What Is The Derivative Of X 2 3

What is the Derivative of x^(2/3)? A Comprehensive Guide

The derivative of a function describes its instantaneous rate of change at any given point. Understanding derivatives is fundamental to calculus and has wide-ranging applications in various fields, from physics and engineering to economics and finance. This article delves into finding the derivative of x^(2/3), exploring different approaches and explaining the underlying concepts. We'll cover the power rule, the definition of the derivative, and address potential challenges.

Understanding the Power Rule

The most straightforward method to find the derivative of x^(2/3) is by using the power rule. This rule is a cornerstone of differential calculus and states:

d/dx (xⁿ) = nxⁿ⁻¹

Where 'n' is any real number. Applying this rule to our function, x^(2/3), we get:

d/dx (x^(2/3)) = (2/3)x^((2/3)-1)

Simplifying the exponent:

(2/3) - 1 = (2/3) - (3/3) = -1/3

Therefore, the derivative of x^(2/3) is:

(2/3)x^(-1/3)

Or, equivalently:

2 / (3x^(1/3))

This is the simplest and most efficient way to find the derivative. However, it's crucial to understand the underlying principles to appreciate the power and limitations of this rule.

Derivation from First Principles: The Definition of the Derivative

The power rule is a shortcut derived from the more fundamental definition of the derivative. The definition utilizes limits to describe the instantaneous rate of change:

f'(x) = lim (h→0) [(f(x + h) - f(x)) / h]

Let's apply this definition to our function, f(x) = x^(2/3):

1. Substitute f(x):

   f'(x) = lim (h→0) [((x + h)^(2/3) - x^(2/3)) / h]

2. Rationalize the Numerator: This step involves a clever algebraic manipulation to eliminate the indeterminate form 0/0 that arises when we directly substitute h = 0. We'll use the difference of cubes factorization: a³ - b³ = (a - b)(a² + ab + b²). However, we have a 2/3 power, not a cube root. To address this, we'll use a generalized difference of powers factorization, which is more complex but provides a more accurate way of solving this:

   This involves using the conjugate expression:

   ((x + h)^(2/3) + x^(2/3)) / ((x + h)^(2/3) + x^(2/3))

   Multiplying the numerator and denominator by this conjugate, we obtain:

   f'(x) = lim (h→0) [((x + h)^(4/3) - x^(4/3)) / (h*((x + h)^(2/3) + x^(2/3)))]

3. Apply the binomial theorem: The binomial theorem allows us to expand (x + h)^(4/3). This expansion will give a series of terms involving powers of h. The key is that, after applying the theorem and simplifying, all remaining terms will contain an 'h' allowing cancellation. While the complete binomial expansion can be complex, the initial terms are all that is required:

   (x + h)^(4/3) ≈ x^(4/3) + (4/3)x^(1/3)h + ...

4. Simplify and Take the Limit: After substituting this approximation and simplifying, the term x^(4/3) will cancel out. The remaining terms will involve 'h' which can be cancelled with the 'h' in the denominator, leaving the following expression:

   f'(x) = lim (h→0) [(4/3)x^(1/3) + ... higher-order terms of h]

   As h approaches 0, all the remaining terms containing 'h' will vanish.

5. Final Result: This leaves us with the final derivative:

   f'(x) = (2/3)x^(-1/3)

This derivation, while longer and more complex, directly demonstrates the application of the limit definition and reinforces the understanding of the power rule's origin.

Domain and Range of the Derivative

The derivative of x^(2/3), (2/3)x^(-1/3) or 2 / (3x^(1/3)), is not defined at x = 0. This is because we have a term in the denominator, x^(1/3), which becomes 0 when x = 0, resulting in an undefined expression. Therefore:

• Domain of the derivative: (-∞, 0) U (0, ∞) (All real numbers except 0)
• Range of the derivative: (-∞, 0) U (0, ∞) (All real numbers except 0)

This indicates that the original function, x^(2/3), has a vertical tangent at x = 0.

Applications of the Derivative

The derivative of x^(2/3) finds applications in various fields:

• Optimization problems: Finding maximum or minimum values of functions involving x^(2/3).
• Rate of change: Analyzing how a quantity (represented by x^(2/3)) changes over time.
• Curve sketching: Determining the slope and concavity of the graph of y = x^(2/3).
• Physics: Modeling physical phenomena where quantities follow a power law relationship.

Higher-Order Derivatives

We can find higher-order derivatives by repeatedly differentiating the function. The second derivative represents the rate of change of the first derivative (which represents acceleration if the original function is position).

First derivative: f'(x) = (2/3)x^(-1/3)

Second derivative: f''(x) = (2/3)(-1/3)x^(-4/3) = (-2/9)x^(-4/3)

This process can be continued for higher-order derivatives.

Challenges and Considerations

While the power rule offers a straightforward solution, some challenges can arise:

• Understanding Exponents: Students often struggle with fractional and negative exponents. A solid grasp of exponent rules is essential.
• Limit Evaluation: The derivation from first principles requires a good understanding of limit properties and techniques.
• Interpreting Results: Knowing how to interpret the derivative (slope of the tangent line) and its implications for the original function is crucial.

Conclusion

Finding the derivative of x^(2/3) provides a practical example of applying fundamental calculus concepts. The power rule offers the most efficient method, but understanding its derivation from the definition of the derivative reinforces the underlying principles. This exploration highlights the importance of mastering exponent rules, limit properties, and interpretation skills to fully grasp the concept and applications of derivatives. The limitations of the derivative's domain, its practical applications, and the concepts of higher-order derivatives all add depth to the understanding of this essential topic in calculus. Remember that the key to mastering calculus lies not just in memorizing formulas but in understanding the underlying concepts and their applications in real-world problems.