Problem 25 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \(y=\log _{4} x+\log _{4} x^{2}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{3}{x \log_{e} 4} \).

Step 1: Simplify Logarithmic Expression

The given function is \( y = \log_{4} x + \log_{4} x^2 \). First, use the logarithmic property \( \log_b m + \log_b n = \log_b (mn) \) to combine the terms. Therefore, \( y = \log_{4} (x \cdot x^2) = \log_{4} (x^3) \).

Step 2: Change of Base Formula

Apply the change of base formula, \( \log_{b} a = \frac{\log_{c} a}{\log_{c} b} \), where \(c\) is any positive number (often \(e\) or 10). Here, we use \(e\) to get \( y = \frac{\log_{e} x^3}{\log_{e} 4} \).

Step 3: Differentiate Using the Chain Rule

Differentiate \( y = \frac{\log_{e} x^3}{\log_{e} 4} \) with respect to \(x\). The derivative of \(\log_{e} x^3\) is \(\frac{3}{x}\) using the chain rule. So, \( \frac{dy}{dx} = \frac{1}{\log_{e} 4} \cdot \frac{3}{x} \).

Step 4: Finalize the Derivative

Combine the constants to write the derivative as \( \frac{dy}{dx} = \frac{3}{x \cdot \log_{e} 4} \).

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse operations of exponential functions. An exponential function with a base, such as 4, will have a corresponding logarithmic function, denoted as \(\log_{4}(x)\). This means that \(\log_{4}(x)\) answers the question: To what power should the base 4 be raised to produce x? For instance, if \(4^2 = 16\), then \(\log_{4}(16) = 2\).
Understanding logarithms is crucial in mathematics as they simplify complex calculations, particularly in cases involving multiplicative processes by turning them into additive ones. This principle plays a key role in solving problems involving logarithms, such as the one you're tackling in this exercise.

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of a composition of functions. It states that the derivative of a composite function \( f(g(x)) \) is the derivative of \( f \) evaluated at \( g(x) \), times the derivative of \( g \) with respect to \( x \). Mathematically, it's expressed as:

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

In our problem, the chain rule is applied when differentiating \( \log_{e}(x^3) \). Here, \(g(x) = x^3\) and the derivative \( g'(x) \) is \(3x^2\), leading us to \( \frac{3}{x} \) upon differentiation of the logarithmic function with respect to \( x \).
The chain rule is essential when dealing with nested or composed functions, allowing you to "unpack" the layers of the function individually.

Change of Base Formula

The change of base formula allows you to evaluate logarithms with different bases by converting them into a commonly used base, such as e (natural log) or 10 (common log). This formula is:

• \( \log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)} \)

This is particularly useful in situations where your calculator only supports natural logarithms (\(\ln\)) or base 10 logs (\(\log\)).
In the solution to our exercise, we used the change of base formula to convert \( \log_{4}(x^3) \) to \( \frac{\log_{e}(x^3)}{\log_{e}(4)} \). This allows the derivative process to be carried out using natural logs, which are more mathematically tractable in calculus.

Properties of Logarithms

The properties of logarithms are a set of rules that simplify the manipulation of logarithmic expressions. Some key properties include:

• Product Property: \( \log_b(mn) = \log_b(m) + \log_b(n) \)
• Quotient Property: \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \)
• Power Property: \( \log_b(m^n) = n\log_b(m) \)

In the original exercise, we utilized the product property to simplify the expression \( \log_{4}(x) + \log_{4}(x^2) \) into \( \log_{4}(x^3) \). These properties are incredibly useful as they enable you to condense or expand logarithmic expressions, making calculations simpler and more straightforward.
Understanding these properties lays the groundwork for handling more intricate logarithmic equations and can substantially streamline your problem-solving approach.