Problem 77 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

\(F'(x) = \frac{1}{2x}\).

Step 1: Identify the Composite Function

We are given two functions: the derivative of one function, \(f'(x) = \frac{x}{x^2 + 1}\), and another function \(g(x) = \sqrt{3x - 1}\). The composite function is \(F(x) = f(g(x))\). Our objective is to find \(F'(x)\).

Step 2: Apply the Chain Rule

To find the derivative of a composite function, we use the chain rule: \[ F'(x) = f'(g(x)) \cdot g'(x) \]. This means we need to evaluate \(f'\) at \(g(x)\) and multiply it by the derivative of \(g(x)\).

Step 3: Calculate \(g'(x)\)

Compute the derivative of \(g(x) = \sqrt{3x - 1}\). Using the chain rule: \[ g'(x) = \frac{1}{2\sqrt{3x - 1}} \cdot 3 = \frac{3}{2\sqrt{3x - 1}} \].

Step 4: Substitute \(g(x)\) into \(f'(x)\)

Substitute \(g(x) = \sqrt{3x - 1}\) into \(f'(x)\): \[ f'(g(x)) = \frac{g(x)}{(g(x))^2 + 1} = \frac{\sqrt{3x - 1}}{3x - 1 + 1} = \frac{\sqrt{3x - 1}}{3x} \].

Step 5: Compute \(F'(x)\)

Combine the results from Steps 3 and 4: \[ F'(x) = f'(g(x)) \cdot g'(x) = \frac{\sqrt{3x - 1}}{3x} \cdot \frac{3}{2\sqrt{3x - 1}} \]. Simplifying, we find: \[ F'(x) = \frac{1}{2x} \].

Step 6: Final Simplification

Confirm the final simplified expression for \(F'(x)\): \[ F'(x) = \frac{1}{2x} \]. Since all terms simplify nicely, we have reached the final result.

Key Concepts

Composite Functions

When dealing with functions like in our exercise, understanding composite functions is key. A composite function, often symbolized as \( F(x) = f(g(x)) \), is a function where the output of one function \( g(x) \) becomes the input for another function \( f(x) \).
This technique is particularly common in calculus when we aim to understand complex relationships by breaking them down into simpler parts.
In the exercise provided, \( F(x) \) is a composite function where \( g(x) = \sqrt{3x - 1} \) is the input for \( f(x) \). One main goal is to find the derivative, or rate of change, of this composite function using existing derivatives of the constituent functions.

Composite functions allow us to model and solve more complicated problems.

They are used in a wide range of disciplines including physics, engineering, and economics to represent increasingly complex systems or models.

By understanding the role of composite functions, you can effectively approach complex calculus problems and solutions.

Derivative

In calculus, derivatives are a measure of how a function changes as its input changes. They provide important information about the function's behavior, such as its rates of increase or decrease.
In our problem, the derivative \( f'(x) = \frac{x}{x^2 + 1} \) is provided. The key is to understand not just what this derivative represents, but how it fits into the bigger picture of taking the derivative of the composite function \( F(x) = f(g(x)) \).
Derivatives serve as a foundation for understanding changes and are useful in numerous real-world applications, allowing us to calculate things like velocity, acceleration, and optimizing functions.

• A derivative helps us determine the slope of a function's tangent at any given point.
• It is the basis for more advanced techniques, like integration and differential equations.

Understanding derivatives are essential not just as standalone concepts but as integral parts of calculus, helping us solve complex problems like the one seen here. Knowing this allows you to harness their power across various math and science disciplines.

Calculus Problem Solving

Effective calculus problem-solving often involves combining multiple techniques and concepts, as seen in this exercise.
The chain rule is one such crucial technique for finding the derivative of composite functions. In our exercise, the chain rule dictates that the derivative of the composite function \( F'(x) \) is the product of \( f'(g(x)) \) and \( g'(x) \).
Using the chain rule requires:

• Identifying the outer function \( f(x) \) and inner function \( g(x) \).
• Finding the derivative of \( g(x) \), here \( g'(x) = \frac{3}{2\sqrt{3x - 1}} \).
• Applying \( g(x) \) into \( f'(x) \) and multiplying by \( g'(x) \) to simplify and find \( F'(x) \).

Breaking down each step, using the chain rule, ensures accuracy and efficiency.
This strategy helps you systematically tackle not only textbook exercises but also any real-world problem by methodically applying the rules of calculus to deconstruct and solve composite functions.