Problem 14 - University Calculus: Early Transcendentals

Question

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: Identify the components of the function

The given function is \( y = \frac{\cos x}{x} + \frac{x}{\cos x} \). This is a sum of two separate functions: \( \frac{\cos x}{x} \) and \( \frac{x}{\cos x} \). We need to differentiate each of these components individually.

Step 2: Differentiate the first component

The first component is \( \frac{\cos x}{x} \). Apply the quotient rule which states that for \( u = \cos x \) and \( v = x \), the derivative \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Here, \( \frac{du}{dx} = -\sin x \) and \( \frac{dv}{dx} = 1 \). Therefore, the derivative is: \[ \frac{x(-\sin x) - \cos x \cdot 1}{x^2} = \frac{-x\sin x - \cos x}{x^2}. \]

Step 3: Differentiate the second component

The second component is \( \frac{x}{\cos x} \). Again, apply the quotient rule where \( u = x \) and \( v = \cos x \). Here, \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -\sin x \). The derivative is: \[ \frac{\cos x(1) - x(-\sin x)}{\cos^2 x} = \frac{\cos x + x \sin x}{\cos^2 x}. \]

Step 4: Combine the derivatives

Combine the derivatives of both components to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-x\sin x - \cos x}{x^2} + \frac{\cos x + x \sin x}{\cos^2 x}. \]

Step 5: Simplify the final expression, if possible

There's no immediate simplification for the combined derivatives due to the different denominators. Therefore, the derivative of the given function is already expressed correctly in two separate terms. Let this be the final answer.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental concept in calculus used to find the derivative of a function that is the ratio of two differentiable functions. When we have a function written as a division, like \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), we need a systematic way to differentiate it.
To apply the Quotient Rule, follow these steps:

• Identify the numerator function \( u \) and the denominator function \( v \).
• Find the derivative of \( u \), denoted \( \frac{du}{dx} \).
• Find the derivative of \( v \), denoted \( \frac{dv}{dx} \).
• Use the formula for the derivative of a quotient: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

This rule provides a straightforward way to deal with fractions of differentiable functions, helping in determining the rate of change of the output with respect to the input. Remember, practice makes perfect, so try applying it to various types of problems to gain confidence!

Trigonometric Functions

Trigonometric functions are fundamental to calculus, thanks to their periodic nature and appearances in various fields such as physics and engineering. The basic trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)).
Here are some key properties:

• They are periodic, meaning they repeat values in regular intervals. For example, \( \sin \) and \( \cos \) have a period of \( 2\pi \).
• They relate to angles within a right triangle, defining relationships between the angles and sides.
• Trigonometric functions can be extended beyond the typical angle inputs, being evaluated for any real number, often using the unit circle.

Understanding these functions is crucial because they form the foundation upon which we apply calculus techniques. This includes evaluating limits, finding derivatives, and analyzing integrals involving trigonometric expressions.

Derivatives of Trigonometric Functions

Derivatives of trigonometric functions are essential for understanding how these functions behave under changes. Knowing these derivatives aids in solving problems related to oscillations, waves, and other periodic phenomena.
The basic derivatives to keep in mind are:

• \( \frac{d}{dx}(\sin x) = \cos x \)
• \( \frac{d}{dx}(\cos x) = -\sin x \)
• \( \frac{d}{dx}(\tan x) = \sec^2 x \)

When applying these derivatives within the context of other functions, like applying them jointly with the Quotient Rule, they often become parts of more complex derivative expressions. For instance, as seen in the given exercise, the derivative of \( \cos x \) is used when differentiating the quotient \( \frac{\cos x}{x} \). Understanding the derivatives of these basic trigonometric functions allows for deeper insights into the behavior of more intricate mathematical models and equations.