Problem 110
Question
Refer to the following: In calculus, to find the derivative of a function of the form \(y=k^{x}\) where \(k\) is a constant, we apply logarithmic differentiation. The first step in this process consists of writing \(y=k^{x}\) in an equivalent form using the natural logarithm. Use the properties of this section to write an equivalent form of the following implicitly defined functions. $$y=4^{x} \cdot 3^{x+1}$$
Step-by-Step Solution
Verified Answer
The function can be written as \( \ln y = x \ln 12 + \ln 3 \).
1Step 1: Express the function as a single power
Our goal is to use logarithmic differentiation to find the derivative. First, let's express the given function as a single power. The function is given as \( y = 4^x \cdot 3^{x+1} \). This can be rewritten using the property of exponents: \( y = 4^x \cdot 3^{x+1} = 4^x \cdot 3^x \cdot 3 \), which simplifies to \( y = (4 \cdot 3)^x \cdot 3 \), or \( y = 12^x \cdot 3 \).
2Step 2: Apply the natural logarithm
We now take the natural logarithm of both sides to facilitate differentiation. Thus, applying logarithms gives: \( \ln y = \ln(12^x \cdot 3) \).
3Step 3: Use the logarithm product and power rules
Next, use the properties of logarithms to simplify the expression. The logarithm of a product can be expressed as the sum of logarithms, and the logarithm of a power can be expressed as a product. Therefore, \( \ln y = x \ln 12 + \ln 3 \).
Key Concepts
Derivative CalculationExponential FunctionsNatural LogarithmProperties of Logarithms
Derivative Calculation
Differentiation allows us to understand how a function changes as its input changes. It is a fundamental tool in calculus used to calculate the rate of change or slope of a function. In the context of logarithmic differentiation, it provides a mechanism to differentiate more complicated functions, such as those that are products, quotients, or powers of other functions.
For example, given a function like \( y = 4^x \cdot 3^{x+1} \), finding its derivative directly can be challenging. By using logarithmic differentiation, we can simplify the process. This involves taking the natural logarithm of both sides, which helps break down the function into more manageable parts. Doing so, we transform \( y = 4^x \cdot 3^{x+1} \) into an equivalent form, making it easier to apply the derivative rules.
By simplifying the expression through properties of logarithms and then differentiating each term separately, we find the derivative of complex exponential functions more efficiently.
For example, given a function like \( y = 4^x \cdot 3^{x+1} \), finding its derivative directly can be challenging. By using logarithmic differentiation, we can simplify the process. This involves taking the natural logarithm of both sides, which helps break down the function into more manageable parts. Doing so, we transform \( y = 4^x \cdot 3^{x+1} \) into an equivalent form, making it easier to apply the derivative rules.
By simplifying the expression through properties of logarithms and then differentiating each term separately, we find the derivative of complex exponential functions more efficiently.
Exponential Functions
Exponential functions are those in which the variable appears in the exponent. The general form is \( k^x \), where \( k \) is a constant, and \( x \) is the exponent. In the exercise, we dealt with \( 4^x \cdot 3^{x+1} \), which are both exponential functions.
Exponential functions have unique properties which make them stand out in calculus. They grow (or decay) at an increasing rate, and they have a constant ratio of increase at each interval. They are commonly used to model growth situations in the real world like population, investments, and more.
In derivative calculations involving such functions, expressing them as a single power can simplify the differentiation process. For example, the expression \( 4^x \cdot 3^{x+1} \) can be written as \( 12^x \cdot 3 \), simplifying the function into a more standard form that's easier to work with.
Exponential functions have unique properties which make them stand out in calculus. They grow (or decay) at an increasing rate, and they have a constant ratio of increase at each interval. They are commonly used to model growth situations in the real world like population, investments, and more.
In derivative calculations involving such functions, expressing them as a single power can simplify the differentiation process. For example, the expression \( 4^x \cdot 3^{x+1} \) can be written as \( 12^x \cdot 3 \), simplifying the function into a more standard form that's easier to work with.
Natural Logarithm
The natural logarithm, denoted by \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It is especially useful in calculus for simplifying differentiation and integration involving exponential functions. Taking the natural logarithm of both sides of an equation helps to linearize complex expressions involving exponents.
In the context of our example, by applying the natural logarithm to the function \( y = 12^x \cdot 3 \), we obtain \( \ln y = \ln (12^x \cdot 3) \). This step makes it simpler to apply differentiation rules, as we transform multiplication into addition and powers into multipliers via the properties of logarithms.
Once this transformation is executed, applying the differentiation rules to each term becomes straightforward, allowing for easier calculation of derivatives for functions that initially seem complex.
In the context of our example, by applying the natural logarithm to the function \( y = 12^x \cdot 3 \), we obtain \( \ln y = \ln (12^x \cdot 3) \). This step makes it simpler to apply differentiation rules, as we transform multiplication into addition and powers into multipliers via the properties of logarithms.
Once this transformation is executed, applying the differentiation rules to each term becomes straightforward, allowing for easier calculation of derivatives for functions that initially seem complex.
Properties of Logarithms
Logarithms have several properties that make them valuable tools in calculus, especially when simplifying expressions before differentiating them. Key properties include:
Understanding and applying these properties are vital for manipulating expressions and ultimately arriving at accurate derivative calculations.
- The product property: \( \log_b (MN) = \log_b M + \log_b N \).
- The power property: \( \log_b (M^k) = k \log_b M \).
- The quotient property: \( \log_b \frac{M}{N} = \log_b M - \log_b N \).
Understanding and applying these properties are vital for manipulating expressions and ultimately arriving at accurate derivative calculations.
Other exercises in this chapter
Problem 109
Refer to the following: In calculus, to find the derivative of a function of the form \(y=k^{x}\) where \(k\) is a constant, we apply logarithmic differentiatio
View solution Problem 109
Determine whether each statement is true or false. The graphs of \(y=\log x\) and \(y=\ln x\) have the same \(x\) -intercept (1,0).
View solution Problem 110
Determine whether each statement is true or false. The graphs of \(y=\log x\) and \(y=\ln x\) have the same vertical asymptote, \(x=0\).
View solution Problem 111
State the domain, range, and \(x\) -intercept of the function \(f(x)=-\ln (x-a)+b\) for \(a\) and \(b\) real positive numbers.
View solution