Problem 8 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\left(x^{2}+1\right) \sec x $$

Step-by-Step Solution

Answer

\( f^{\prime}(x) = 2x \sec x + (x^2 + 1) \sec x \tan x \)

Step 1: Identify the Components

The function given is a product of two functions: 1. The first part is \( u(x) = x^2 + 1 \).2. The second part is \( v(x) = \sec x \).

Step 2: Apply the Product Rule

To differentiate the product of two functions \( f(x) = u(x)v(x) \), use the product rule: \( f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x) \).

Step 3: Differentiate \( u(x) \)

Differentiate \( u(x) = x^2 + 1 \): \( u^{\prime}(x) = 2x \).

Step 4: Differentiate \( v(x) \)

Differentiate \( v(x) = \sec x \): \( v^{\prime}(x) = \sec x \tan x \).

Step 5: Substitute into Product Rule Formula

Using the product rule, substitute the derivatives:\[ f^{\prime}(x) = (2x)(\sec x) + (x^2 + 1)(\sec x \tan x) \].

Step 6: Simplify the Expression

Expand and combine like terms to simplify:- First term: \(2x \sec x\).- Second term: \((x^2 + 1) \sec x \tan x\). The expression cannot be simplified further efficiently, so leave it as: \[ f^{\prime}(x) = 2x \sec x + (x^2 + 1) \sec x \tan x \].

Key Concepts

Product Rule

In calculus, the product rule is an essential tool for differentiation. It helps us find the derivative of a product of two functions. If you have two functions multiplied together, such as \( f(x) = u(x) v(x) \), the product rule states that the derivative \( f'(x) \) will be:

• \( f'(x) = u'(x) v(x) + u(x) v'(x) \)

In the given exercise, where \( f(x) = (x^2 + 1) \sec x \), we treat \( x^2 + 1 \) and \( \sec x \) as two separate functions. Applying the product rule allows us to differentiate each part separately and then combine the results. This process simplifies the differentiation of complex functions, making it manageable and structured.
Learning the product rule is vital because functions often comprise multiple components, especially in real-world problems. With practice, using the product rule becomes second nature.

Trigonometric Derivatives

Trigonometric functions appear frequently in calculus, and knowing how to differentiate them is crucial. The function \( \sec x \), which is part of our exercise, has a specific derivative. Let's explore what the derivative of a few key trigonometric functions look like:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).
• The derivative of \( \sec x \) is \( \sec x \tan x \).

Knowing these derivatives by heart is essential. For our function \( v(x) = \sec x \), the derivative is \( v'(x) = \sec x \tan x \). This makes calculating the derivative straightforward when using the product rule.
Trigonometric derivatives are foundational for handling calculus problems involving oscillatory behaviors. This knowledge helps understand more complex phenomena in fields like physics and engineering.

Calculus

Calculus is a branch of mathematics focused on derivatives and integrals. It's all about understanding how things change. In the context of the exercise, we are dealing with derivatives, which help in finding the rate of change of a function.

• Derivatives give us the slope of a function at any given point.
• By differentiating, we can determine how one variable changes with another.

When we solve the exercise, we calculate how the function \( f(x) = (x^2 + 1) \sec x \) changes as \( x \) changes. The derivative \( f'(x) \) represents this rate of change.Understanding these principles of calculus opens up a new way to tackle real-world problems. From calculating speeds to predicting economic trends, calculus offers valuable insight into dynamic systems. As you delve deeper, you'll uncover the profound importance of these mathematical concepts in various scientific disciplines.