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 Rule to Use

We will use the product rule to differentiate the given function, since it involves a product of two functions, \(y = \cosh(3x) \sinh(x)\). The product rule states that if you have two functions \(u(x)\) and \(v(x)\), their derivative is \((u \cdot v)' = u' \cdot v + u \cdot v'\).

Step 2: Differentiate Each Function Separately

- Differentiate \(u(x) = \cosh(3x)\) with respect to \(x\). Use the chain rule: \(u'(x) = \sinh(3x) \cdot 3\).- Differentiate \(v(x) = \sinh(x)\) with respect to \(x\): \(v'(x) = \cosh(x)\).

Step 3: Apply the Product Rule

Substitute the derivatives into the product rule formula: \[(y)' = u' \cdot v + u \cdot v' = (\sinh(3x) \cdot 3) \cdot \sinh(x) + \cosh(3x) \cdot \cosh(x)\].

Step 4: Simplify the Expression

Multiply and simplify the expression: \[(y)' = 3 \sinh(3x) \sinh(x) + \cosh(3x) \cosh(x)\].

Key Concepts

Product Rule

In calculus, the product rule is a key technique used to differentiate functions that are multiplied together. When you have two functions, say \( u(x) \) and \( v(x) \), the product rule helps to find the derivative of their product. This is essential when dealing with expressions that do not separate into simpler, singular terms. The formula for the product rule is:

• \( (u \, v)' = u' \, v + u \, v' \)

This means that to differentiate \( u(x) \) times \( v(x) \), you differentiate \( u \), multiply by \( v \), then add \( u \) times the derivative of \( v \).

In our specific exercise, the functions involved are \( u(x) = \cosh(3x) \) and \( v(x) = \sinh(x) \). By applying the product rule, you can systematically evaluate the entire derivative of \( \cosh(3x) \sinh(x) \).

Practicing with the product rule prepares you for increasingly complex differentiation problems and provides a method for tackling multiple components in an equation. It serves as a building block for understanding how functions combine at a deeper level.

Hyperbolic Functions

Hyperbolic functions, just like trigonometric functions, play a crucial role in calculus, modeling, and many areas of mathematics. The main hyperbolic functions are \( \sinh(x) \) and \( \cosh(x) \). They are defined using exponential functions:

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

These functions appear in many mathematical applications ranging from geometry to complex analysis. In calculus, they're particularly interesting because they have derivatives resembling those of trigonometric functions.

For example, the derivative of \( \sinh(x) \) is \( \cosh(x) \), and vice versa, the derivative of \( \cosh(x) \) is \( \sinh(x) \). This property makes them easy to differentiate and useful for tackling certain integration problems effortlessly.

In the given exercise, understanding these functions allows us to apply the chain and product rules accurately since they simplify differentiation. Mastering these functions expands your toolkit and prepares you for more advanced manipulations in calculus and related fields.

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiating compositions of functions. Essentially, it provides a way to differentiate a function based on its constituent components. Mathematically, if you have a composite function \( f(g(x)) \), the chain rule states that the derivative is:

• \( f'(g(x)) \cdot g'(x) \)

This tells us to differentiate the outer function \( f \) evaluated at the inner function \( g \), and then multiply by the derivative of the inner function.

In our case, \( u(x) = \cosh(3x) \) requires the chain rule. Differentiating it involves recognizing \( g(x) = 3x \) as the inner function. Therefore, the derivative \( u'(x) \) is \( \sinh(3x) \cdot 3 \) due to the chain rule application.

The chain rule is ubiquitous in calculus because many realistic functions are compositions of other simpler functions. By mastering the chain rule, you empower yourself to tackle a wide array of derivative problems that cannot be solved using simpler operations.