Chapter 3: Problem 10
In Exercises \(1-12,\) find \(d y / d x\) $$ y=\frac{\cos x}{x}+\frac{x}{\cos x} $$
Short Answer
Expert verified
\( \frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} + \frac{\cos x + x \sin x}{\cos^2 x} \)
Step by step solution
01
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 \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \).
02
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:\[\frac{d}{dx}\left(\frac{\cos x}{x}\right) = \frac{x(-\sin x) - \cos x(1)}{x^2} = \frac{-x \sin x - \cos x}{x^2}\]
03
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:\[\frac{d}{dx}\left(\frac{x}{\cos x}\right) = \frac{\cos x(1) - x(-\sin x)}{\cos^2 x} = \frac{\cos x + x \sin x}{\cos^2 x}\]
04
Combine the Derivatives
The derivative \( \frac{dy}{dx} \) is the sum of the derivatives of the two terms:\[\frac{dy}{dx} = \frac{-x \sin x - \cos x}{x^2} + \frac{\cos x + x \sin x}{\cos^2 x}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)
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 \( \cos x \).
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.