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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.3.117

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
117. y = (√t)ᵗ

Guida verificata passo dopo passo
1
Rewrite the function to a form that is easier to differentiate using logarithmic differentiation. Start by expressing the function as \(y = (\sqrt{t})^{t} = (t^{1/2})^{t} = t^{t/2}\).
Take the natural logarithm of both sides to simplify the exponentiation: \(\ln y = \ln \left(t^{t/2}\right)\).
Use the logarithm power rule to bring down the exponent: \(\ln y = \frac{t}{2} \ln t\).
Differentiate both sides with respect to \(t\). Remember that \(y\) is a function of \(t\), so use implicit differentiation on the left side: \(\frac{1}{y} \frac{dy}{dt} = \frac{d}{dt} \left( \frac{t}{2} \ln t \right)\).
Apply the product rule to differentiate the right side: \(\frac{d}{dt} \left( \frac{t}{2} \ln t \right) = \frac{1}{2} \ln t + \frac{t}{2} \cdot \frac{1}{t}\). Then multiply both sides by \(y\) to solve for \(\frac{dy}{dt}\).

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Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions where the variable appears both in the base and the exponent. By taking the natural logarithm of both sides, the expression simplifies, allowing the use of implicit differentiation and product or chain rules more easily.
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Chain Rule

The chain rule is a fundamental differentiation rule used when differentiating composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
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Properties of Logarithms

Properties of logarithms, such as log(a^b) = b log(a), help simplify expressions before differentiation. Applying these properties transforms complicated exponentials into products, making differentiation more straightforward.
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