Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=x \cosh ^{-1}(3 x) $$

Step-by-Step Solution

Answer

The derivative is \( D_x y = \cosh^{-1}(3x) + \frac{3x}{\sqrt{9x^2 - 1}} \).

Step 1: Understand the Function

The function given is \( y = x \cosh^{-1}(3x) \). Here, \( \cosh^{-1} \) denotes the inverse hyperbolic cosine function.

Step 2: Apply the Product Rule

Since the expression is a product of two functions, \( u = x \) and \( v = \cosh^{-1}(3x) \), we will use the product rule for differentiation: \( D_x(uv) = u'v + uv' \).

Step 3: Differentiate \( u = x \)

The derivative of \( u = x \) with respect to \( x \) is \( u' = 1 \).

Step 4: Differentiate \( v = \cosh^{-1}(3x) \)

The derivative of \( v = \cosh^{-1}(3x) \) with respect to \( x \) involves using the chain rule. First, recall the derivative of \( \cosh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{u^2 - 1}} \). Here, \( u = 3x \). Then, by applying the chain rule, \( v' = \frac{d}{dx}\left(\cosh^{-1}(3x)\right) = \frac{1}{\sqrt{(3x)^2 - 1}} \cdot \frac{d}{dx}(3x) = \frac{3}{\sqrt{9x^2 - 1}} \).

Step 5: Substitute Back into the Product Rule Formula

Using the results from Steps 3 and 4, substitute back into the product rule formula:\[ D_x(x \cosh^{-1}(3x)) = 1 \cdot \cosh^{-1}(3x) + x \cdot \frac{3}{\sqrt{9x^2 - 1}} \].

Step 6: Simplify the Expression

Combine the terms obtained from the product rule:\[ D_x y = \cosh^{-1}(3x) + \frac{3x}{\sqrt{9x^2 - 1}} \].

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate expressions where two functions are multiplied together. If you're working with the functions \( u(x) \) and \( v(x) \), the derivative of their product, \( u(x)v(x) \), is determined using the formula:

• \( D_x(uv) = u'v + uv' \)

This formula shows that you have to differentiate each function separately and then combine those derivatives. Let's break it down with an example:

• Differentiate \( u(x) \), to get \( u'(x) \)
• Differentiate \( v(x) \), to get \( v'(x) \)
• Substitute into \( u'v + uv' \)

In our exercise, \( u(x)=x \) and \( v(x)=\cosh^{-1}(3x) \). Applying the product rule is crucial because it allows us to effectively separate and manage the differentiation tasks.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as \( \cosh^{-1}(x) \), play a significant role in mathematics similar to how inverse trigonometric functions do. Particularly, \( \cosh^{-1}(x) \) denotes the inverse of the hyperbolic cosine function and is used when you have outputs from the hyperbolic cosine and want to find the original input.When differentiating an inverse hyperbolic function like \( \cosh^{-1}(u) \), you need to use a specific formula:

• \( \frac{d}{du}(\cosh^{-1}(u)) = \frac{1}{\sqrt{u^2 - 1}} \)

This formula is important because it allows us to handle complex expressions involving inverse hyperbolic functions while differentiating. In the exercise, we have \( u = 3x \), requiring the application of this derivative formula to find \( v' \). Understanding how these inverse functions and their derivatives work helps to illuminate their behavior under calculus operations.

Chain Rule

The chain rule is another powerful differentiation tool for finding the derivative of composite functions. When one function is nested inside another, the chain rule tells us how to take the derivative of such functions. In simple terms, if you have a function \( y = f(g(x)) \), the derivative is given by:

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

This means you first differentiate the outer function \( f \) with respect to its input \( g(x) \), and then multiply it by the derivative of the inner function \( g(x) \).In our example, the chain rule is applied when differentiating \( v = \cosh^{-1}(3x) \). Here, we have \( f(u) = \cosh^{-1}(u) \) and \( g(x) = 3x \). Applying the chain rule helps us find \( v' \) as:

• \( v' = \frac{d}{dx}(\cosh^{-1}(3x)) = \frac{1}{\sqrt{(3x)^2 - 1}} \cdot \frac{d}{dx}(3x) = \frac{3}{\sqrt{9x^2 - 1}} \)

The chain rule elegantly handles the complexity of composite functions and is an indispensable tool in differential calculus.