Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problema 7.7.65

Capitolo 7, Problema 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)

Guida verificata passo dopo passo

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.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.

Video consigliato:

4:49

Inverse Cosine

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.

Video consigliato:

5:26

Graphs of Logarithmic Functions

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.

Video consigliato:

Percorso guidato

5:10

Finding the Domain and Range of a Graph

Pratica correlata

Domanda del libro di testo

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.

120. y = x^(sin x)

views

Domanda del libro di testo

13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x→∞? Give reasons for your answer.

views

Domanda del libro di testo

In Exercises 139–142, find the length of each curve.

141. y = ln(cos(x)) from x = 0 to x = π/4.

views

Domanda del libro di testo

Evaluate the integrals in Exercises 31–78.

69. ∫dy/(y√(4y²-1))