Problem 11 - Calculus Early Transcendentals

Question

Find \(f^{\prime}(x)\) $$ f(x)=\sec x \tan x $$

Step-by-Step Solution

Answer

The derivative is \(f^{\prime}(x) = \sec x \tan^2 x + \sec^3 x\).

Step 1: Understand the Function

We are given the function \(f(x) = \sec x \tan x\). To differentiate this function, we'll use the product rule since it is a product of two functions: \(u(x) = \sec x\) and \(v(x) = \tan x\).

Step 2: Recall the Product Rule

The product rule states that if \(f(x) = u(x) v(x)\), then \(f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)\). We will apply this rule using the functions \(u(x)\) and \(v(x)\).

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

To find \(u^{\prime}(x)\), differentiate \(u(x) = \sec x\). The derivative of \(\sec x\) is \(\sec x \tan x\). So, \(u^{\prime}(x) = \sec x \tan x\).

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

To find \(v^{\prime}(x)\), differentiate \(v(x) = \tan x\). The derivative of \(\tan x\) is \(\sec^2 x\). So, \(v^{\prime}(x) = \sec^2 x\).

Step 5: Apply the Product Rule

Substitute the derivatives into the product rule formula: \(f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)\). Using the values, we have \(f^{\prime}(x) = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x)\).

Step 6: Simplify the Expression

Simplify \(f^{\prime}(x) = \sec x \tan^2 x + \sec^3 x\). This expression is already quite simplified, but you can factor out a \(\sec x\) to get \(f^{\prime}(x) = \sec x (\tan^2 x + \sec^2 x)\), if preferred.

Key Concepts

Product Rule

The product rule is a vital technique in calculus, particularly when you're dealing with the differentiation of functions that are multiplied together. Whether you have two or more functions being multiplied, the product rule provides a straightforward method to find the derivative.
In essence, if you have a function expressed as the product of two functions, say \(f(x) = u(x) \times v(x)\), the product rule states that the derivative \(f^{\prime}(x)\) can be found using the formula:

• \(f^{\prime}(x) = u^{\prime}(x) \cdot v(x) + u(x) \cdot v^{\prime}(x)\)

This formula is very handy as it allows us to handle the differentiation separately for each function, \(u(x)\) and \(v(x)\), before combining the results to get the final derivative.
In our specific case with \(f(x) = \sec x \tan x\), we split the function into \(u(x) = \sec x\) and \(v(x) = \tan x\), and then find the derivatives of each function to substitute them back into the product rule formula. This approach simplifies the process and helps us solve the problem methodically.

Trigonometric Functions

Trigonometric functions like \(\sec x\) and \(\tan x\) are fundamental in calculus and often require special attention when differentiating. Here's a quick breakdown of the derivatives for some commonly encountered trigonometric functions:

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

Understanding these derivatives is crucial as they frequently appear in calculus problems involving oscillations, waves, and other periodic behavior.
When working with trigonometric functions, always remember to refer back to their fundamental identities. For example, knowing that \(\tan^2 x + 1 = \sec^2 x\) can help you simplify expressions further.
In our problem, both \(\sec x\) and \(\tan x\) play a role in defining the function we aim to differentiate. Being comfortable with their properties and derivatives makes finding \(f^{\prime}(x)\) more manageable.

Differentiation Techniques

Differentiation techniques are the tools you use to compute derivatives, which are central in calculus for analyzing rates of change. The product rule is part of a larger toolkit that also includes:

• The power rule, for differentiating expressions of the form \(x^n\).
• The chain rule, for dealing with composite functions.
• Basic rules for higher derivatives, like for \(\sin x\), \(\cos x\), and other trigonometric functions.

Each technique is used under specific circumstances to break down more complex differentiation problems into manageable parts.
In the example problem, the combination of the product rule and knowledge of trigonometric derivatives allowed us to successfully differentiate \(f(x)\ = \sec x \tan x\).
By leveraging these fundamental differentiation techniques, you can tackle any problem involving derivatives, thereby gaining a deeper understanding of the behavior of functions in calculus.