Problem 74 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{3}(1+\theta \ln 3)$$

Step-by-Step Solution

Answer

\(\frac{1}{1+\theta \ln 3}\) is the derivative.

Step 1: Recognize the form and apply the change of base formula

The function given is in the form of a logarithm with a base other than 10 or e. We rewrite the function using the change of base formula for logarithms. Thus, we have \( y = \log_{3}(1+\theta \ln 3) = \frac{\ln(1+\theta \ln 3)}{\ln 3} \). This makes it easier to find the derivative using the properties of natural logarithms.

Step 2: Differentiate using the chain and quotient rules

To differentiate \( y = \frac{\ln(1+\theta \ln 3)}{\ln 3} \), use the constant multiple rule and the chain rule in conjunction. The derivative with respect to \( \theta \) will be \( \frac{1}{\ln 3} \cdot \frac{1}{1+\theta \ln 3} \cdot \ln 3 \). This simplifies to \( \frac{1}{1+\theta \ln 3} \).

Step 3: Simplify the expression

Simplifying the expression \( \frac{1}{1+\theta \ln 3} \) gives us the final derivative. Ensure that all constants and variables are correctly restated to reflect any simplifications made during differentiation.

Key Concepts

Chain Rule

When faced with the task of finding the derivative of a composition of functions, the Chain Rule is your go-to tool. Essentially, it allows us to take the derivative of a nested function. To grasp the chain rule, think of it as peeling an onion: you differentiate the outer layer, then work your way to the inner layer. Breaking it down:

• Identify the outer function and the inner function. Often, the inner function is enclosed in a bracket or inside a composition.
• Differentiate the outer function with the inner function still inside.
• Multiply the result by the derivative of the inner function.

In our original problem, we use the chain rule to differentiate the natural logarithm of an expression involving \(\theta\). By identifying the outer function as the logarithm and the inner function as \(1+\theta \ln 3\), we simplify the differentiation process. Thus, finding the derivative involves applying the chain rule to each layer.

Quotient Rule

The Quotient Rule becomes crucial when dealing with derivatives of ratios of two functions. Simply put, it's applied when you have a division of two expressions. The formula for the quotient rule is:\[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\]where \(u\) and \(v\) are functions of the independent variable. To efficiently use this:

• Identify the numerator \(u\) and the denominator \(v\).
• Calculate the derivative of \(u\), denoted as \(u'\), and the derivative of \(v\), \(v'\).
• Apply the quotient rule formula by substituting these derivatives.

In our exercise, since \(\ln(1+\theta \ln 3)\) is divided by the constant \(\ln 3\), we simplify our work by noting that constant denominators reduce the problem to applying derivative rules to the numerator alone. This minimizes the use of the quotient rule itself, streamlining the calculation process.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique for simplifying the process of differentiating complex products or quotients. It leverages the properties of logarithms to transform multiplication and division into addition and subtraction. Here's why and how you use it:

• Take the natural logarithm of both sides of an equation where necessary.
• This turns products into sums and quotients into differences thanks to logarithm laws.

In many scenarios, especially when variables are entangled in multipliers or exponents, differentiating directly would be challenging. By replacing direct differentiation with natural logarithms, you untangle complex expressions. In our case, the presence of a logarithm function in the original equation shows how logarithmic differentiation simplifies the derivative operation, particularly when combined with the chain rule.

Change of Base Formula

The Change of Base Formula is a handy logarithmic identity that allows converting logarithms from one base to another, typically to the natural logarithm. This is expressed as:\[\log_b(a) = \frac{\ln(a)}{\ln(b)}\]This transformation is particularly valuable because it allows us to work with natural logarithms (\(\ln\)), which are often easier to differentiate due to their well-established properties. To effectively apply this in calculus:

• Identify when a logarithmic function isn't in base 10 or \(e\), as in our exercise.
• Convert the logarithm using the change of base formula.
• Proceed with differentiation using more familiar natural logarithm rules.

In the exercise discussed, rewriting log base 3 to a natural log form immediately simplifies the differentiation task, harmonizing it with the chain and constant multiple rules for a cleaner result.