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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.7.15
Capitolo 7, Problema 7.7.15

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

Guida verificata passo dopo passo
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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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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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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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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
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