Understanding the Differentiation Problem

The question asks us to find the derivative of the function \( y = \log_5 (\log x^2) \) with respect to \( x \). This involves differentiating a composite function, requiring the use of the chain rule and properties of logarithms.

The function is a logarithm with base 5, where the argument is another logarithm, specifically \( \log x^2 \). The base of the inner logarithm \( \log x^2 \) is not explicitly given, but based on the options provided, it is implied to be the natural logarithm (\( \ln \)) or a common logarithm (\( \log_{10} \)). The structure of the correct answer strongly suggests that \( \log \) refers to the natural logarithm (\( \ln \)). We will proceed with this assumption.

So, we interpret the function as \( y = \log_5 (\ln x^2) \).

Step-by-Step Differentiation Using Chain Rule

We need to find \( \frac{dy}{dx} \). We will use the chain rule and the formula for differentiating logarithmic functions. The chain rule states that if \( y = f(u) \) and \( u = g(x) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Our function is \( y = \log_5 (\ln x^2) \). Let's simplify the argument of the inner logarithm first. Assuming \( x > 0 \) (which is required for \( \ln x \) to be defined, and implicitly for \( \ln(\ln x^2) \) to be defined, as \( \ln x^2 > 0 \) implies \( x^2 > 1 \), so \( |x| > 1 \)), we have \( \ln x^2 = 2 \ln x \).

So the function becomes \( y = \log_5 (2 \ln x) \).

Applying the Change of Base Formula

To differentiate a logarithm with base 5, it's helpful to convert it to a more standard base, like the natural logarithm (ln). The change of base formula is \( \log_b a = \frac{\ln a}{\ln b} \).

Applying this to our function:

\( y = \log_5 (2 \ln x) = \frac{\ln (2 \ln x)}{\ln 5} \)

Differentiating the Function

Now we differentiate \( y \) with respect to \( x \):

\( \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\ln (2 \ln x)}{\ln 5} \right) \)

Since \( \ln 5 \) is a constant, we can take it out of the differentiation:

\( \frac{dy}{dx} = \frac{1}{\ln 5} \cdot \frac{d}{dx} (\ln (2 \ln x)) \)

Now we need to differentiate \( \ln (2 \ln x) \). This is a composite function. Let \( u = 2 \ln x \). Then we need to differentiate \( \ln u \). The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \).

By the chain rule, \( \frac{d}{dx} (\ln (2 \ln x)) = \frac{d}{du} (\ln u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \).

Substitute back \( u = 2 \ln x \):

\( \frac{d}{dx} (\ln (2 \ln x)) = \frac{1}{2 \ln x} \cdot \frac{d}{dx} (2 \ln x) \)

Now differentiate \( 2 \ln x \) with respect to \( x \):

\( \frac{d}{dx} (2 \ln x) = 2 \cdot \frac{d}{dx} (\ln x) = 2 \cdot \frac{1}{x} = \frac{2}{x} \)

Substitute this result back:

\( \frac{d}{dx} (\ln (2 \ln x)) = \frac{1}{2 \ln x} \cdot \frac{2}{x} = \frac{2}{2x \ln x} = \frac{1}{x \ln x} \)

Combining the Parts

Now substitute this result back into the expression for \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{1}{\ln 5} \cdot \frac{1}{x \ln x} = \frac{1}{x (\ln 5) (\ln x)} \)

Assuming \( \log \) in the options refers to the natural logarithm (\( \ln \)), the result is \( \frac{1}{x (\log 5) (\log x)} \).

Comparing with Options

Let's compare our derived differentiation result with the given options:

• Option 1: \( \frac{2}{x^2 (\log 5) (\log x^2)} \)
• Option 2: \( \frac{2}{x (\log 5) (\log x)} \)
• Option 3: \( \frac{1}{x (\log 5) (\log x^2)} \)
• Option 4: \( \frac{1}{x (\log 5) (\log x)} \)

Our calculated result, \( \frac{1}{x (\ln 5) (\ln x)} \), matches Option 4 when \( \log \) is interpreted as \( \ln \).

Final Result of Differentiation

The differentiation of \( \log_5 (\log x^2) \) with respect to \( x \) is \( \frac{1}{x (\log 5) (\log x)} \), assuming \( \log \) denotes the natural logarithm.

Revision Table: Key Calculus Concepts

Additional Information on Logarithm Differentiation

Differentiating functions involving logarithms is a common task in calculus. Understanding the properties of logarithms and the chain rule is crucial. When dealing with logarithms of bases other than \( e \), it is often easiest to convert them to the natural logarithm using the change of base formula before differentiating. Remember that the domain of \( \ln x \) is \( x > 0 \). For a term like \( \ln x^2 \), the domain is \( x \neq 0 \), but \( \ln x^2 = 2 \ln |x| \). However, when \( \ln x^2 \) appears inside another logarithm, like \( \log_5(\ln x^2) \), the argument \( \ln x^2 \) must be positive, which means \( x^2 > 1 \). This implies \( |x| > 1 \). In standard calculus problems like this, we often assume the domain is restricted such that all terms are well-defined and simplify \( \ln x^2 \) as \( 2 \ln x \) assuming \( x > 1 \).

The general formula for the derivative of \( \log_b(g(x)) \) using the chain rule is:

\( \frac{d}{dx} (\log_b(g(x))) = \frac{1}{g(x) \ln b} \cdot g'(x) \)

In our problem, \( b = 5 \) and \( g(x) = \ln x^2 \). We found \( g'(x) = \frac{d}{dx} (\ln x^2) = \frac{2}{x} \) (assuming \( x>0 \)). So, applying the formula:

\( \frac{dy}{dx} = \frac{1}{(\ln x^2) \ln 5} \cdot \frac{2}{x} \)

Using \( \ln x^2 = 2 \ln x \), we get:

\( \frac{dy}{dx} = \frac{1}{(2 \ln x) \ln 5} \cdot \frac{2}{x} = \frac{2}{2x (\ln x) (\ln 5)} = \frac{1}{x (\ln x) (\ln 5)} \)

This confirms our previous step-by-step result.