Problem 30 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{2x(3x+2)} \).

Step 1: Simplify the Expression

Rewrite the given function in a simpler form. We have:\[ y = \log_5 \left( \sqrt{\left( \frac{7x}{3x+2} \right)^{\ln 5}} \right) \]First, express the square root as an exponent:\[ y = \log_5 \left( \left( \frac{7x}{3x+2} \right)^{(\ln 5)/2} \right) \]Use the property of logarithms that states \( \log_b(a^c) = c \cdot \log_b(a) \):\[ y = \frac{\ln 5}{2} \cdot \log_5 \left( \frac{7x}{3x+2} \right) \]

Step 2: Change Logarithm Base

Use the change of base formula to convert the base of the logarithm to natural logarithms as it simplifies differentiation:\[ \log_5 \left( \frac{7x}{3x+2} \right) = \frac{\ln\left( \frac{7x}{3x+2} \right)}{\ln 5} \]Substitute back into the expression for \(y\):\[ y = \frac{\ln 5}{2} \cdot \frac{\ln\left( \frac{7x}{3x+2} \right)}{\ln 5} \]This simplifies to:\[ y = \frac{1}{2} \ln\left( \frac{7x}{3x+2} \right) \]

Step 3: Differentiate Simplified Expression

Differentiate \( y = \frac{1}{2} \ln\left( \frac{7x}{3x+2} \right) \) with respect to \( x \). Use the chain rule and the derivative of \( \ln(u) \) which is \( \frac{1}{u} \frac{du}{dx} \):Let \( u = \frac{7x}{3x+2} \).Then, \( \frac{du}{dx} = \frac{(3x+2) \cdot 7 - 7x \cdot 3}{(3x+2)^2} = \frac{21x + 14 - 21x}{(3x+2)^2} = \frac{14}{(3x+2)^2} \).The derivative of \( y \) is:\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{\frac{7x}{3x+2}} \cdot \frac{14}{(3x+2)^2} \]Simplifies to:\[ \frac{dy}{dx} = \frac{7(3x+2)}{2(7x)(3x+2)^2} = \frac{7}{2(7x)(3x+2)} = \frac{1}{2x(3x+2)} \]

Step 4: Final Expression

So, the derivative \( \frac{dy}{dx} \) is:\[ \frac{dy}{dx} = \frac{1}{2x(3x+2)} \]

Key Concepts

Logarithm

Logarithms are an essential mathematical concept, especially when dealing with exponential functions. A logarithm essentially asks the question: "What power must this base be raised to, to obtain a certain number?" This is expressed in the form:

• For example, if we say \( \log_b(a) = c \), it means that \( b^c = a \).

In the exercise, the logarithm is originally in base 5, which is represented as \( \log_5(\cdot) \). When solving calculus problems, having the logarithm in a base that is more often used, like base \( e \) or base 10, assists in simplifying the differentiation process.
The logarithmic property used to simplify the expression is \( \log_b(a^c) = c \cdot \log_b(a) \). This allows expressing complex exponents or powers in a simpler, linear form. Another property that helps in converting bases is the change of base formula, which will be discussed in detail in another section.

Chain Rule

The chain rule is a fundamental tool for differentiating composite functions. When a function is composed of multiple functions nested together, the chain rule helps to break them apart for differentiation.

• The standard expression of the chain rule is: \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).

In the context of the original exercise, the chain rule was used to differentiate \( \ln(u) \). By recognizing that the logarithm function \( \ln(u) \) could be differentiated, the internal function or "inside" part \( u = \frac{7x}{3x+2} \) was set apart for differentiation.
To find \( \frac{du}{dx} \), the chain rule requires finding derivatives of these internal components. Consequently, the straightforward derivative of the natural logarithm \( \ln(u) \), which is \( \frac{1}{u} \cdot \frac{du}{dx} \), is applied. This breaks the derivation into manageable pieces, efficiently moving through complex logarithmic and algebraic transformations.

Change of Base Formula

The change of base formula is a significant mathematical utility for working with logarithms. It allows us to express a logarithm in one base in terms of another base, typically using base 10 or base \( e \). This is particularly useful in calculus, where differentiating logarithms in the natural base \( e \) simplifies the process.

• The formula is expressed as \( \log_b(a) = \frac{\ln(a)}{\ln(b)} \) when converting to natural logs.

In the exercise at hand, the initial logarithm is in base 5, \( \log_5 \). By using the change of base formula, it is rewritten in terms of natural logarithms, allowing easier application of derivative rules since \( \ln(x) \) is a straightforward derivative.
The conversion helps streamline the differentiation process: the base conversion reduces the expression to involve \( \ln \) operations, which standardizes the derivatives involved, thus transforming the function into a simpler form that is less cumbersome to handle analytically.