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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.7.31
Chapter 7, Problem 7.7.31

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
31. y = cos⁻¹(x) - x sech⁻¹(x)

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Identify the function to differentiate: \(y = \cos^{-1}(x) - x \ \text{sech}^{-1}(x)\), where \(\cos^{-1}(x)\) is the inverse cosine function and \(\text{sech}^{-1}(x)\) is the inverse hyperbolic secant function.
Recall the derivative formulas for the inverse functions involved: - For \(y = \cos^{-1}(x)\), the derivative is \(\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}\). - For \(y = \text{sech}^{-1}(x)\), the derivative is \(\frac{dy}{dx} = -\frac{1}{x \sqrt{1 - x^2}}\) (valid for \(0 < x < 1\)).
Apply the product rule to the term \(x \ \text{sech}^{-1}(x)\): If \(u = x\) and \(v = \text{sech}^{-1}(x)\), then \(\frac{d}{dx}(uv) = u'v + uv'\). Calculate \(u' = 1\) and use the derivative of \(v\) from the previous step.
Write the derivative of \(y\) as: \(\frac{dy}{dx} = \frac{d}{dx} \left( \cos^{-1}(x) \right) - \frac{d}{dx} \left( x \ \text{sech}^{-1}(x) \right)\), which becomes \(\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} - \left( 1 \cdot \text{sech}^{-1}(x) + x \cdot \left(-\frac{1}{x \sqrt{1 - x^2}}\right) \right)\).
Simplify the expression by distributing the negative sign and combining like terms, keeping the derivative in terms of \(x\) without evaluating the final numeric value.

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Key Concepts

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

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions like arccos(x) have specific derivative formulas. For example, the derivative of arccos(x) is -1 / √(1 - x²), valid for |x| < 1. Understanding these derivatives is essential for differentiating expressions involving inverse trig functions.
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Derivatives of Other Inverse Trigonometric Functions

Derivative of Inverse Hyperbolic Functions

Inverse hyperbolic functions such as sech⁻¹(x) have distinct derivative rules. The derivative of sech⁻¹(x) is -1 / (|x|√(1 - x²)) for appropriate domains. Recognizing and applying these formulas is crucial when differentiating terms involving inverse hyperbolic functions.
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Derivatives of Other Inverse Trigonometric Functions

Product Rule for Differentiation

When differentiating a product of two functions, use the product rule: (fg)' = f'g + fg'. This rule is necessary here because the term x sech⁻¹(x) is a product of x and sech⁻¹(x). Applying the product rule correctly ensures accurate differentiation of such expressions.
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The Product Rule
Related Practice
Textbook Question

Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.


f(x) = 1/x², x > 0

Textbook Question

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.

63. tanh⁻¹(-1/2)

Textbook Question

Evaluate the integrals in Exercises 97–110.

109. ∫ (dx / (x log₁₀x))

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Textbook Question

Evaluate the integrals in Exercises 33–54.

∫8e^(x+1) dx

Textbook Question

In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.

y = ln(e^(θ)/(1+e^θ))

Textbook Question

Evaluate the integrals in Exercises 53–76.

69. ∫dx/((2x-1)√((2x-1)²-4))

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