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Chapter 7: Problem 83

Find the derivative of \(y\) with respect to the given independent variable. \begin{equation}y=3^{\log _{2} t}\end{equation}

Short Answer

Expert verified
The derivative is \( \frac{dy}{dt} = \frac{3^{\log _{2} t} \ln 3}{t \ln 2} \).

Step by step solution

01

Identify the function form

The given function is in the form \( y = a^{u} \), where \( a = 3 \) and \( u = \log_{2} t \). This indicates it's an exponential function with base 3 and exponent \( \log_{2} t \).
02

Apply the derivative of an exponential function

For a function of the form \( y = a^u \), the derivative \( \frac{dy}{dt} \) is given by \( \frac{dy}{du} \cdot \frac{du}{dt} \), where \( \frac{dy}{du} = a^{u} \ln a \).
03

Find du/dt

First, calculate \( \frac{du}{dt} \). Since \( u = \log_{2} t \), it can be rewritten using the change of base formula: \( u = \frac{\ln t}{\ln 2} \). Thus, \( \frac{du}{dt} = \frac{1}{t \ln 2} \) using the derivative of \( \ln t \), which is \( \frac{1}{t} \).
04

Combine results to find dy/dt

Combine the results: use \( \frac{dy}{du} = 3^{\log_{2} t} \ln 3 \) and \( \frac{du}{dt} = \frac{1}{t \ln 2} \). Therefore, \( \frac{dy}{dt} = 3^{\log _{2} t} \ln 3 \cdot \frac{1}{t \ln 2} \), simplify to \( \frac{dy}{dt} = \frac{3^{\log _{2} t} \ln 3}{t \ln 2} \).
05

Simplify the result (optional)

Further simplification is not required as \( \frac{dy}{dt} = \frac{3^{\log _{2} t} \ln 3}{t \ln 2} \) is in a simplified form, expressing the derivative explicitly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function
An exponential function is a type of mathematical function where a constant base is raised to a variable exponent. In the exercise, the function is given as \( y = 3^{\log_{2} t} \).
  • Here, the constant base is \( 3 \) and the variable exponent is \( \log_{2} t \).
  • The value of the base, \( a \), is crucial because it defines the growth rate of the function; larger bases grow faster.
  • Exponential functions are commonly used to model growth processes, such as population growth or compound interest, where quantities increase rapidly over time.
An important property of these functions is that their derivatives involve themselves, making them unique compared to polynomials or rational functions. This makes them both fascinating and powerful tools in calculus and mathematical modeling.
Logarithmic Differentiation
Logarithmic differentiation is a technique used to differentiate functions where the standard rules of differentiation are cumbersome. It's particularly useful when dealing with complex or nested exponential functions.
In the example function \( y = 3^{\log_{2} t} \), logarithmic differentiation helps to iterate through the layers of functions.
  • First, take the natural logarithm on both sides: \( \ln y = \ln (3^{\log_{2} t}) \).
  • Use logarithmic identities to simplify: \( \ln y = \log_{2} t \cdot \ln 3 \).
  • This allows the identification and separation of the exponent function, making the differentiation easier to execute step by step.
Logarithmic differentiation converts multiplicative complexities into simpler additive forms, harnessing the properties of logarithms to effortlessly tackle challenging differentiation tasks.
Change of Base Formula
The change of base formula is a mathematical technique enabling the conversion of logarithms from one base to another. It states: \[ \log_b a = \frac{\ln a}{\ln b} \]This formula is extremely useful when working with logarithms in derivatives, as it simplifies the differentiation process.

In the given exercise, \( u = \log_{2} t \), we employ this formula by changing it to natural logarithms as follows:
  • Rewrite \( \log_{2} t = \frac{\ln t}{\ln 2} \).
  • This manipulation converts base 2 logarithms into natural logarithms, which are easier to differentiate due to well-known derivative rules.
  • This conversion is crucial for calculating \( \frac{du}{dt} \) because the derivative of \( \ln t \) is simply \( \frac{1}{t} \).
The change of base formula streamlines the differentiation task by providing a pathway from complex bases to the familiar natural base \( e \). This is especially advantageous for those tackling logarithmic differentiation in calculus.

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