Problem 10 - Thomas Calculus

Question

In Exercises \(1-12,\) find \(d y / d x\) $$ y=\frac{\cos x}{x}+\frac{x}{\cos x} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} + \frac{\cos x + x \sin x}{\cos^2 x} \)

Step 1: Express the Derivatives

The given function is composed of two terms: 1. \( \frac{\cos x}{x} \)2. \( \frac{x}{\cos x} \)We'll differentiate each term separately using the quotient rule, which states that if you have a function \( \frac{u}{v} \), then its derivative is given by \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \).

Step 2: Derive the First Term

For the first term \( \frac{\cos x}{x} \):- Let \( u = \cos x \) and \( v = x \).- The derivative \( \frac{du}{dx} = -\sin x \) and \( \frac{dv}{dx} = 1 \).Using the quotient rule:\[\frac{d}{dx}\left(\frac{\cos x}{x}\right) = \frac{x(-\sin x) - \cos x(1)}{x^2} = \frac{-x \sin x - \cos x}{x^2}\]

Step 3: Derive the Second Term

For the second term \( \frac{x}{\cos x} \):- Let \( u = x \) and \( v = \cos x \).- The derivative \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -\sin x \).Using the quotient rule:\[\frac{d}{dx}\left(\frac{x}{\cos x}\right) = \frac{\cos x(1) - x(-\sin x)}{\cos^2 x} = \frac{\cos x + x \sin x}{\cos^2 x}\]

Step 4: Combine the Derivatives

The derivative \( \frac{dy}{dx} \) is the sum of the derivatives of the two terms:\[\frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} + \frac{\cos x + x \sin x}{\cos^2 x}\]

Key Concepts

Quotient Rule

The Quotient Rule in calculus is a technique used to differentiate functions that are expressed as a division of two other functions. If you have a function of the form \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions, the derivative \( \frac{d}{dx} \left( \frac{u}{v} \right) \) can be found using the formula:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

This formula helps to determine the rate of change of one function relative to another when they are in a quotient. The Quotient Rule is particularly helpful for complicated expressions where other differentiation methods, like the Product Rule, are not applicable. It requires us to:- Differentiate the numerator \( u \)- Differentiate the denominator \( v \)- Plug these derivatives into the quotient rule formulaUsing the Quotient Rule effectively requires careful organization of components, making sure to subtract correctly and divide by the square of the denominator.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, especially when working with derivatives. In this context, the primary trigonometric functions involved are cosine (\( \cos x \)) and sine (\( \sin x \)). Derivatives of these functions are essential for solving calculus problems:

• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \sin x \) is \( \cos x \).

These rules are direct results from the behavior of trigonometric curves and their periodic nature. When you differentiate a trigonometric function inside another function, especially in quotients like \( \frac{\cos x}{x} \), it’s important to use these basic derivatives as stepping stones. Trigonometric functions often appear in calculus problems, not just due to their practical significance in modeling waves and oscillations, but also because they introduce periodicity into otherwise linear problems.

Differentiation Techniques

Differentiation techniques in calculus are methods used to find the rate of change of a function. These include basic rules like the power rule, product rule, and indeed, the quotient rule. Additionally, differentiation can involve:

• Chain Rule: Used when a function is composed of other functions.
• Implicit Differentiation: Helpful for equations not solved for one variable.
• Logarithmic Differentiation: Useful for functions involving products or powers, often simplifying complex derivatives.

In our original exercise, the application of the Quotient Rule is crucial to handle trigonometric terms divided by other functions. Mastering these techniques enables you to deconstruct complex expressions and solve derivatives effectively. Understanding when and how to use these techniques can distinguish a challenging calculus problem from a manageable one, making them essential skills for any student or professional working with mathematical models.