Problem 39 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \cosh (x) / x $$

Step-by-Step Solution

Answer

The derivative is \( \frac{x \sinh(x) - \cosh(x)}{x^2} \).

Step 1: Identify the Quotient Rule for Derivatives

To differentiate the expression \( \frac{\cosh(x)}{x} \), we apply the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Here, \( u = \cosh(x) \) and \( v = x \).

Step 2: Differentiate the Numerator

Differentiate the numerator \( \cosh(x) \) with respect to \( x \). The derivative of \( \cosh(x) \) is \( \sinh(x) \). So, \( \frac{du}{dx} = \sinh(x) \).

Step 3: Differentiate the Denominator

Differentiate the denominator \( x \) with respect to \( x \). The derivative of \( x \) is \( 1 \). So, \( \frac{dv}{dx} = 1 \).

Step 4: Apply the Quotient Rule

Substitute the derivatives into the quotient rule formula: \[ \frac{d}{dx}\left(\frac{\cosh(x)}{x}\right) = \frac{x \cdot \sinh(x) - \cosh(x) \cdot 1}{x^2} \].

Step 5: Simplify the Expression

Simplify the expression to get the final derivative: \[ \frac{x \cdot \sinh(x) - \cosh(x)}{x^2} \]. This is the derivative of the given expression with respect to \( x \).

Key Concepts

Differentiation

Differentiation is one of the fundamental concepts in calculus, involved in finding how a function changes at any given point. It helps in understanding the rate at which quantities change; essentially, it is the mathematics of change. The process of differentiation allows us to find the derivative, which gives the slope of the tangent line to the curve at any particular point.
When differentiating functions, rules such as the product rule, chain rule, and quotient rule become invaluable tools. These rules simplify the process and provide systematic ways to tackle complex problems. In the specific case of the quotient rule, it helps differentiate a division of two functions.

• Quotient Rule: This rule is used when we need to differentiate ratios of two functions.
• Formula: To apply it, we consider a function as a ratio of two differntiable functions, say \(u(x)\) and \(v(x)\). The derivative is then found using:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}.\]

In essence, differentiation is pivotal in calculus as it provides tools to model and understand changes, allowing deeper insight into the behavior of functions.

Hyperbolic Functions

Hyperbolic functions, much like their trigonometric counterparts, are functions that appear frequently in calculus, physics, and engineering. They include functions like \( \cosh(x) \) and \( \sinh(x) \), which are analogous to the cosine and sine functions but are defined using the exponential function. These functions have interesting properties that make them useful in various scientific fields.

• Definition: The hyperbolic cosine, \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• Sinh and Cosh: The hyperbolic sine function, \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).

The important characteristic of hyperbolic functions is how they behave like exponential functions while possessing similarities with trigonometric functions. When differentiating these functions:

• The derivative of \( \cosh(x) \) is \( \sinh(x) \).
• The derivative of \( \sinh(x) \) is \( \cosh(x) \).

These functions are significant in solving various real-world problems, including those in engineering and complex mathematical models.

Calculus Problem Solving

Calculus problems often require a strategic approach for effective solving, especially when involving differentiation and the use of specialized rules like the quotient rule. Successfully tackling such problems depends on understanding the fundamental concepts and methodically applying the appropriate techniques.
When solving calculus problems:

• Understand the Function: Start by identifying the functions you are working with. Determine whether a rule, such as the quotient rule, is necessary.
• Apply Differentiation Rules: Depending on the form of the function, you may need to use different rules such as the product rule or chain rule, in addition to the quotient rule.
• Simplify: Once the derivative is obtained, always simplify the expression where possible. This can help in gaining better insights and can often reveal further properties about the function.

By meticulously applying calculus principles, students can effectively solve problems and gain a richer understanding of different mathematical concepts. This practical skill is not only crucial in mathematics but also extends to physics, engineering, and other domains that rely on mathematical modeling.