 12 \times 0.01 = 0.12 \).

This tells us that for the small change in \( x \), the function\( 's \) value is approximated to change by 0.12 units. This approximation is particularly useful when dealing with very small changes, as it provides a quick way to estimate \( \Delta y \) without full computation.