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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.7.27

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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
3:17
Inverse Tangent

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.
추천 영상:
05:18
The Product Rule

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.
추천 영상:
05:02
Intro to the Chain Rule