Problem 14 - Calculus - AP Edition

Question

Find \(f^{\prime}(x)\). $$f(x)=\frac{\sec x}{1+\tan x}$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{\sec x ( \tan x + \tan^2 x - \sec^2 x )}{(1 + \tan x)^2} \).

Step 1: Rewrite the Function

The function is given as \( f(x) = \frac{\sec x}{1 + \tan x} \). To differentiate it, we should first identify this as a quotient of two functions: the numerator \( g(x) = \sec x \) and the denominator \( h(x) = 1 + \tan x \).

Step 2: Use the Quotient Rule

To find the derivative of a quotient \( \frac{g(x)}{h(x)} \), use the quotient rule which states: \((\frac{g}{h})' = \frac{g'h - gh'}{h^2}\).

Step 3: Differentiate the Numerator

Find the derivative of the numerator: \( g'(x) = (\sec x)' = \sec x \tan x \). This comes from the derivative of \( \sec x \) being \( \sec x \tan x \).

Step 4: Differentiate the Denominator

Find the derivative of the denominator: \( h'(x) = (1 + \tan x)' = \sec^2 x \). This is because the derivative of \( \tan x \) is \( \sec^2 x \).

Step 5: Apply the Quotient Rule

Substitute into the quotient rule: \(\frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}\). So, \( g'(x) = \sec x \tan x \), \( h(x) = 1+\tan x \), \( g(x) = \sec x \), and \( h'(x) = \sec^2 x \).This gives: \[\frac{(\sec x \tan x)(1 + \tan x) - (\sec x)(\sec^2 x)}{(1 + \tan x)^2} \].

Step 6: Simplify the Expression

Simplify the numerator: \((\sec x \tan x)(1 + \tan x) = \sec x \tan x + \sec x \tan^2 x \). Hence,\((\sec x \tan x + \sec x \tan^2 x) - (\sec x \sec^2 x) = \sec x \tan x + \sec x \tan^2 x - \sec^3 x \).Combine terms:\[ f'(x) = \frac{\sec x ( \tan x + \tan^2 x - \sec^2 x )}{(1 + \tan x)^2} \].

Key Concepts

Quotient Rule

When dealing with derivatives of functions expressed as a division, the quotient rule is indispensable. It helps us find the derivative of a quotient of two functions. For a function of the form \(\frac{g(x)}{h(x)}\), the derivative is found using:

• Numerator: The derivative of the top function \(g'(x)\)
• Denominator: The derivative of the bottom function \(h'(x)\)
• Quotient Formula: \(\left(\frac{g}{h}\right)' = \frac{g'h - gh'}{h^2}\)

In simpler terms, you take the derivative of the numerator and multiply it by the denominator. Then, subtract the product of the numerator and the derivative of the denominator. Finally, divide everything by the square of the original denominator. This method allows the use of existing derivatives to form a new derivative, streamlining calculations.

Trigonometric Functions

Trigonometric functions like \(\sec x\) and \(\tan x\) are pivotal in calculus. These functions are integral in various math principles, including differentiation. Differentiating trigonometric functions requires knowing their specific derivatives.

• For \(\sec x\), the derivative is \(\sec x \tan x\).
• For \(\tan x\), the derivative is \(\sec^2 x\).

These derivatives are derived from fundamental trigonometric identities and rules. Familiarity with them will help you solve a wide range of calculus problems. It is also essential to understand that simplifying such expressions often involves employing trigonometric identities and manipulations.

Differentiation Techniques

Differentiation is one of the core operations in calculus. Various techniques help us address different types of functions, like products, quotients, and compositions. In this exercise, we applied the quotient rule as part of differentiation techniques. Here's why it's crucial:

• **Identify the Function Type:** Recognize if the function can be differentiated via the product rule, chain rule, or quotient rule.
• **Apply the Rule Correctly:** Accurately applying these rules requires understanding each element's role (i.e., numerator vs. denominator in quotient).
• **Simplify Often:** After applying differentiation rules, simplifying the expressions can clarify the function's behavior or result.

Understanding and mastering these techniques not only aids in solving textbook exercises but also enhances your problem-solving toolkit in calculus, giving you the ability to tackle a broader scope of mathematical problems.