Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\cosh 3 x \sinh x $$

Step-by-Step Solution

Answer

The derivative is \( 3\sinh(3x)\sinh(x) + \cosh(3x)\cosh(x) \).

Step 1: Identify the Derivative Rule

The function given is a product of two hyperbolic functions, \( \cosh(3x) \) and \( \sinh(x) \). We will use the product rule of differentiation. Recall that if \( y = u \cdot v \), then the derivative \( D_{x} y \) is \( u'v + uv' \).

Step 2: Differentiate \( \cosh(3x) \) with Respect to \( x \)

Using the chain rule, differentiate \( \cosh(3x) \). The derivative of \( \cosh(x) \) is \( \sinh(x) \), so for \( \cosh(3x) \), it is \( 3 \cdot \sinh(3x) \). Therefore, \( u' = 3\sinh(3x) \).

Step 3: Differentiate \( \sinh(x) \) with Respect to \( x \)

The derivative of \( \sinh(x) \) is \( \cosh(x) \). Therefore, \( v' = \cosh(x) \).

Step 4: Apply the Product Rule

Now substitute \( u = \cosh(3x) \), \( v = \sinh(x) \), \( u' = 3\sinh(3x) \), and \( v' = \cosh(x) \) into the product rule formula: \[ D_{x} (\cosh(3x)\sinh(x)) = (3\sinh(3x)) \cdot \sinh(x) + (\cosh(3x)) \cdot \cosh(x) \]

Step 5: Simplify the Expression

Simplify the expression obtained: \[ 3\sinh(3x) \sinh(x) + \cosh(3x) \cosh(x) \]. There are no further simplifications possible, so this is the final result.

Key Concepts

Product Rule

The product rule is an essential concept in calculus, which allows us to differentiate products of functions. It is incredibly useful when dealing with expressions where one function is multiplied by another. In essence, the product rule states that if you have a function in the form of \( y = u(x) \cdot v(x) \), then the derivative of \( y \) with respect to \( x \), denoted \( D_x(y) \), is calculated as:

• \( u'v + uv' \)

where \( u' \) and \( v' \) are the derivatives of \( u(x) \) and \( v(x) \) respectively.
The purpose of this rule is to enable differentiation when functions are intertwined by multiplication, allowing partial differentiation of each function one at a time. It simplifies deriving complex functions by breaking them down into manageable sub-parts, making it easier to work through their individual derivatives before combining the results back together.

Chain Rule

The chain rule is another fundamental theorem of differentiation, which deals with the composition of functions. It's particularly important when you have functions inside other functions, known as nested functions. When you're asked to differentiate such a composition, represented by \( y = f(g(x)) \), the chain rule provides the formula:

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

This rule expresses that you first differentiate the outer function, keeping the inner function unchanged, and then multiply this by the derivative of the inner function, \( g(x) \).
The chain rule is crucial in instances where differentiating the outer function directly isn't possible without considering how an inner function might affect it. This approach effectively unpacks the nested structure, allowing for differentiation step-by-step.

Hyperbolic Functions

Hyperbolic functions, analogous to the trigonometric functions like sine and cosine, derive from exponential functions rather than circular functions. The primary hyperbolic functions are the hyperbolic sine \( \sinh(x) \) and hyperbolic cosine \( \cosh(x) \). These functions are defined as:

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

Their derivatives follow specific patterns. For instance, the derivative of \( \sinh(x) \) is \( \cosh(x) \), while the derivative of \( \cosh(x) \) is \( \sinh(x) \), reflecting their alternating nature.
Understanding these properties aids in differentiating complex functions that include hyperbolic terms, as these often appear in calculus problems involving exponential growth, disciplines such as engineering, and in solving differential equations. Hyperbolic functions become particularly handy when dealing with equations that involve hyperbolas, or when working with models that grow exponentially.