Problem 24 - Brief Applied Calculus

Question

Find the derivative of each function. $$ g(x)=\sqrt[3]{x}-\frac{1}{x} $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} + \frac{1}{x^2} \).

Step 1: Identify the Functions to Differentiate

The given function is \( g(x) = \sqrt[3]{x} - \frac{1}{x} \). We can rewrite this for differentiation purposes as \( g(x) = x^{\frac{1}{3}} - x^{-1} \). Now, we're going to differentiate each component separately.

Step 2: Differentiate \( x^{\frac{1}{3}} \)

To differentiate \( x^{\frac{1}{3}} \), we use the power rule for differentiation: \( \frac{d}{dx}[x^n] = nx^{n-1} \). Here, \( n = \frac{1}{3} \), so: \[\frac{d}{dx}[x^{\frac{1}{3}}] = \frac{1}{3}x^{-\frac{2}{3}}.\]

Step 3: Differentiate \( x^{-1} \)

Next, we differentiate \( x^{-1} \) using the power rule: \( \frac{d}{dx}[x^n] = nx^{n-1} \). Here \( n = -1 \), so: \[\frac{d}{dx}[x^{-1}] = -1x^{-2} = -x^{-2}.\]

Step 4: Combine the Derivatives

Combine the derivatives of each part to find the derivative of the entire function. Thus, \( g'(x) = \frac{1}{3}x^{-\frac{2}{3}} + x^{-2} \).

Step 5: Rewrite the Final Derivative

The derivative can be rewritten in a compact form using fractions: \[g'(x) = \frac{1}{3}x^{-\frac{2}{3}} + x^{-2} = \frac{1}{3\sqrt[3]{x^2}} + \frac{1}{x^2}.\]

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus for finding the derivative of polynomial expressions. It simplifies the differentiation process by handling expressions in the form of \( x^n \), where \( n \) is any real number.
To apply the power rule, multiply the exponent \( n \) by the variable’s coefficient (if any), and then subtract one from the exponent. This results in:

• \( \frac{d}{dx}[x^n] = nx^{n-1} \)

This technique allows you to quickly and efficiently find derivatives without needing to apply more complex rules.
For instance, in the function \( x^{\frac{1}{3}} \), applying the power rule gives us \( \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} \). Similarly, for \( x^{-1} \), the derivative is \( -x^{-2} \).
The clarity and speed of the power rule make it a go-to method when differentiating polynomial terms.

Differentiation Techniques

Differentiation is a core aspect of calculus, used to find rates of change or slopes of curves. Various techniques exist, each suited for different forms of functions.
The primary techniques include:

• Power Rule: Used for polynomial expressions, as just described. It makes finding derivatives straightforward when working with exponents.
• Product Rule: Useful when differentiating two multiplicative functions.
• Quotient Rule: Perfect for division of functions.
• Chain Rule: Essential for composite functions; it helps take the derivative of nested functions.

In this exercise, the power rule was key to find the derivative of the rewritten function \( g(x) = x^{\frac{1}{3}} - x^{-1} \).
Knowing when and how to apply these techniques is crucial for solving differentiation tasks efficiently.

Calculus Functions

Calculus functions can come in many forms, from simple polynomials to more complex expressions involving roots and exponential terms.
To successfully handle calculus problems, one should understand:

• How to manipulate algebraic expressions.
• The significance of rewriting complex expressions into simpler polynomial terms, as shown when \( \sqrt[3]{x} \) is rewritten as \( x^{\frac{1}{3}} \).
• The relationship between derivatives and the original function — derivatives offer a snapshot of the function's rate of change at any point \( x \).

In our problem, starting with \( g(x) = \sqrt[3]{x} - \frac{1}{x} \) and moving to \( g(x) = x^{\frac{1}{3}} - x^{-1} \) illustrates the process of simplifying expressions before applying differentiation techniques.
This strategy not only aids in more efficient computation but also enriches the understanding of how different calculus functions behave.