Problem 8 - Calculus Early Transcendentals

Question

On what interval is the formula \(d / d x\left(\tanh ^{-1} x\right)=1 /\left(x^{2}-1\right)\) valid?

Step-by-Step Solution

Answer

Answer: The formula is valid for x values in the interval \((-1, 1)\).

Step 1: Determine the domain of \(\tanh^{-1}(x)\)

The hyperbolic tangent function, \(\tanh(x)\), is defined as: \(\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}\) To find the domain of \(\tanh^{-1}(x)\), we need to determine the range of \(\tanh(x)\). Let's analyze the behavior of \(\tanh(x)\) as \(x\) approaches positive and negative infinity. As \(x \rightarrow \infty\), the denominator grows exponentially faster, effectively canceling out any fractions in the numerator: \(\lim_{x\to\infty} \tanh(x) = 1\) As \(x \rightarrow -\infty\), the denominator again grows exponentially faster, canceling out any fractions in the numerator: \(\lim_{x\to-\infty} \tanh(x) = -1\) Therefore, the range of \(\tanh(x)\) is \((-1, 1)\). Since the inverse function "swaps" domain and range, the domain of \(\tanh^{-1}(x)\) is \((-1, 1)\). This means that the given formula for the derivative of \(\tanh^{-1}(x)\) is only valid for \(x\) values within this interval.

Step 2: State the valid interval for the derivative formula

The valid interval for the formula \(d / dx \left(\tanh^{-1}(x)\right) = 1/\left(x^{2} - 1\right)\) is the same as the domain of the inverse hyperbolic tangent function: \((-1, 1)\)

Key Concepts

Derivative Formula

The derivative of an inverse hyperbolic function can be a bit tricky to understand at first. Specifically, for the inverse hyperbolic tangent function, denoted as \( \tanh^{-1}(x) \), the derivative formula is given by:

• \( \frac{d}{dx}(\tanh^{-1}(x)) = \frac{1}{1 - x^2} \)

This formula tells us how the inverse hyperbolic tangent function changes as \( x \) changes. It's essential to remember that this derivative is calculated under specific conditions, specifically within the domain of the inverse function.

Notice that the formula \( \frac{1}{1 - x^2} \) involves a denominator that can become zero, which leads us to the next point regarding when this formula can actually be used safely: the domain of \( \tanh^{-1}(x) \). Understanding these conditions is critical to preventing mathematical errors.

Domain Analysis

Domain analysis helps us identify the set of input values \( x \) for which a given function is defined. For \( \tanh^{-1}(x) \), understanding the domain is crucial because it determines where the derivative formula is valid.

First, we observe the hyperbolic tangent function, \( \tanh(x) \), which behaves like a stretch of the trigonometric tangent but with exponential function properties.

• As \( x \to \infty \), \( \tanh(x) \to 1 \)
• As \( x \to -\infty \), \( \tanh(x) \to -1 \)

Hence, \( \tanh(x) \) maps real numbers to the interval \((-1, 1)\). Consequently, the inverse function \( \tanh^{-1}(x) \) has its domain restricted to where \( \tanh(x) \) produces valid results—so, only real numbers within \((-1, 1)\).

This analysis is critical because it means the derivative \( \frac{1}{1-x^2} \) is valid only for \( x \) within \((-1, 1)\), as working outside this interval would result in undefined mathematical expressions (like division by zero when \( x=\pm 1 \)).

Hyperbolic Tangent Function

The hyperbolic tangent function, denoted as \( \tanh(x) \), is integral to understanding the inverse hyperbolic tangent function \( \tanh^{-1}(x) \). The hyperbolic tangent is defined using exponential functions:

• \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

This definition highlights its exponential nature, setting it apart from the standard trigonometric tangent.

The behavior of \( \tanh(x) \) asymptotically approaches 1 as \( x \) grows positively infinite and -1 as it decreases negatively infinite. This characteristic generates a range of \((-1, 1)\) for \( \tanh(x) \), which is crucial for determining the domain of \( \tanh^{-1}(x) \). Furthermore, this function displays properties similar to other hyperbolic functions:

• Smoothness and continuity across all real numbers
• Symmetry about the origin

Understanding \( \tanh(x) \) provides insight into its inverse and the conditions under which its derivative formula \( \frac{1}{1-x^2} \) holds true.