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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.7.15

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
15. y = 2√t tanh(√t)

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Identify the function to differentiate: \(y = 2 \sqrt{t} \tanh(\sqrt{t})\). Notice that \(y\) is a product of two functions of \(t\): \(2 \sqrt{t}\) and \(\tanh(\sqrt{t})\).
Recall the product rule for derivatives: if \(y = u(t) v(t)\), then \(\frac{dy}{dt} = u'(t) v(t) + u(t) v'(t)\). Here, let \(u(t) = 2 \sqrt{t}\) and \(v(t) = \tanh(\sqrt{t})\).
Find \(u'(t)\): Since \(u(t) = 2 t^{1/2}\), use the power rule to get \(u'(t) = 2 \times \frac{1}{2} t^{-1/2} = t^{-1/2}\).
Find \(v'(t)\): Since \(v(t) = \tanh(\sqrt{t})\), apply the chain rule. First, recall that \(\frac{d}{dx} \tanh(x) = \operatorname{sech}^2(x)\). Let \(g(t) = \sqrt{t} = t^{1/2}\), so \(v(t) = \tanh(g(t))\). Then, \(v'(t) = \operatorname{sech}^2(g(t)) \cdot g'(t)\), where \(g'(t) = \frac{1}{2} t^{-1/2}\).
Combine all parts using the product rule: \(\frac{dy}{dt} = u'(t) v(t) + u(t) v'(t) = t^{-1/2} \tanh(\sqrt{t}) + 2 \sqrt{t} \times \operatorname{sech}^2(\sqrt{t}) \times \frac{1}{2} t^{-1/2}\). Simplify the expression as needed.

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주요 개념

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

Derivative of Composite Functions (Chain Rule)

The chain rule is used to differentiate composite functions, where one function is inside another. It states that the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x). This is essential when differentiating expressions like √t or tanh(√t), which involve nested functions.
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The Chain Rule for 3+ Functions

Derivative of Hyperbolic Functions

Hyperbolic functions such as tanh(x) have specific derivatives; for example, the derivative of tanh(x) is sech²(x). Knowing these derivatives allows you to differentiate terms like tanh(√t) correctly, especially when combined with the chain rule.
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Asymptotes of Hyperbolas

Product Rule for Differentiation

The product rule is used when differentiating the product of two functions. It states that the derivative of u(t)v(t) is u'(t)v(t) + u(t)v'(t). Since y = 2√t * tanh(√t) is a product of two functions of t, applying the product rule is necessary.
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The Product Rule