Problem 10 - Calculus

Question

Find the derivatives of the functions in exercise. Even if \(n\) is negative or a fraction, the derivative of \(x^{n}\) is \(n x^{n-1}\). $$ \frac{2}{3} x^{3 / 2}+\frac{2}{5} x^{5 / 2} $$

Step-by-Step Solution

Answer

The derivative is \( x^{1/2} + x^{3/2} \).

Step 1: Identify the Functions

We begin by identifying the individual functions in the given expression. The expression provided is \( \frac{2}{3} x^{3/2} + \frac{2}{5} x^{5/2} \). We can see two functions: \( \frac{2}{3} x^{3/2} \) and \( \frac{2}{5} x^{5/2} \).

Step 2: Apply the Power Rule

The power rule for differentiation states that the derivative of \( x^n \) is \( nx^{n-1} \). We apply this to each function separately.

Step 3: Differentiate the First Term

For the first term \( \frac{2}{3} x^{3/2} \), apply the power rule: Take the exponent, \( \frac{3}{2} \), multiply it by the coefficient \( \frac{2}{3} \), and reduce the exponent by 1. The derivative is \( \frac{2}{3} \times \frac{3}{2} x^{3/2 - 1} = x^{1/2} \).

Step 4: Differentiate the Second Term

For the second term \( \frac{2}{5} x^{5/2} \), apply the power rule: Take the exponent, \( \frac{5}{2} \), multiply it by the coefficient \( \frac{2}{5} \), and reduce the exponent by 1. The derivative is \( \frac{2}{5} \times \frac{5}{2} x^{5/2 - 1} = x^{3/2} \).

Step 5: Combine the Derivatives

Now that we have found derivatives for each term, we combine them to get the final derivative of the entire function. Thus, the derivative is \( x^{1/2} + x^{3/2} \).

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus, especially when dealing with derivatives. It is a straightforward technique used to find the derivative of a function in the form of a power of x. To apply the power rule, follow these simple steps:

• Identify the power of the variable, typically denoted as \(n\).
• Multiply the power \(n\) by the coefficient of \(x^n\).
• Subtract 1 from the power \(n\), resulting in a new power of \(x^{n-1}\).

Let's illustrate with an example: For a function \(f(x) = 3x^4\), applying the power rule gives \(f'(x) = 4 \times 3x^{4-1} = 12x^3\). By following these steps, differentiating powers of x becomes manageable. This rule simplifies the process and is applicable even when exponents are fractions or negative numbers, making it incredibly versatile.

Differentiation

Differentiation is the process of finding the derivative of a function. It measures how a function's output value changes as its input changes. In calculus, differentiation provides the rate at which one quantity changes with respect to another. This technique is fundamental in various applications like physics, engineering, and economics. To differentiate a function:

• Identify the terms which need differentiation.
• Apply relevant rules, such as the power rule, to each term individually.
• Combine the differentiated parts to form the derivative of the entire function.

Differentiation results in a new function, the derivative, which can be used to find the slope or rate of change at any given point in the original function's domain. This process is essential for understanding how functions behave and for solving real-world problems.

Exponents in Calculus

Exponents play a crucial role in calculus, particularly in differentiation and integration. In calculus, exponents are not just limited to whole numbers. They often include fractions and negative numbers, which broaden the applicability of calculus in various scenarios. When differentiating terms with exponents:

• Follow the power rule, which works for any real number exponent \(n\), including fractions and negative numbers.
• Fractional exponents often appear in expressions such as \(x^{3/2}\), representing roots and powers simultaneously.
• Negative exponents, like \(x^{-2}\), reflect reciprocal relationships.

Understanding how to handle these different types of exponents is key to mastering calculus operations. They provide the flexibility to describe more complex phenomena, allowing calculus to be used in more sophisticated applications.