Problem 78 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \begin{equation}y=\log _{5} \sqrt{\left(\frac{7 x}{3 x+2}\right)^{\ln 5}}\end{equation}

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{3}{2x(3x+2)} \).

Step 1: Simplify the Expression

Begin by simplifying the expression inside the logarithm. We have \( y = \log_{5} \sqrt{\left( \frac{7x}{3x + 2} \right)^{\ln 5}} \). Recognize that the square root can be expressed as a power of \( \frac{1}{2} \): \( y = \log_{5} \left( \left( \frac{7x}{3x + 2} \right)^{\ln 5 * \frac{1}{2}} \right) \). This further simplifies to: \( y = \log_{5} \left( \frac{7x}{3x + 2} \right)^{\frac{\ln 5}{2}} \).

Step 2: Use Logarithmic Identity

Apply the logarithmic identity \( \log_{b}(A^n) = n \log_{b}(A) \). Thus, \( y = \frac{\ln 5}{2} \log_{5} \left( \frac{7x}{3x + 2} \right) \).

Step 3: Change of Base Formula

Use the change of base formula for logarithms: \( \log_{b}(A) = \frac{\ln(A)}{\ln(b)} \). Therefore, \( \log_{5}(A) = \frac{\ln(A)}{\ln(5)} \). Apply it: \( y = \frac{\ln 5}{2} \cdot \frac{\ln \left( \frac{7x}{3x + 2} \right)}{\ln 5} \).

Step 4: Simplify Expression Again

The expression \( y = \frac{1}{2} \ln \left( \frac{7x}{3x + 2} \right) \) now is much simpler. Differentiate this expression with respect to \( x \).

Step 5: Differentiate Using Chain and Quotient Rules

The derivative \( \frac{d}{dx} \left( \ln \left( \frac{7x}{3x + 2} \right) \right) \) is obtained by using the chain rule and the quotient rule. The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \), where \( u = \frac{7x}{3x + 2} \).

Step 6: Apply Quotient Rule

Apply the quotient rule: \( u = \frac{7x}{3x + 2} \). \( u' = \frac{(3x+2)\cdot 7 - 7x\cdot 3}{(3x+2)^2} = \frac{21}{(3x+2)^2} \). Substitute \( u \) and \( u' \) back into: \( \frac{1}{2} \cdot \frac{1}{u} \cdot u' = \frac{21}{2} \cdot \frac{1}{\frac{7x}{3x+2}} \cdot \frac{1}{(3x+2)^2} \).

Step 7: Simplify Final Derivative

Combine all parts: \( y' = \frac{21}{2} \cdot \left( \frac{3x+2}{7x} \right) \cdot \frac{1}{(3x+2)^2} = \frac{21}{14x(3x+2)} = \frac{3}{2x(3x+2)} \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a very useful technique, especially when dealing with complex functions that involve logarithms and products or quotients of functions. The main idea is to take the natural logarithm of both sides of the equation you wish to differentiate. This leverages the properties of logarithms, which often make the differentiation process easier.

Here's why logarithmic differentiation is helpful:

• It simplifies the process of differentiating powers, products, and quotients.
• By converting products into sums and powers into multipliers, it reduces complexity.
• Logarithmic differentiation is particularly beneficial when dealing with functions raised to another function's power.

In this exercise, by applying logarithmic differentiation, complex exponents and roots were simplified using logarithmic identities to facilitate the differentiation.

Quotient Rule

The Quotient Rule helps us find the derivative of a function that is defined as the division of two differentiable functions. If you have a function expressed as a quotient, \( \frac{f(x)}{g(x)} \), the Quotient Rule is essential.

The formula is given by:\[\left( \frac{f}{g} \right)' = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\]In this exercise, we applied the Quotient Rule to the component \( \frac{7x}{3x + 2} \). This involved finding the derivative of the numerator (7x), and subtracting the product of the numerator and the derivative of the denominator \( (3x+2) \), all over the square of the denominator. The calculation was crucial for continuing with the logarithmic differentiation process.

Using the Quotient Rule simplifies the differentiation of rational functions seamlessly.

Chain Rule

The Chain Rule is indispensable when you need to differentiate a composite function. It allows us to deal with functions inside other functions, which is a typical case when using logarithms.

The Chain Rule states:\[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\]Essentially, it dictates that to differentiate a composite function \(f(g(x))\), you first differentiate the outer function \(f\) at \(g(x)\), and then multiply by the derivative of the inner function \(g(x)\).

In our exercise, the Chain Rule was used to find the derivative of the natural log function involving \(\frac{7x}{3x+2}\). Here, the outer function is \( \ln(u) \) and the inner function \(u\) is \(\frac{7x}{3x+2}\). We first differentiated the natural log, and then multiplied by the derivative of \(u\), obtained using the Quotient Rule.

Logarithmic Identities

Logarithmic identities provide algebraic properties that can transform expressions involving logarithms, making them easier to work with during differentiation. These identities are more than shortcuts — they form the backbone of logarithm manipulation.

Some of the key logarithmic identities used include:

• \( \log_b(a^n) = n \cdot \log_b(a) \): Converts powers inside a log into a multiplier outside.
• Change of Base Formula: \( \log_b(a) = \frac{\ln(a)}{\ln(b)} \): Useful for changing the base of a logarithm to the natural logarithm (ln), which is especially handy for calculus.

By strategically applying these identities, you'll simplify complex expressions before differentiating them. In the given problem, recognizing these identities allowed for converting logarithmic expressions into their simplest form, preparing them for differentiation. Using the formula \( \log_b(x) = \frac{\ln(x)}{\ln(b)} \), we transitioned to natural logarithms, which makes it easier to apply calculus techniques effectively.