Problem 36 - Technical Mathematics with Calculus

Question

Find the derivative of each function. Verify some of your results by calculator. As usual, the letters \(a, b, c, \ldots\) represent constants. Derivative of a Sum. $$y=x^{2 / 3}-a^{2 / 3}$$

Step-by-Step Solution

Answer

The derivative of the function \(y = x^{2/3} - a^{2/3}\) is \(y' = \frac{2}{3}x^{-1/3}\).

Step 1: Identify the Function Components

Recognize that the function consists of a sum of two terms: the variable term, which is a power function of x, given by \(x^{2/3}\), and a constant term, which is \(-a^{2/3}\). The constant term is a constant with respect to x, so its derivative will be zero.

Step 2: Differentiate the Variable Term

Use the power rule for differentiation, which states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). Apply this rule to differentiate the variable term \(x^{2/3}\) with respect to x to find its derivative.

Step 3: Apply the Power Rule

Applying the power rule to the term \(x^{2/3}\), we get the derivative as \(\frac{2}{3}x^{(2/3)-1} = \frac{2}{3}x^{-1/3}\).

Step 4: Differentiate the Constant Term

The derivative of any constant, such as \(-a^{2/3}\), with respect to x is zero because constants do not change as x changes.

Step 5: Combine the Derivatives

Combine the derivatives of the individual terms to obtain the derivative of the entire function. Since the derivative of \(-a^{2/3}\) is zero, it does not contribute to the final result.

Key Concepts

Power Rule for Differentiation

Understanding the power rule for differentiation is vital for any student studying calculus. Simply put, the power rule is a shortcut that helps us find the derivative of power functions, where the function is of the form f(x) = x^n for any real number n. When applying the power rule, the derivative f'(x) is found by multiplying the exponent n by the power function and then subtracting one from the exponent.

To illustrate, consider a function f(x) = x^m, where m is any real number. The derivative, f'(x), of this function using the power rule will be mx^{m-1}. It simplifies the process of differentiation by avoiding the limit definition of the derivative, which can be much more cumbersome.

For example, when we apply this to x^{2/3}, we multiply the exponent 2/3 by the function and subtract one from the exponent to get (2/3)x^{(2/3)-1} = (2/3)x^{-1/3}. This method is extremely efficient and is a foundational tool used in calculus.

Derivative of a Sum

In calculus, the derivative of a sum term can be found by applying the rule that the derivative of a sum is the sum of the derivatives. This means that for any functions u(x) and v(x), the derivative d/dx of u(x) + v(x) is simply u'(x) + v'(x).

This rule makes it easier to break down complex functions into simpler parts and take derivatives piece by piece. For example, if you were given y = f(x) + g(x), you would find f'(x) and g'(x) separately, and then add them together for your final answer. In the case of our example problem y = x^{2/3} - a^{2/3}, we apply the derivative separately to each term and combine them at the end. Since the power rule already gives us the derivative of x^{2/3}, and we know from the constant term differentiation that the derivative of -a^{2/3} is zero, the final derivative of the function is the sum of these individual derivatives.

Constant Term Differentiation

When working with differentiation, it's important to recognize that the derivative of a constant is always zero. This is because a constant does not vary with respect to the variable being differentiated, and hence its rate of change is always zero. In mathematical terms, if c is a constant and x is the variable, then d/dx(c) = 0.

This concept is crucial when differentiating functions that involve both variable terms and constants. In our example problem, while the term x^{2/3} requires the power rule to find its derivative, the constant term -a^{2/3} has a derivative of zero. This significantly simplifies the differentiation process, as any constant term can be disregarded when taking the derivative of a function. In a polynomial, for instance, this means we only need to find the derivatives of the terms that actually involve the variable x.