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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 7.3.85

In Exercises 59–86, find the derivative of y with respect to the given independent variable.
85. y = log₂(8t^(ln 2))

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Recall the change of base formula for logarithms: \(\log_a b = \frac{\ln b}{\ln a}\). Use this to rewrite \(y = \log_2 \left(8t^{\ln 2}\right)\) as \(y = \frac{\ln \left(8t^{\ln 2}\right)}{\ln 2}\).
Apply the logarithm property \(\ln(ab) = \ln a + \ln b\) to expand the numerator: \(\ln \left(8t^{\ln 2}\right) = \ln 8 + \ln \left(t^{\ln 2}\right)\).
Use the power rule for logarithms: \(\ln \left(t^{\ln 2}\right) = (\ln 2) \cdot \ln t\). So the expression becomes \(y = \frac{\ln 8 + (\ln 2) \cdot \ln t}{\ln 2}\).
Since \(\ln 8\) and \(\ln 2\) are constants, separate the terms: \(y = \frac{\ln 8}{\ln 2} + \frac{(\ln 2) \cdot \ln t}{\ln 2}\). Simplify the second term by canceling \(\ln 2\).
Now differentiate \(y\) with respect to \(t\). The first term is constant, so its derivative is zero. For the second term, differentiate \(\ln t\) using \(\frac{d}{dt} \ln t = \frac{1}{t}\). Combine these results to find \(\frac{dy}{dt}\).

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