Problem 39 - Essential Calculus Early Transcendentals

Question

Find the derivative. Simplify where possible. $$ y=x \sinh ^{-1}(x / 3)-\sqrt{9+x^{2}} $$

Step-by-Step Solution

Answer

\( y' = \frac{x}{3\sqrt{9-x^2}} + \sinh^{-1}(x/3) - \frac{x}{\sqrt{9+x^2}} \).

Step 1: Identify Functions to Differentiate

The function \( y \) is composed of two parts: \( f(x) = x \sinh^{-1}(x/3) \) and \( g(x) = -\sqrt{9+x^2} \). We'll differentiate each part separately and then sum the derivatives.

Step 2: Differentiate \( f(x) = x \sinh^{-1}(x/3) \)

Use the product rule for differentiation: \( u = x \) and \( v = \sinh^{-1}(x/3) \). The derivative of \( u \) is \( 1 \) and the derivative of \( v \) is \( \frac{1}{\sqrt{1-(x/3)^2}} \cdot \frac{1}{3} \). Therefore, \( f'(x) = x \cdot \frac{1}{3\sqrt{1-(x/3)^2}} + \sinh^{-1}(x/3) \).

Step 3: Differentiate \( g(x) = -\sqrt{9+x^2} \)

Use the chain rule where \( h(x) = \sqrt{9+x^2} \). The derivative is \( g'(x) = -\frac{x}{\sqrt{9+x^2}} \).

Step 4: Combine the Derivatives

Combine the derivatives from Step 2 and Step 3. Thus, the derivative \( y' = f'(x) + g'(x) = \frac{x}{3\sqrt{1-(x/3)^2}} + \sinh^{-1}(x/3) - \frac{x}{\sqrt{9+x^2}} \).

Step 5: Simplify the Expression

Examine terms and simplify if possible. After simplification, the derivative \( y' \) is expressed as \( y' = \frac{x}{3\sqrt{9-x^2}} + \sinh^{-1}(x/3) - \frac{x}{\sqrt{9+x^2}} \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but they relate to a hyperbola rather than a circle. They often appear in calculus problems, especially when dealing with certain integrals and derivatives. In this exercise, we deal with the inverse hyperbolic sine function, represented as \( \sinh^{-1}(x) \). The derivative of the inverse hyperbolic sine function is an important tool and is given by:

• \( \frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}} \)

However, in our problem, it is expressed as a function of \( x/3 \), hence the derivative becomes \( \frac{1}{\sqrt{1-(x/3)^2}} \) multiplied by the derivative of \( x/3 \), which is \( 1/3 \). Remember, mastering these derivatives enhances your understanding and ability to solve calculus problems involving hyperbolic functions.

Product Rule

The product rule is an essential differentiation technique used when finding the derivative of a product of two functions. If you have a function \( y = u \cdot v \), its derivative is given by:

• \( y' = u' \cdot v + u \cdot v' \)

In our exercise, \( f(x) = x \cdot \sinh^{-1}(x/3) \) necessitates the product rule. Here, \( u \) is \( x \), whose derivative is \( 1 \), and \( v \) is \( \sinh^{-1}(x/3) \), for which we found the derivative. Applying the product rule correctly is key to solving problems involving the differentiation of products of functions.

Chain Rule

The chain rule is another fundamental differentiation technique. It applies when dealing with composite functions. If you have a function \( y = f(g(x)) \), the chain rule states:

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

In our problem, we use the chain rule to differentiate \( g(x) = -\sqrt{9+x^2} \). Here, the outer function is \( f(u) = -\sqrt{u} \) and the inner function is \( g(x) = 9 + x^2 \). Calculating these derivatives step by step allows you to tackle even the most intricate parts of a calculus problem involving composite functions. Understanding the chain rule deeply will greatly enhance your calculus toolbox.

Simplifying Expressions

Simplification is often the final step in calculus problems, reducing complex expressions into more manageable forms. After finding the derivatives, you combine them and simplify where possible to achieve the final solution. In the original exercise, you note the changes from:

• \( \frac{x}{3\sqrt{1-(x/3)^2}} + \sinh^{-1}(x/3) - \frac{x}{\sqrt{9+x^2}} \)

to a more simplified expression. Simplification involves algebraic manipulation, including combining like terms and reducing fractions. The goal is to make the resulting expression easy to interpret and calculate. Mastery of simplification techniques makes your problem-solving process smoother and helps you present your final answer with clarity.