Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\operatorname{coth} 4 x \sinh x $$

Step-by-Step Solution

Answer

\(D_x y = -4 \sinh x \csch^2 4x + \coth 4x \cosh x\).

Step 1: Understand the function

We need to find the derivative of the function \(y = \coth 4x \sinh x\) with respect to \(x\). The function is a product of two components: \(\coth 4x\) and \(\sinh x\).

Step 2: Use the Product Rule

The product rule for differentiation states that if \(y = u(x) \cdot v(x)\), then the derivative \(D_x y\) is given by \(u'(x)v(x) + u(x)v'(x)\). Here, let \(u(x) = \coth 4x\) and \(v(x) = \sinh x\).

Step 3: Differentiate \(\coth 4x\)

The derivative of \(\coth 4x\) with respect to \(x\) is \(u'(x) = -4 \csch^2 4x\), using the chain rule, where \(\csch(4x) = \frac{1}{\sinh(4x)}\).

Step 4: Differentiate \(\sinh x\)

The derivative of \(\sinh x\) is \(v'(x) = \cosh x\). This is a basic hyperbolic derivative.

Step 5: Apply the Product Rule

Substituting the derivatives into the product rule formula, we get:\[ D_x y = (-4 \csch^2 4x) \cdot \sinh x + \coth 4x \cdot \cosh x \].

Step 6: Simplify the expression

Simplify the expression:1. \((-4 \csch^2 4x) \cdot \sinh x\) remains as is.2. \(\coth 4x \cdot \cosh x\) encapsulates the original hyperbolic tangent relationship.Thus, the expression becomes:\[-4 \sinh x \cdot \csch^2 4x + \coth 4x \cdot \cosh x \].

Key Concepts

Product Rule

The product rule is a fundamental principle in calculus used to find the derivative of a product of two functions. If you have a function that can be expressed as the product of two separate functions, like \(y = u(x) \cdot v(x)\), then the derivative is more than just the derivative of each component. It requires a specific computation.

• According to the product rule, the derivative \(D_x y\) is given by the formula: \(u'(x)v(x) + u(x)v'(x)\).
• This means you first multiply the derivative of the first function by the second function as it is, then add it to the product of the first function and the derivative of the second function.
• For the function \(y = \coth 4x \sinh x\), we apply the rule by setting \(u(x) = \coth 4x\) and \(v(x) = \sinh x\), and their derivatives are calculated separately.

Using the product rule systematically ensures you find the correct derivative for such products, capturing all interactions between the two functions.

Hyperbolic Functions

Hyperbolic functions, like the regular trigonometric functions, have their own set of unique rules and identities. They are denoted as \(\sinh x\), \(\cosh x\), \(\tanh x\), \(\coth x\), etc. These functions relate to the hyperbola, similar to how trigonometric functions relate to the circle.

• The functions \(\sinh x\) and \(\cosh x\) serve as the building blocks, defined as \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\).
• Other hyperbolic functions, like \(\coth x\), are derived from these primary functions, with \(\coth x = \frac{\cosh x}{\sinh x}\).
• Hyperbolic functions have derivatives that mirror their trigonometric counterparts, such as \(\frac{d}{dx}\sinh x = \cosh x\), which is essential in differentiation tasks.

Understanding these derivatives helps in simplifying expressions and finding derivatives efficiently, especially in complex applications.

Differentiation

Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to a variable. This key concept in calculus is essential for analyzing functions within a wide range of fields, from physics to economics.

• The derivative of a function \(f(x)\) at any point provides the slope of the tangent line to the graph of the function at that point.
• This slope indicates how much \(f(x)\) changes for a small change in \(x\).
• In our function \(y = \coth 4x \sinh x\), differentiation involves finding how each part of our expression changes, applying both the product rule and the understanding of hyperbolic derivatives.

Mastering differentiation techniques are fundamental to navigating complex mathematical and scientific challenges.

Chain Rule

The chain rule is a powerful tool for differentiating composite functions, where one function is nested within another. It allows us to find the derivative of these compositions accurately.

• The chain rule states that if \(y = f(g(x))\), then the derivative \(D_x y\) is found by multiplying the derivative of the outer function \(f\) by the derivative of the inner function \(g\).
• In our function, differentiating \(\coth 4x\) involves the chain rule because it is a composition: \(4x\) is inside the \(\coth\) function.
• We find the derivative of \(\coth x\) and multiply it by the derivative of the internal function \(4x\), resulting in \(-4 \csch^2 4x\).

By using the chain rule, we effectively decompose complex functions and simplify their differentiation, turning seemingly daunting problems into manageable tasks.