Problem 38 - Essential Calculus Early Transcendentals

Question

Find the derivative. Simplify where possible. \(y=x \tanh ^{-1} x+\ln \sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

The derivative is \( \tanh^{-1}(x) \).

Step 1: Differentiate Using Product Rule for First Term

The given function is \( y = x \tanh^{-1}(x) + \ln \sqrt{1-x^2} \). Start by differentiating the first term \( x \tanh^{-1}(x) \). Use the product rule: \( u'v + uv' \) where \( u = x \) and \( v = \tanh^{-1}(x) \). Thus, \( u' = 1 \) and \( v' = \frac{1}{1-x^2} \). Thus, the derivative of the first term is: \( 1 \cdot \tanh^{-1}(x) + x \cdot \frac{1}{1-x^2} \).

Step 2: Differentiate Using Chain Rule for Second Term

Now, differentiate the second term \( \ln \sqrt{1-x^2} \), which can be rewritten as \( \frac{1}{2} \ln (1-x^2) \). Apply the chain rule. The derivative of \( \ln(1-x^2) \) is \( \frac{-2x}{1-x^2} \). Thus, multiplying by \( \frac{1}{2} \), the derivative is \( \frac{-x}{1-x^2} \).

Step 3: Combine the Derivatives

Now, combine the derivatives from the two steps. The derivative of the first term is \( \tanh^{-1}(x) + \frac{x}{1-x^2} \). The derivative of the second term is \( \frac{-x}{1-x^2} \). Adding these, we have: \( \tanh^{-1}(x) + \frac{x}{1-x^2} - \frac{x}{1-x^2} \).

Step 4: Simplify the Expression

Simplify the expression from the previous step. The terms \( \frac{x}{1-x^2} \) and \( -\frac{x}{1-x^2} \) cancel each other out. Thus, the derivative simplifies to \( \tanh^{-1}(x) \).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of the inverse classical trigonometric functions. These functions introduce a deeper layer of mathematical exploration in calculus and analysis. The function \( \tanh^{-1} x \) is the inverse hyperbolic tangent. Its domain includes all real numbers between -1 and 1, excluding the endpoints. This makes it quite different from the regular tangent function.

The derivative of \( \tanh^{-1} x \) is given by \( \frac{1}{1-x^2} \). Understanding this derivative is crucial when differentiating expressions involving hyperbolic functions because it often appears in conjunction with the product and chain rules. Always ensure that the argument \( x \) remains within the function's domain limits to avoid undefined expressions.

Product Rule

The product rule is an essential tool in calculus for differentiating products of two functions. It states that the derivative of a product \( uv \) is given by \( u'v + uv' \). This rule allows us to break down complex expressions into simpler parts for differentiation.

• Set one part of the function as \( u \), compute its derivative \( u' \).
• Set the other part as \( v \), find its derivative \( v' \).
• Multiply and add them as per the product rule.

In our exercise, \( u = x \) and \( v = \tanh^{-1} x \), leading to the application of the product rule to simplify and find the derivative effectively. This helps to systematically manage both parts of the product, making the differentiation process straightforward.

Chain Rule

The chain rule is another fundamental calculus principle used when differentiating composite functions. It is particularly useful when a function is a combination of multiple outer and inner functions, such as \( \ln \sqrt{1 - x^2} \).

• Identify the outer function and differentiate it first.
• Multiply the result by the derivative of the inner function.

In the given exercise, the second term, \( \ln \sqrt{1-x^2} \), is rewritten as \( \frac{1}{2}\ln(1-x^2) \) for easier differentiation. The derivative of the outer function, \( \ln(u) \), is \( \frac{1}{u} \), and the inner function \( (1-x^2) \) differentiates to \( -2x \). Applying the chain rule ensures that every component is properly formulated and calculated, leading to a precise and simplified derivative.

Simplification in Calculus

Simplification is the process of condensing a mathematical expression to its most reduced form. In calculus, this usually involves removing redundant terms and making an expression as clear as possible. The purpose is to make the expression easier to interpret and understand.

In the exercise given, after applying both the product and chain rules, simplification plays a crucial role. The terms \( \frac{x}{1-x^2} \) and \( -\frac{x}{1-x^2} \) in the expression cancel each other out, showcasing how simplification can remove complexity and reveal the true form of the derivative. Simplification also helps in reducing errors, and aids in further calculations by providing results that are easier to work with.