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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 7.3.83

In Exercises 59–86, find the derivative of y with respect to the given independent variable.
83. y = 3^(log₂ t)

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1
Identify the function given: \(y = 3^{\log_{2} t}\). Here, the base of the exponent is 3, and the exponent is \(\log_{2} t\).
Recall that when differentiating an exponential function with a variable exponent, it is often helpful to rewrite the function using the natural exponential and logarithm: \(y = e^{\ln(3^{\log_{2} t})}\).
Use the logarithm power rule to simplify the exponent inside the natural exponential: \(\ln(3^{\log_{2} t}) = \log_{2} t \cdot \ln(3)\), so \(y = e^{\log_{2} t \cdot \ln(3)}\).
Differentiate \(y\) with respect to \(t\) using the chain rule: \(\frac{dy}{dt} = y \cdot \frac{d}{dt} (\log_{2} t \cdot \ln(3))\). Since \(\ln(3)\) is constant, focus on differentiating \(\log_{2} t\).
Recall that \(\log_{2} t = \frac{\ln t}{\ln 2}\), so \(\frac{d}{dt} \log_{2} t = \frac{1}{t \ln 2}\). Substitute this back to get \(\frac{dy}{dt} = 3^{\log_{2} t} \cdot \ln(3) \cdot \frac{1}{t \ln 2}\).

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