In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
27. y = (1 - θ)tanh⁻¹(θ)

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Identify the function given: \(y = (1 - \theta) \tanh^{-1}(\theta)\), where \(\tanh^{-1}(\theta)\) is the inverse hyperbolic tangent function of \(\theta\).

Recognize that \(y\) is a product of two functions of \(\theta\): \(u = (1 - \theta)\) and \(v = \tanh^{-1}(\theta)\). To find \(\frac{dy}{d\theta}\), apply the product rule: \(\frac{dy}{d\theta} = u'v + uv'\).

Compute the derivative of the first function: \(u' = \frac{d}{d\theta}(1 - \theta) = -1\).

Compute the derivative of the second function: \(v' = \frac{d}{d\theta} \tanh^{-1}(\theta) = \frac{1}{1 - \theta^2}\), which is the standard derivative formula for the inverse hyperbolic tangent.

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the product rule formula: \(\frac{dy}{d\theta} = (-1) \cdot \tanh^{-1}(\theta) + (1 - \theta) \cdot \frac{1}{1 - \theta^2}\). This expression represents the derivative of \(y\) with respect to \(\theta\).

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Key Concepts

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

Inverse Hyperbolic Tangent Function (tanh⁻¹)

The inverse hyperbolic tangent function, denoted as tanh⁻¹(x), is the inverse of the hyperbolic tangent function. It is defined for values of x between -1 and 1 and has the derivative d/dx[tanh⁻¹(x)] = 1/(1 - x²). Understanding its domain and derivative is essential for differentiating expressions involving tanh⁻¹.

Product Rule of Differentiation

The product rule is used to differentiate functions that are products of two or more functions. If y = u(x)v(x), then dy/dx = u'(x)v(x) + u(x)v'(x). Applying this rule correctly is crucial when differentiating y = (1 - θ) * tanh⁻¹(θ), where both factors depend on θ.

Chain Rule and Variable Differentiation

The chain rule helps differentiate composite functions by multiplying the derivative of the outer function by the derivative of the inner function. Additionally, recognizing the variable with respect to which differentiation is performed (here, θ) ensures correct application of derivative formulas and variable treatment.

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Related Practice

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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