Problem 41 - College Algebra and Calculus: An Applied Approach

Question

find \(f^{\prime}(x)\). $$ f(x)=(x+4)\left(2 x^{2}-1\right) $$

Step-by-Step Solution

Answer

The derivative of the function is \( f^{\prime}(x) = 6x^{2}+16x-1\)

Step 1: Identify the functions

Identify the two functions which are being multiplied together, treating \(x+4\) as the first function (\(u\)) and \(2x^{2}-1\) as the second function (\(v\)).

Step 2: Determine the derivatives

Calculate the derivatives of \(u\) and \(v\). For \(u=(x+4)\), the derivative \(u^{\prime}\) = 1. For \(v = (2x^{2}-1)\), use the power rule to find that \(v^{\prime} = 4x\).

Step 3: Apply the Product rule

Apply the product rule, which states that \((uv)^{\prime} = u^{\prime}v + uv^{\prime}\). After substituting \(u\), \(v\), \(u^{\prime}\), and \(v^{\prime}\) into this equation, we get \( f^{\prime}(x) = 1*(2x^{2}-1) + (x+4)*4x \).

Step 4: Simplify

Simplify \( f^{\prime}(x)\) to find a more convenient form. After applying the distributive property and combining like terms, we get \( f^{\prime}(x) = 2x^{2}-1+4x^{2}+16x = 6x^{2}+16x-1\).

Key Concepts

Derivative

The concept of a derivative is central in calculus. It measures how a function changes as its input changes. Think of it as the slope of the function at any given point, representing the rate of change or the velocity of that point. The notation for the derivative is often written as \( f'(x) \) for a function \( f(x) \).

In the exercise we're considering, \( f(x) = (x+4)(2x^{2}-1) \), the derivative \( f'(x) \) tells us how this function changes as \( x \) varies. By finding the derivative, we determine the best linear approximation for the function at every point on its curve.

Power Rule

The Power Rule is a fundamental tool in differentiation which makes finding derivatives straightforward. It states that if you have a function of the form \( x^n \), its derivative is \( nx^{n-1} \). For example, the derivative of \( x^2 \) is \( 2x \).

This rule comes in handy when working with algebraic functions, especially polynomials. In our problem, when calculating the derivative of the product \( (x+4)(2x^2-1) \), the power rule was applied to the second function \( v = 2x^{2}-1 \). Using the power rule, its derivative \( v' \) became \( 4x \). This simplification is crucial for performing calculations efficiently and accurately.

Differentiation

Differentiation is the process of finding a derivative of a function. It's a way of assessing how a function is changing at any point. The process can involve several rules, such as the product rule, which are key to solving more complex problems.

In the exercise, we are differentiating a product of two functions: \( u = x+4 \) and \( v = 2x^2-1 \). The product rule is essential here, which is a formula used when taking the derivative of a product of two functions. It is expressed as \( (uv)' = u'v + uv' \). By following this steps, we arrive at the derivative \( f'(x) \) as shown in the solution.

Algebraic Functions

Algebraic functions are built from basic algebraic operations, such as addition, subtraction, multiplication, and division, involving variable expressions and coefficients. These include polynomials like the ones seen in this exercise.

The exercise features a product of algebraic functions: \( f(x) = (x+4)(2x^2-1) \). Understanding the structure of these functions helps us to simplify and find their derivatives efficiently. By recognizing each component, such as \( x+4 \) and \( 2x^2 - 1 \), and applying rules like the power rule and product rule, deriving simpler forms such as \( f'(x) = 6x^2 + 16x - 1 \) becomes more manageable.

Mastery of algebraic functions expands one's capabilities in calculus, enabling the solving of intricate problems with confidence.