Problem 68 - Thomas Calculus in SI Units

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

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

Step 1: Understand the Function

The given function is expressed as \(y = \log_{3}(1+\theta \ln 3)\). It represents a logarithmic function with base 3. The variable \(\theta\) is the independent variable, and we are looking for its derivative.

Step 2: Change of Base Formula

In order to differentiate the function more easily, change the base of the logarithm from 3 to \(e\), the natural logarithm base. Use the change of base formula: \(\log_{a}(x) = \frac{\ln x}{\ln a}\). This gives us \( y = \frac{\ln(1+\theta \ln 3)}{\ln 3} \).

Step 3: Differentiate the Function

Differentiate the expression \( y = \frac{\ln(1+\theta \ln 3)}{\ln 3} \) using the chain rule. The derivative of \(\ln(u)\) with respect to \(\theta\) is \(\frac{1}{u} \frac{du}{d\theta}\). Here, let \(u = 1+\theta \ln 3\), therefore \(\frac{du}{d\theta} = \ln 3\). This yields \(\frac{d}{d\theta}\left(\frac{\ln(1+\theta \ln 3)}{\ln 3}\right) = \frac{1}{\ln 3} \cdot \frac{1}{1+\theta \ln 3} \cdot \ln 3\).

Step 4: Simplify the Derivative

Simplify the expression \(\frac{1}{\ln 3} \cdot \frac{1}{1+\theta \ln 3} \cdot \ln 3\) by canceling out \(\ln 3\). This results in \(\frac{1}{1+\theta \ln 3}\).

Step 5: State the Final Derivative

The derivative of the original function \(y\) with respect to \(\theta\) is \(\frac{1}{1+\theta \ln 3}\).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that involve logarithms or where the function is a product, quotient, or power of several functions. This method is useful for simplifying the differentiation of complex expressions.
To apply logarithmic differentiation, you typically follow these steps:

• Take the natural logarithm (ln) of both sides of the equation. This simplifies multiplication and division into addition and subtraction, making differentiation easier.
• Differentiation then follows using common rules, such as the chain rule and product rule, as necessary.

In the context of our exercise, we used this method to make dealing with the base-3 logarithm simpler by converting it into a natural logarithm. By doing so, the differentiation process was made straightforward, as we eventually used the chain rule to find the derivative.

Chain Rule

The chain rule is one of the most important differentiation techniques in calculus. It allows you to find the derivative of composite functions, which are functions composed of one function inside another. The chain rule is expressed as:\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
This formula shows that to find the derivative of a composite function \(y\) with respect to \(x\), you differentiate \(y\) with respect to an intermediate variable \(u\), then multiply by the derivative of \(u\) with respect to \(x\).
In our problem, we let \(u = 1 + \theta \ln 3\) to implement the chain rule. The derivative of \(\ln(u)\) was first calculated with respect to \(u\), and then we found the derivative of \(u\) with respect to \(\theta\). These steps allow us to simplify the differentiation process significantly, solving for the derivative more easily.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is approximately 2.71828. Natural logarithms are widely used in calculus due to their simple differentiation properties.
When dealing with logarithms in calculus, the natural logarithm is often preferred. This is because the derivative of \(\ln(x)\) is a straightforward expression: \(\frac{d}{dx}[\ln(x)] = \frac{1}{x}\).
In the exercise provided, we moved from a base-3 logarithm to a natural logarithm using the change of base formula with \(\ln\). This effectively allowed us to differentiate more easily, taking advantage of the simpler \(\ln\) rules rather than the cumbersome base-3 derivatives. This technique shows the powerful utility of natural logarithms in simplifying differentiation tasks.