91Ó°ÊÓ

Derivatives play a crucial role in calculus, particularly when identifying critical points of functions. A derivative is a mathematical tool used to calculate the rate at which a function's value changes with respect to changes in another variable. In simpler terms, it tells us how a function is changing at any given point. For our particular function \( f(x) = x^3 - 3ax^2 + 3a^2x - a^3 \), we start by taking the derivative with respect to \( x \).

To find the derivative, we apply the

• "Power Rule," which provides an easy way to differentiate functions of the form \( x^n \), resulting in \( n \cdot x^{n-1} \).

Applying this rule to the terms yields the derivative \( f'(x) = 3x^2 - 6ax + 3a^2 \). The values obtained from the derivative allow us to determine how the original function is behaving at any particular \( x \)-value. Specifically, it helps us find the points where the function changes its increasing or decreasing nature, known as critical points.

A quadratic equation is a second-degree polynomial equation in a single variable \( x \) with the form \( ax^2 + bx + c = 0 \). In our problem, the derivative \( f'(x) = 3x^2 - 6ax + 3a^2 \) is simplified by factoring out the 3, resulting in \( x^2 - 2ax + a^2 = 0 \).

To find the critical points, we must solve this quadratic equation. Recognizing the standard form, we can apply the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

where \( a = 1 \), \( b = -2a \), and \( c = a^2 \) per our simplified equation. Through substitution, we find \( x = a \). Since the discriminant (\( b^2 - 4ac \)) equals zero, there is one unique solution, indicating a double root at this critical point.

The Power Rule is a fundamental principle in calculus used to find derivatives of power functions. It simplifies the differentiation process by stating that if \( f(x) = x^n \), then the derivative \( f'(x) \) is \( n \cdot x^{n-1} \).

When we apply the Power Rule to the function \( f(x) = x^3 - 3ax^2 + 3a^2x - a^3 \), it allows us to handle each term independently:

• For \( x^3 \), the derivative is \( 3x^2 \).
• For \(-3ax^2 \), applying the rule gives \(-6ax\).
• The term \(3a^2x\) results in a derivative of \(3a^2\).
• Lastly, the constant term \(-a^3\) simply vanishes, as constants have a derivative of zero.

These calculations culminate in the derivative \( f'(x) = 3x^2 - 6ax + 3a^2 \), illustrating the power of this rule in simplifying functions to identify their behavior efficiently.