Problem 26 - Calculus and its applications

Question

Find each derivative. $$ \frac{d}{d x}\left(\sqrt[3]{x}+\frac{4}{\sqrt{x}}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Function

First, let's express the function in a form that's easier to differentiate. Instead of \( \sqrt[3]{x} \), we will write \( x^{1/3} \). Similarly, we rewrite \( \frac{4}{\sqrt{x}} \) as \( 4x^{-1/2} \). The function becomes \( x^{1/3} + 4x^{-1/2} \).

Step 2: Differentiate Each Term Separately

Now, apply the power rule to differentiate each term. The power rule states that \( \frac{d}{dx} x^n = nx^{n-1} \). Differentiating \( x^{1/3} \) gives: \( \frac{1}{3}x^{-2/3} \). Differentiating \( 4x^{-1/2} \) gives: \( 4\left( -\frac{1}{2} x^{-3/2} \right) = -2x^{-3/2} \).

Step 3: Combine the Derivatives

Combine the derivatives from Step 2. The derivative of the function \( x^{1/3} + 4x^{-1/2} \) is \[ \frac{1}{3}x^{-2/3} - 2x^{-3/2} \].

Key Concepts

Power Rule

The Power Rule is a fundamental concept in calculus used for finding the derivative of expressions that involve powers of variables. It simplifies the differentiation process, especially when dealing with polynomial functions or terms with fractional exponents.
To use the Power Rule, we follow this straightforward formula: if you have a term in the form of \( x^n \), its derivative is \( n \cdot x^{n-1} \). This means you multiply the original power by the coefficient (if there’s any), then reduce the exponent by one.

• For example, to differentiate \( x^{1/3} \), apply the rule: the derivative becomes \( \frac{1}{3}x^{-2/3} \).

This method also works for larger exponents or negative exponents. Understanding this rule enables us to handle a wide range of calculus problems efficiently and is often combined with other rules like the Chain Rule or Product Rule for more complex functions.

Fractional Exponents

Fractional exponents, also known as rational exponents, play a crucial role in calculus as they make challenging expressions more approachable for differentiation and integration.
When a term involves a fractional exponent, such as roots, it is written as \( x^{m/n} \). Here, \( m \) represents the power to which the base is raised, while \( n \) corresponds to the root being taken. Hence, \( x^{1/3} \) is analogous to \( \sqrt[3]{x} \), simplifying its differentiation through the Power Rule.

• For instance, applying the Power Rule to \( x^{1/3} \), the derivative becomes \( \frac{1}{3} x^{-2/3} \), showcasing how fractional exponents can ease the differentiation process.

Using fractional exponents offers a streamlined approach to dealing with roots and powers, enhancing our ability to apply calculus techniques effectively.

Derivative of a Sum

The Derivative of a Sum is another fundamental calculus concept. It states that the derivative of a sum of functions is just the sum of their individual derivatives. This property makes differentiation significantly easier when dealing with multiple terms in an expression.
For any functions \( u(x) \) and \( v(x) \), the rule can be expressed as: \( \frac{d}{dx} [u(x) + v(x)] = \frac{d}{dx} u(x) + \frac{d}{dx} v(x) \).
When solving problems, like differentiating \( \sqrt[3]{x} + \frac{4}{\sqrt{x}} \), this rule allows us to consider each term separately.

• For example, differentiating \( \sqrt[3]{x} \) after rewriting it as \( x^{1/3} \), and \( \frac{4}{\sqrt{x}} \) as \( 4x^{-1/2} \), shows how the derivative of each step culminates in the solution: \( \frac{1}{3} x^{-2/3} - 2x^{-3/2} \).

Understanding this concept simplifies the process of handling complex expressions and reinforces the application of differential calculus principles.