Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\( D_x y = \cosh x \cosh 4x + 4 \sinh x \sinh 4x \).

Step 1: Identify Product Rule

The expression given is in the form of a product, which means we will use the product rule for differentiation. The product rule states that if you have a function in the form of \( u(x) \cdot v(x) \), then the derivative \( D_x (u(x) \cdot v(x)) = u'(x)v(x) + u(x)v'(x) \). For our problem, let \( u(x) = \sinh x \) and \( v(x) = \cosh 4x \).

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

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

Step 3: Differentiate \( v(x) = \cosh 4x \)

First, apply the chain rule to \( \cosh 4x \). The derivative of \( \cosh x \) is \( \sinh x \). Therefore, by the chain rule, \( v'(x) = 4 \cdot \sinh (4x) \) because the inner function \( 4x \) has a derivative of 4.

Step 4: Apply the Product Rule

Using the results from the previous steps, apply the product rule: \( D_x(y) = u'(x)v(x) + u(x)v'(x) = (\cosh x)(\cosh 4x) + (\sinh x)(4 \sinh 4x) \).

Step 5: Simplify the Expression

The final answer is \( D_x y = \cosh x \cosh 4x + 4 \sinh x \sinh 4x \). This is the derivative of the given function.

Key Concepts

Product Rule

When working with calculus, you often encounter products of functions that need differentiation. This is where the "Product Rule" comes into play. The Product Rule is a fundamental derivative rule that is used to find the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), their product is \( u(x) \cdot v(x) \). To differentiate this product, apply the formula:

• \( D_x (u(x) \cdot v(x)) = u'(x)v(x) + u(x)v'(x) \)

Essentially, you differentiate each function individually, multiply it by the other function, and then sum them together. This technique allows us to break down complex derivatives into manageable parts so we can solve them effectively.
In our problem, \( u(x) = \sinh x \) and \( v(x) = \cosh 4x \). Applying the Product Rule, we first derive these functions separately and then use the formula to find the derivative of the product.

Chain Rule

Sometimes, functions are composed within each other. This is where the "Chain Rule" becomes extremely helpful. The Chain Rule is used when you need to differentiate a composite function, which essentially means a function inside another function. It is formulated as:

• \( D_x (f(g(x))) = f'(g(x)) \cdot g'(x) \)

Here, differentiate the outer function \( f \) with respect to the inner function \( g(x) \), and then multiply by the derivative of the inner function \( g(x) \). Using the chain rule, you can "dig" into each layer of a composite function and differentiate effectively.
For our specific problem, \( v(x) = \cosh 4x \) is a composite function. The outer function is \( \cosh \) and the inner function is \( 4x \). The derivative of the outer function \( \cosh x \) is \( \sinh x \). You multiply this by the derivative of the inner function \( 4x \), which is 4, to get \( v'(x) = 4 \cdot \sinh(4x) \).

Hyperbolic Functions

Hyperbolic functions, such as \( \sinh \) and \( \cosh \), are analogs to the trigonometric functions but for the hyperbola instead of the circle. They are frequently encountered in various areas of calculus and have properties and derivatives that make them similar, yet distinct, from trigonometric functions.

• The function \( \sinh x = \frac{e^x - e^{-x}}{2} \) and resembles the sine function.
• The function \( \cosh x = \frac{e^x + e^{-x}}{2} \) and is similar to the cosine function.

Hyperbolic functions are crucial in solving many calculus problems involving exponentials and can model various physical phenomena.
The derivatives of hyperbolic functions are:

• \( D_x (\sinh x) = \cosh x \)
• \( D_x (\cosh x) = \sinh x \)

In our exercise, understanding these derivatives was essential for applying both the product and chain rules effectively.

Derivative Simplification

After finding the derivative of a function, simplifying it is often necessary for a clean, understandable result. "Derivative Simplification" involves combining like terms, reducing complex expressions, and arranging the components of derivatives into the simplest form possible.

• This step is vital for clarity, particularly in mathematical proofs or when solving problems requiring further calculations.

In our example, after applying the product rule, we obtained the expression:

• \( D_x y = \cosh x \cosh 4x + 4 \sinh x \sinh 4x \)

While this is a valid expression, observing its structure can reveal simplifications or connections to known identities. Simple, reorganized results are not only easier to interpret but also facilitate identifying patterns or further manipulating derivatives if needed within larger, more complex problems.