Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\tanh x \sinh 2 x $$

Step-by-Step Solution

Answer

The derivative is \( \text{sech}^2 x \sinh 2x + 2 \tanh x \cosh 2x \).

Step 1: Identify the product rule for differentiation

The given function is the product of two simpler functions: \( y = u imes v \) with \( u = \tanh x \) and \( v = \sinh 2x \). To differentiate \( y \) with respect to \( x \), we will use the product rule which states: \( D_x(uv) = u'v + uv' \).

Step 2: Differentiate \( \tanh x \)

The derivative of \( u = \tanh x \) is \( u' = \text{sech}^2 x \). This is because \( \text{sech}^2 x \) is the derivative of the hyperbolic tangent function, \( \tanh x \).

Step 3: Differentiate \( \sinh 2x \)

The function \( v = \sinh 2x \) has a derivative \( v' = 2 \cosh 2x \). This is derived from applying the chain rule to the \( \sinh \) function.

Step 4: Apply the product rule

Substitute \( u' = \text{sech}^2 x \), \( v = \sinh 2x \), \( u = \tanh x \), and \( v' = 2 \cosh 2x \) into the product rule formula: \( D_x y = (\text{sech}^2 x)(\sinh 2x) + (\tanh x)(2 \cosh 2x) \).

Step 5: Simplify the expression

Combine terms: \( D_x y = \text{sech}^2 x \sinh 2x + 2 \tanh x \cosh 2x \). This is the simplified expression for the derivative.

Key Concepts

Tanh Hyperbolic Function

The hyperbolic tangent function, noted as \( \tanh x \), is a fundamental hyperbolic function frequently used in calculus and applications involving hyperbolic equations. It is similar in shape to the ordinary tangent function, but it operationally behaves quite differently.

### Definition and Properties

• The hyperbolic tangent function is defined as: \( \tanh x = \frac{\sinh x}{\cosh x} \).
• It maps real numbers to the open interval (-1, 1), making it a great tool for scenarios where bounded output is needed.
• Graphically, \( \tanh x \) exhibits an S-shaped curve, asymptotically approaching 1 and -1 as \( x \) moves towards positive and negative infinity respectively.

One of the notable aspects of \( \tanh x \) is its derivative. The derivative \( \text{sech}^2 x \) indicates the function's slope relative to changes in \( x \), where \( \text{sech} x \) is the hyperbolic secant function. This reveals that \( \tanh x \) grows steadily without abrupt changes in the input. Recognizing this behavior is important for students learning calculus differentiation techniques as it helps in understanding the stability of functions.

Sinh Hyperbolic Function

The \( \sinh \) function, or hyperbolic sine function, is another core hyperbolic function. It shares similarities with the usual sine function but is distinctly hyperbolic in form.

### Understanding Sinh

• The hyperbolic sine function is described by: \( \sinh x = \frac{e^x - e^{-x}}{2} \).
• It is an odd function, meaning \( \sinh(-x) = -\sinh(x) \).
• Unlike the periodic sine function, \( \sinh x \) continuously increases and decreases with an exponential shape.

When differentiating \( \sinh x \), you will apply the chain rule, especially for composite arguments, such as \( \sinh 2x \). The derivative of \( \sinh x \) is \( \cosh x \), highlighting another key relationship between hyperbolic functions. Therefore, for \( \sinh 2x \), the derivative becomes \( 2\cosh 2x \), which emphasizes how the hyperbolic cosine function \( \cosh x \) is integral in handling derivatives.

Calculus Differentiation Techniques

Differentiation is the fundamental calculus operation used to find the rate at which a function changes at any point. Various rules, such as the product rule and the chain rule, simplify the process of differentiating complex expressions.

### Key Differentiation Rules

• Product Rule: If a function \( y \) is the product of two functions, \( u(x) \) and \( v(x) \), i.e., \( y = u \times v \), the derivative follows the formula: \( D_x(uv) = u'v + uv' \).
• Chain Rule: Useful for composing functions within other functions, allowing the differentiation of nested functions. This rule states that: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

These rules were applied in the original exercise to derive the function \( y = \tanh x \sinh 2x \). The differentiation task involved applying the product rule and recognizing the derivatives of both \( \tanh x \) and \( \sinh 2x \). Understanding and using these differentiation techniques is crucial for solving complex calculus problems efficiently and accurately.