Problem 216 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } f(x)=\log _{x^{2}}(\ln x), \text { then find } f^{\prime}(e) \text { . } $$

Step-by-Step Solution

Answer

The derivative of the function at \(x = e\) is \(1/2\).

Step 1: Rewrite the function

We start by rewriting the function as\( f(x) = \ln(\ln x) / \ln( x^2 ) \). We did this because the logarithm using base \(x^{2}\) can be transformed into a quotient of natural logarithms according to the rule \( \log_b(a) = \ln(a) / \ln(b) \).

Step 2: Differentiate the function

We differentiate this function using the quotient rule, which states that the derivative of \(u/v\) where \(u\) and \(v\) are functions of \(x\) is \(u'v - uv' / v^2\). Here, \(u = \ln(\ln x)\) and \(v = \ln(x^2)\). Let's get \(u'\) and \(v'\). \(u' = 1/ (\ln x) * 1/x = 1/xlnx\), and \(v' = 2/x\). Hence, \(f'(x) = (u'v - uv')/v^2 = [(1 / (xlnx))(lnx^2) - (\ln(\ln x))(2/x)]/((ln(x^2))^2)\). This simplifies to \(f'(x) = (2/xlnx - 2 \ln(\ln x)/x^2) / (4\ln(x)^2/ x^2)\).

Step 3: Find f'(e)

Finally, we find \(f'(e)\) by replacing \(x\) with \(e\) in \(f'(x)\). Since base of natural logarithm, \(\ln(e) = 1\), the derivative at \(x = e\) is given by \(f'(e) = (2/e -2/e^2) / (4/ e^2) = 1/2\).

Key Concepts

Logarithmic Differentiation

Logarithmic Differentiation can simplify complex differentiation problems. When faced with functions involving logarithms with unusual bases, rewriting them in terms of natural logarithms allows for easier manipulation.

In the original exercise, the function \( f(x) = \log_{x^2}(\ln x) \) was rewritten using natural logarithms: \( f(x) = \frac{\ln(\ln x)}{\ln(x^2)} \).

Here's why this transformation works:

• The base-change formula: \( \log_b(a) = \frac{\ln(a)}{\ln(b)} \) allows expressing logarithms with any base in terms of logarithms with base \(e\).
• This conversion is crucial because it provides a standard form that is easier to differentiate using calculus rules.

Logarithms in different bases can often lead to cumbersome calculations. By converting them to natural logs, we streamline the process and make differentiation manageable.

Quotient Rule

The Quotient Rule is a handy tool for differentiating functions that are expressed as a quotient of two other functions. If you have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the derivative \( f'(x) \) can be found using:

\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]

In our exercise, \( u = \ln(\ln x) \) and \( v = \ln(x^2) \). To find \( f'(x) \), we:

• Differentiate \( u(x) \) to find \( u'(x) = \frac{1}{x \ln x} \).
• Differentiate \( v(x) \) to find \( v'(x) = \frac{2}{x} \).
• Plug into the quotient rule: \( f'(x) = \frac{(\frac{1}{x \ln x})(\ln(x^2)) - (\ln(\ln x))(\frac{2}{x})}{(\ln(x^2))^2} \).

This approach helps break down the differentiation process into manageable steps and is particularly useful for intricate functions involving division.

Derivative Evaluation

Evaluating the derivative at a specific point is often the final step in solving a differentiation problem. For the exercise, after applying the quotient rule, we needed to evaluate \( f'(x) \) at \( x = e \).

Here's the step-by-step process:

• The natural logarithm of \( e \), \( \ln(e) \), is 1, which simplifies calculations.
• At \( x = e \), the expression becomes \( f'(e) = \frac{2/e - 2/e^2}{4/e^2} \).
• Simplifying this expression gives \( f'(e) = \frac{1}{2} \).

By substituting \( x = e \) into the differentiated expression, the specific behavior of the function at this point is revealed. Keep in mind that evaluating derivatives at specific values can provide insights into the function's behavior at particular points, which is crucial in analysis and applications.