Problem 179 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given functions. $$y=\frac{\sec x}{x}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{\sec x (x \tan x - 1)}{x^2} \)

Step 1: Recognize the Quotient Rule

We need to find the derivative of the function \( y = \frac{\sec x}{x} \). This is a quotient of two functions, so we'll use the Quotient Rule for derivatives, which is given by: \[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]where \( u = \sec x \) and \( v = x \).

Step 2: Differentiate the Numerator

Find the derivative of the numerator \( u = \sec x \). The derivative is:\[ u' = \frac{d}{dx}(\sec x) = \sec x \cdot \tan x \].

Step 3: Differentiate the Denominator

Find the derivative of the denominator \( v = x \). The derivative is:\[ v' = \frac{d}{dx}(x) = 1 \].

Step 4: Apply the Quotient Rule

Substitute \( u, u', v, \) and \( v' \) into the Quotient Rule formula:\[ \frac{d}{dx}\left(\frac{\sec x}{x}\right) = \frac{(\sec x \cdot \tan x) \cdot x - \sec x \cdot 1}{x^2} \].

Step 5: Simplify the Expression

Simplify the expression from the Quotient Rule:\[ = \frac{\sec x \cdot \tan x \cdot x - \sec x}{x^2} \]\[ = \frac{x \sec x \tan x - \sec x}{x^2} \].Factor out \( \sec x \):\[ = \frac{\sec x(x \tan x - 1)}{x^2} \].

Key Concepts

Derivative

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. When finding the derivative of a function, you're essentially finding the slope of its tangent line at any given point. This is crucial for understanding how functions behave.

In our problem, we deal with the derivative of a quotient, which involves two functions: the numerator and the denominator. The derivative gives insight into the rate of change of the quotient itself. In the example with the function \( y = \frac{\sec x}{x} \), we use the Quotient Rule to find its derivative.

Remember, the Quotient Rule is specifically designed for handling divisions of two functions. It's a handy tool that simplifies the process of finding derivatives for these kinds of functions. This makes it possible to calculate the rate at which \( y \) changes with respect to \( x \), even when \( y \) is expressed as a fraction.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and secant are not just for studying triangles. They are also vital in calculus for exploring periodic phenomena and other applications. In this problem, the secant function \( \sec x = \frac{1}{\cos x} \) arises naturally.

When differentiating trigonometric functions, you will encounter specific formulas. For instance, the derivative of \( \sec x \) is \( \sec x \cdot \tan x \). This result stems from the product and chain rules for differentiation, which relate the complexity of trigonometric functions to their behavior as their input changes.

Understanding these derivatives enables you to solve a variety of problems, especially those that involve oscillating or wave-like behaviors, which are modeled by trigonometric functions. By mastering these concepts, you'll better comprehend how derivatives apply to different functions, including those that oscillate or repeat.

Calculus Problem Solving

Solving calculus problems often means applying learned rules and methods to find solutions. Here, we used the Quotient Rule for derivatives, an essential technique when dealing with fractions of functions. The steps involve identifying your "u" (numerator) and "v" (denominator) before substituting them into the rule.

In practice, solving such problems requires:

• Identifying the type of function involved (e.g., a quotient of two functions).
• Choosing and applying the appropriate differentiation rule (e.g., Quotient Rule).
• Simplifying the resulting expression for a clear solution.

Developing these skills in calculus will help in analyzing more complex functions and understanding their behavior in-depth. It also sharpens your problem-solving skills by requiring you to break down a problem systematically and solve it step by step.