Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\ln (\sinh x) $$

Step-by-Step Solution

Answer

The derivative is \( \coth(x) \).

Step 1: Recall the Derivative Formula

To find the derivative of a function involving a natural logarithm, such as \(y = \ln(u)\), remember that the derivative is \(\frac{1}{u} \cdot \frac{du}{dx}\). Here, \(u = \sinh(x)\).

Step 2: Differentiate \(\sinh(x)\)

We need the derivative of \(\sinh(x)\). Recall that the derivative of \(\sinh(x)\) is \(\cosh(x)\). This gives us \(\frac{du}{dx} = \cosh(x)\).

Step 3: Apply the Chain Rule

Using the derivative formula for logarithmic functions \(\frac{1}{u} \cdot \frac{du}{dx}\), substitute \(u = \sinh(x)\) and \(\frac{du}{dx} = \cosh(x)\). Thus, the derivative is \(\frac{1}{\sinh(x)} \cdot \cosh(x)\).

Step 4: Simplify the Expression

To simplify the expression \(\frac{1}{\sinh(x)} \cdot \cosh(x)\), recall the hyperbolic identity \(\coth(x) = \frac{\cosh(x)}{\sinh(x)}\). Therefore, the derivative \(D_{x} y = \coth(x)\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, as opposed to a circle. The most common hyperbolic functions are hyperbolic sine ( sinh ) and hyperbolic cosine ( cosh ). These functions have properties similar to their trigonometric counterparts.
The hyperbolic sine function, \( sinh(x) \), is defined as:

• \(\ sinh(x) = \frac{e^x - e^{-x}}{2} \)

This function closely resembles the sine function' behavior but for the hyperbolic setting. Then there's the hyperbolic cosine function, \( cosh(x) \), defined by:

• \(\ cosh(x) = \frac{e^x + e^{-x}}{2} \)

Both functions, \( sinh(x) \) and \( cosh(x) \), satisfy the identity:

• \( cosh^2(x) - sinh^2(x) = 1 \)

Understanding these helps in calculus, especially when differentiating or integrating expressions involving hyperbolic functions.

Natural Logarithm

The natural logarithm, denoted as \( ln(x) \), is the logarithm to the base of the constant \( e \) (approximately 2.718). It is a fundamental concept in calculus.
The natural logarithm has several useful properties:

• The derivative of \( ln(x) \) is \( \frac{1}{x} \).
• The natural logarithm and exponential function are inverses: \( e^{ln(x)} = x \).
• ln(ab) = ln(a) + ln(b).
• ln(\( \frac{a}{b} \)) = ln(a) - ln(b).

These properties are especially helpful when working on logarithmic differentiation or integration. It's crucial to get comfortable with these logarithmic rules when tackling calculus problems involving logarithmic functions.

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. It states that if a function \( y \) can be expressed as a composite of two functions such that \( y = f(g(x)) \), then its derivative is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

This means, firstly, you take the derivative of the outer function while keeping the inner function intact. Then, multiply it by the derivative of the inner function.
Applying the chain rule correctly is crucial when dealing with nested functions. In the given exercise, the function \( y = ln(sinh(x)) \) demonstrates this. The outer function is \( ln(u) \) where \( u = sinh(x) \). So we first take the derivative of \( ln(u) \) which is: \( \frac{1}{u} \) and then multiply by the derivative of \( u = sinh(x) \), which is \( cosh(x) \). This gives us the final derivative.

Derivative Simplification

Once the derivative of a function is calculated, simplifying it to the most comprehensive form often helps in further applications such as solving integrals or setting up equations. Simplification involves using known identities or properties.
In the context of hyperbolic functions, one can use identities such as:

• \( coth(x) = \frac{cosh(x)}{sinh(x)} \)

This is what was done in the final step of the original solution. After applying the chain rule, the derivative was obtained as \( \frac{1}{sinh(x)} \cdot cosh(x) \). By utilizing the identity for \( coth(x) \), the expression simplifies to:

• \( D_{x} y = coth(x) \)

Such simplifications not only provide a cleaner solution but also express the derivative in terms that might be more useful for further mathematical manipulation or interpretation.