Problem 22 - Thomas Calculus

Question

In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\frac{1+\ln t}{t} $$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{-\ln t}{t^2} \).

Step 1: Apply the Quotient Rule

The given function is a quotient where the numerator is \( 1+\ln t \) and the denominator is \( t \). The quotient rule for derivatives states that if a function has the form \( \frac{u}{v} \), then its derivative is \( \frac{u'v - uv'}{v^2} \). Here, \( u = 1+\ln t \) and \( v = t \).

Step 2: Find the Derivatives of the Numerator and Denominator

For \( u = 1+\ln t \), the derivative \( u' \) is \( \frac{1}{t} \) because the derivative of \( \ln t \) is \( \frac{1}{t} \) and the derivative of a constant is 0. For \( v = t \), the derivative \( v' \) is 1, because \( \frac{d}{dt}(t) = 1 \).

Step 3: Apply the Quotient Rule Formula

Substitute \( u' = \frac{1}{t} \), \( u = 1 + \ln t \), \( v = t \), and \( v' = 1 \) into the quotient rule formula: \[y' = \frac{\left( \frac{1}{t} \right)t - (1 + \ln t)(1)}{t^2}.\]

Step 4: Simplify the Expression

Simplifying the expression from the formula: \[y' = \frac{1 - (1 + \ln t)}{t^2} = \frac{1 - 1 - \ln t}{t^2} = \frac{-\ln t}{t^2}.\]

Key Concepts

Quotient Rule

When you encounter a function that is the division of two other functions, the derivative can't be found by just applying the basic derivative rules for individual terms. The Quotient Rule comes in handy here. Suppose you have a function of the form \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of the same variable. The Quotient Rule states that the derivative of \( \frac{u}{v} \) is given by:

• \( \frac{u'v - uv'}{v^2} \)

This formula helps you find the rate of change of quotient functions.
In our original problem, the numerator \( u = 1 + \ln t \) and the denominator \( v = t \) are our key targets for differentiation.Once you compute the respective derivatives (\( u' \) and \( v' \)), plug them into the formula.
Ensure that you carefully substitute and then solve for the derivative, as handling fractions requires attention to detail in both algebraic steps and differentiation.

Logarithmic Differentiation

Logarithmic Differentiation is not applied directly here but understanding it enriches our insight into how derivatives involving logarithms are handled. It's a method particularly useful for functions where variables and exponents are intertwined, but starters can remember that:

• The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \).

This feature makes calculating derivatives involving log functions a breeze when the natural log \( \ln \) is present.
In the exercise, the presence of \( \ln t \) simplifies the problem substantially, as it means you only need to apply the derivative rule for logarithms directly: turning \( \ln t \) into \( \frac{1}{t} \). This also reflects in the derivative of our function \( 1+\ln t \), where the constant 1 vanishes, leaving us with \( \frac{1}{t} \) alone.
Embrace this understanding not just as a calculation trick, but as an insight into how easily the natural log interacts with calculus differentiation rules.

Simplifying Expressions

After applying differentiation rules, you're often left with complex expressions. Simplifying these expressions is crucial for interpreting results clearly and providing an elegant solution.
After using the Quotient Rule in this exercise, our derivative expression appeared as:

• \( y' = \frac{\left( \frac{1}{t} \right)t - (1 + \ln t)(1)}{t^2} \)

Simplification steps include conducting basic multiplication and cancellation to simplify terms. In this breakdown, the aim is to simplify the fraction:

• Recognize \( \left( \frac{1}{t} \right)t = 1 \)
• Subtract the terms inside the parentheses: \( 1 - 1 - \ln t \)
• Reduce the expression further down to: \( \frac{-\ln t}{t^2} \)

Through these steps, what appears to be a complex mess unravels into a cleaner, more functional form of the derivative. This is essential in both studying calculus and its applications. Always strive to perform simplification to ensure you communicate your results as clearly as possible.
Remember, simplicity is the ultimate sophistication in calculus!