Problem 116 - Calculus and its applications

Question

Differentiate. $$ f(t)=\frac{\ln t^{2}}{t^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{2 - 2 \ln(t^2)}{t^3} \).

Step 1: Recognize the Function Type

The function given is a quotient, where the numerator is \( \ln(t^2) \) and the denominator is \( t^2 \). Therefore, we'll use the quotient rule for differentiation, which is: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}, \] where \( u = \ln(t^2) \) and \( v = t^2 \).

Step 2: Differentiate the Numerator

First, differentiate the numerator \( u = \ln(t^2) \). Using the chain rule, we find: \[ u' = \frac{d}{dt} \left( \ln(t^2) \right) = \frac{2}{t}. \]

Step 3: Differentiate the Denominator

Differentiate the denominator \( v = t^2 \), resulting in: \[ v' = \frac{d}{dt}(t^2) = 2t. \]

Step 4: Apply the Quotient Rule

Now apply the quotient rule using the derivatives found: \[ \left( \frac{\ln(t^2)}{t^2} \right)' = \frac{\left( \frac{2}{t} \right)t^2 - (\ln(t^2))(2t)}{(t^2)^2}. \]

Step 5: Simplify the Expression

Simplify the expression from step 4: \[ = \frac{2t - 2t \ln(t^2)}{t^4}. \] Divide each term by \( t^2 \) to simplify further: \[ = \frac{2 - 2 \ln(t^2)}{t^3}. \]

Key Concepts

Quotient Rule

When we come across a function that is expressed as a ratio of two differentiable functions, the quotient rule is our go-to method for differentiation. It helps us find the derivative of the function by working with the derivatives of its numerator and denominator separately. The general formula for the quotient rule is:

• If a function is in the form \( \frac{u}{v} \), then:
• The derivative is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).

Understanding each component:

• \( u \) is the numerator function.
• \( v \) is the denominator function.
• \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

In the given problem, identifying that our function \( f(t)=\frac{\ln(t^{2})}{t^{2}} \) is a quotient, lets us know that we should use the quotient rule. It helps keep us organized by structurally breaking down the complex expressions into four manageable derivatives.

Chain Rule

The chain rule is a fundamental component in differentiation, especially useful when dealing with composite functions. A composite function is essentially a function within another function. For example, in our problem, the numerator is \( \ln(t^2) \), which is a composite function because \( t^2 \) is inside the natural logarithm function.Here's how the chain rule works:

• Suppose you have a function \( h(x) = f(g(x)) \), then the derivative \( h'(x) \) is found using:
• \( h'(x) = f'(g(x)) \cdot g'(x) \).

This approach means we first differentiate the outer function while keeping the inner function intact, and then multiply by the derivative of the inner function.In our example, when differentiating \( \ln(t^2) \), notice:

• The outer function is \( \ln(x) \), with derivative \( \frac{1}{x} \).
• The inner function is \( t^2 \), with derivative \( 2t \).

Using the chain rule, we calculate \( u' = \frac{2}{t} \) for the numerator, simplifying our differentiation.

Logarithmic Differentiation

Logarithmic differentiation can be a powerful tool, particularly when simplifying the process of differentiating products or quotients involving exponential functions. This technique utilizes the properties of logarithms to make differentiation more manageable.For logarithmic functions, the basic idea is:

• Take the natural logarithm of both sides of an equation.
• Use properties of logarithms to simplify.
• Differentiate implicitly with respect to the variable.

In our exercise, although not directly using logarithmic differentiation, understanding it can still be beneficial. It simplifies the numerator \( \ln(t^2) \) by allowing us to break it down using the property of logarithms: \( \ln(t^2) = 2\ln(t) \).Such manipulation can make differentiation steps cleaner. Remember, while direct logarithmic differentiation wasn't necessary here, knowing its principles ensures we are equipped to tackle more complex expressions in the future with confidence and ease.