Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.7.65

Chapter 7, Problem 7.7.65

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.
sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞
cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1
tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1
sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1
csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1
coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1
Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.

65. sech⁻¹(3/5)

Verified step by step guidance

Identify the formula for the inverse hyperbolic secant function: \(\text{sech}^{-1} x = \ln \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)\), where \(0 < x \leq 1\).

Substitute the given value \(x = \frac{3}{5}\) into the formula: \(\text{sech}^{-1} \left( \frac{3}{5} \right) = \ln \left( \frac{1 + \sqrt{1 - \left( \frac{3}{5} \right)^2}}{\frac{3}{5}} \right)\).

Simplify the expression inside the square root: calculate \(1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25}\).

Take the square root of the simplified expression: \(\sqrt{1 - \left( \frac{3}{5} \right)^2} = \sqrt{\frac{16}{25}}\).

Rewrite the entire expression inside the logarithm and simplify the fraction to get the final logarithmic form without evaluating the numerical value.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Hyperbolic Functions

Inverse hyperbolic functions undo the hyperbolic functions, similar to how inverse trigonometric functions relate to trigonometric functions. They are defined for specific domains and can be expressed using logarithmic formulas, which allow evaluation without relying on geometric interpretations.

Logarithmic Expressions of Inverse Hyperbolic Functions

Each inverse hyperbolic function can be rewritten as a natural logarithm involving algebraic expressions with square roots. This transformation leverages the exponential definitions of hyperbolic functions, enabling easier computation and simplification in calculus problems.

Domain Restrictions and Validity

The domain of each inverse hyperbolic function is crucial to ensure the expressions inside the logarithms and square roots are valid (e.g., non-negative under the root). Understanding these restrictions prevents errors when substituting values and ensures the expressions are mathematically sound.

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