Problem 11 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=-3 x^{-8}+2 \sqrt{x} $$

Step-by-Step Solution

Answer

The derivative is \(f'(x) = 24x^{-9} + x^{-1/2}\).

Step 1: Identify the Individual Functions

The given function is \(f(x) = -3x^{-8} + 2\sqrt{x}\). This function consists of two separate terms: \(-3x^{-8}\) and \(2\sqrt{x}\). Our goal is to find the derivative of each term separately.

Step 2: Differentiate the First Term \(-3x^{-8}\)

Use the power rule for differentiation, \(\frac{d}{dx}[x^n] = nx^{n-1}\). For \(-3x^{-8}\), the derivative is \(-3(-8)x^{-8-1} = 24x^{-9}\).

Step 3: Differentiate the Second Term \(2\sqrt{x}\)

First, rewrite \(\sqrt{x}\) as \(x^{1/2}\). Then apply the power rule: \(\frac{d}{dx}[2x^{1/2}] = 2 \cdot \frac{1}{2}x^{1/2-1} = x^{-1/2}\).

Step 4: Combine the Derivatives

Now that we have the derivatives of each term, we combine them to find \(f'(x)\). Therefore, \(f'(x) = 24x^{-9} + x^{-1/2}\).

Key Concepts

Power Rule

The power rule is a fundamental technique used in calculus for finding the derivative of a power function. It's especially useful when working with polynomial expressions. The power rule states that if you have a function of the form \(x^n\), where \(n\) is any real number, the derivative is \(nx^{n-1}\). This rule simplifies the differentiation process, allowing us to quickly find the slopes of power functions.
Imagine you have a term like \(x^3\); using the power rule, the derivative becomes \(3x^2\). The process involves multiplying the exponent by the coefficient and reducing the exponent by one.

• Formula: \(\frac{d}{dx}[x^n] = nx^{n-1}\)
• Example: If \(f(x) = 5x^4\), then \(f'(x) = 20x^3\).
• Why it works: It reduces complexity when differentiating polynomial equations.

In our exercise, the power rule helped us differentiate each term individually. For \(-3x^{-8}\), applying the rule gave us \(24x^{-9}\), showcasing how negative exponents are handled effortlessly. With the term \(2\sqrt{x}\), rewriting it as \(2x^{1/2}\) simplified its differentiation using this versatile rule.

Differentiation

Differentiation is a key operation in calculus that focuses on finding the rate at which a function changes. It's basically finding the slope of the tangent line at any point of the function. Imagine trying to assess how fast something is moving; differentiation gives you the tool to measure that.
Whenever we differentiate a function, we're calculating its derivative, which tells us the gradient or rate of change. In simpler terms, it answers, "How fast is this changing?"
Here's why differentiation is important:

• It allows us to find maximum and minimum points of functions.
• It's used in physics for finding velocity and acceleration.
• Financial analysts use it to predict market trends.

In our problem, we differentiated the function \(f(x) = -3 x^{-8} + 2 \sqrt{x}\). By differentiating each term separately, we could easily find \(f'(x)\), which describes how \(f(x)\) changes across different values of \(x\). Differentiation enabled us to transform complex algebraic expressions into simpler derivatives like \(24x^{-9} + x^{-1/2}\).

Functions

Functions are the cornerstone of calculus, relating input values to output values in a prescribed way. Think of a function like a machine that takes in a number, performs a computation, and then spits out another number. This concept is simple yet powerful, underpinning many real-world phenomena.
A mathematics function is usually written as \(f(x)\), where \(f\) denotes the function, and \(x\) represents the independent variable. This notation shows how changing the input (\(x\)) affects the output (\(f(x)\)). Understanding how functions behave allows us to model everything from the path of a ball flying through the air to the growth of investments over time.
Some key points to remember about functions:

• They map each input to exactly one output.
• Functions can be linear (straight lines), quadratic (parabolas), exponential, and more.
• The domain is the set of all possible inputs, while the range is the set of possible outputs.

In our textbook problem, when we identified the function \(f(x) = -3 x^{-8} + 2 \sqrt{x}\), we considered it a composition of simpler sub-functions. Each part was differentiated separately to understand how the entire function behaves when \(x\) changes. Recognizing functions and their derivatives is crucial for solving complex calculus problems efficiently.