Problem 76 - Calculus Early Transcendentals

Question

Given that \(f^{\prime}(x)=\sqrt{3 x+4}\) and \(g(x)=x^{2}-1,\) find \(F^{\prime}(x)\) if \(F(x)=f(g(x))\)

Step-by-Step Solution

Answer

\(F'(x) = 2x \sqrt{3x^2 + 1}\).

Step 1: Understanding the Chain Rule

The problem requires us to find the derivative of the composite function \(F(x) = f(g(x))\). To do this, we use the chain rule, which states \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).

Step 2: Calculate \(g'(x)\)

Differentiate \(g(x) = x^2 - 1\) with respect to \(x\). The derivative is \(g'(x) = 2x\) because the derivative of \(x^2\) is \(2x\) and the derivative of a constant is zero.

Step 3: Apply the Chain Rule

Using the chain rule formula \((f(g(x)))' = f'(g(x)) \cdot g'(x)\), substitute \(f'(x) = \sqrt{3x + 4}\) evaluated at \(g(x)\) and \(g'(x) = 2x\).

Step 4: Evaluate \(f'(g(x))\)

Since \(g(x) = x^2 - 1\), substitute \(x^2 - 1\) for \(x\) in \(f'(x) = \sqrt{3x + 4}\) to get \(f'(g(x)) = \sqrt{3(x^2 - 1) + 4} = \sqrt{3x^2 - 3 + 4} = \sqrt{3x^2 + 1}\).

Step 5: Combine Results to Find \(F'(x)\)

Substitute \(f'(g(x)) = \sqrt{3x^2 + 1}\) and \(g'(x) = 2x\) into the chain rule expression: \(F'(x) = \sqrt{3x^2 + 1} \cdot 2x\).

Key Concepts

Composite Function

When we talk about functions in mathematics, a composite function is a very interesting concept. It's like baking a cake using two distinct layers. First, we have a function, let's call it \( f(x) \), which takes input from another function, \( g(x) \). The output of \( g(x) \) becomes the input for \( f(x) \). This combination creates a new function, represented as \( f(g(x)) \).
Let's break it down a bit more:

• \( g(x) \) is the inside function that calculates a value first.
• Then, \( f(x) \) takes that output and uses it to find a new result.

Essentially, you are "composing" these two functions to form a single expression. Understanding this is important because it will help you in finding the derivatives of more complex functions using the chain rule.

Derivative

Derivatives are a fundamental concept in calculus, representing how a function changes at any given point. Think of it as the speedometer in your car—it tells you how fast you're going at any instant. Calculating derivatives enables you to understand the rate of change of a function.
In this particular problem, you are given the derivative of \( f(x) \), denoted as \( f'(x) = \sqrt{3x + 4} \). This expression tells us how \( f(x) \) is changing with respect to \( x \). Similarly, the derivative \( g'(x) = 2x \) is derived from the function \( g(x) = x^2 - 1 \).
These individual derivatives play a crucial role when applying the chain rule in composite functions. Rather than working directly with the function, you're using these derivative functions to find how the output of one affects the other.

Function Composition

Function composition is connecting two or more functions to create a new function. Imagine you're building a machine where the output of one module becomes the input for another. This concept is essential when solving problems involving composite functions.
When you have \( F(x) = f(g(x)) \), you've composed the two functions \( f(x) \) and \( g(x) \). Here, we compute \( g(x) \) first and then pass this result into \( f(x) \).

• First, compute \( g(x) \) with some number or variable.
• Next, use the result as input into \( f(x) \).

Function composition is more than just combining functions; it forms the basis for using the chain rule to find derivatives. It's a layered approach that reflects how operations can be stacked to produce new, insightful results. Understanding this lays the groundwork for easily tackling composite derivative problems.