Problem 77 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{2} r \cdot \log _{4} r$$

Step-by-Step Solution

Answer

The derivative is \(\frac{\ln r}{r (\ln 2)^2}\).

Step 1: Change the Base of the Logarithms

Convert the logarithms into a base that simplifies the use of the derivative. We can use the change of base formula.\[ \log_{2} r = \frac{\ln r}{\ln 2} \quad \text{and} \quad \log_{4} r = \frac{\ln r}{\ln 4} \]Hence, the function becomes:\[ y = \frac{\ln r}{\ln 2} \cdot \frac{\ln r}{\ln 4} = \frac{(\ln r)^2}{\ln 2 \cdot \ln 4} \]

Step 2: Simplify the Constant

Notice that \( \ln 4 = 2 \ln 2 \), therefore:\[ \ln 2 \cdot \ln 4 = (\ln 2)(2 \ln 2) = 2 (\ln 2)^2 \]Now the expression for \( y \) simplifies to:\[ y = \frac{(\ln r)^2}{2 (\ln 2)^2} \]

Step 3: Differentiate the Function

Differentiate \( y = c (\ln r)^2 \), where \( c = \frac{1}{2(\ln 2)^2} \), with respect to \( r \). The derivative of \((\ln r)^2\) is obtained using the chain rule:\[ \frac{d}{dr}(\ln r)^2 = 2(\ln r) \cdot \frac{1}{r} = \frac{2 \ln r}{r} \]Thus, the derivative \( \frac{dy}{dr} \) is:\[ \frac{dy}{dr} = c \cdot \frac{2 \ln r}{r} = \frac{2 \ln r}{r \cdot 2 (\ln 2)^2} = \frac{\ln r}{r (\ln 2)^2} \]

Step 4: Present the Final Answer

The result of the differentiation gives the final derivative of the function:\[ \frac{dy}{dr} = \frac{\ln r}{r (\ln 2)^2} \]

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used when differentiating functions that are complicated products, quotients, or powers. This method simplifies the process by applying logarithms, which turns products into sums and quotients into differences. To begin with logarithmic differentiation, you take the natural logarithm of both sides of the equation. This allows the application of the properties of logarithms, such as:

• Converting products into sums: \( \ln(ab) = \ln a + \ln b \)
• Changing quotients into differences: \( \ln(\frac{a}{b}) = \ln a - \ln b \)
• Transforming powers to coefficients: \( \ln(a^b) = b \ln a \)

After applying these properties, you differentiate implicitly with respect to the independent variable to find the derivative. This technique is particularly useful when you're dealing with functions of a complex nature, like the product of two logarithms, as in our original exercise.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. These are functions where one function is nested inside another.The chain rule states that if you have a composite function \( f(g(x)) \), its derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function:

• \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \)

In the given solution, the expression \((\ln r)^2\) involves an outer power function and an inner logarithmic function. To differentiate, we:

• Identify the outer function as \( u^2 \) and the inner function as \( \ln r \)
• Apply the power rule to get 2u, and then multiply by the derivative of \( \ln r \)

Therefore, the result \( \frac{d}{dr}(\ln r)^2 = 2(\ln r)\frac{1}{r} \) is obtained. This showcases how the chain rule breaks down the differentiation into manageable steps.

Change of Base Formula

The change of base formula is a trick used in logarithms to shift bases into a more manageable form for calculation. This formula is particularly useful when the base of the logarithms in your function is not natural (base \(e\)) or common (base 10). It allows conversion using:\[ \log_{b} a = \frac{\ln a}{\ln b} \]This change of base transformation assists in simplifying differentiation problems by converting different logarithmic bases into natural logarithms. In our context:

• \( \log_{2} r = \frac{\ln r}{\ln 2} \)
• \( \log_{4} r = \frac{\ln r}{\ln 4} \)

Thus, the expression \(y = \log_{2} r \cdot \log_{4} r\) transformed into \( \frac{(\ln r)^2}{\ln 2 \cdot \ln 4} \), which is easier to handle when applying derivative rules.

Simplification of Logarithms

Simplifying logarithms is critical when dealing with complex expressions in calculus. By recognizing properties of logarithms, such simplification makes differentiation more straightforward.One key property used in the example solution is recognizing \( \ln 4 = 2 \ln 2 \). Applying this substitution to the expression \( \frac{(\ln r)^2}{\ln 2 \cdot \ln 4} \) leads to:

• \( \ln 2 \cdot \ln 4 = \ln 2 \cdot (2 \ln 2) = 2 (\ln 2)^2 \)

This realization enables further simplification of \( y \) to \( \frac{(\ln r)^2}{2 (\ln 2)^2} \). Such simplification reduces fraction complexity and streamlines differentiation. It is a testament to how recognizing logarithmic identities and properties can facilitate more efficient calculus operations.