Problem 25 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\frac{\ln t}{t}$$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{1 - \ln t}{t^2} \).

Step 1: Recognize the Structure

The function given is in the form of a quotient: \( y = \frac{u}{v} \), where \( u = \ln t \) and \( v = t \). We will use the quotient rule to find the derivative.

Step 2: Apply the Quotient Rule

The quotient rule states that the derivative of \( \frac{u}{v} \) is \( \frac{u'v - uv'}{v^2} \). First, find \( u' \) and \( v' \):- \( u' = \frac{d}{dt}(\ln t) = \frac{1}{t} \).- \( v' = \frac{d}{dt}(t) = 1 \).Now apply the quotient rule: \[ y' = \frac{(\frac{1}{t})t - (\ln t)(1)}{t^2} \].

Step 3: Simplify the Expression

Simplify the expression:- The numerator becomes \( 1 - \ln t \).Thus, the derivative is \[ y' = \frac{1 - \ln t}{t^2} \].

Key Concepts

Quotient Rule

The Quotient Rule is an essential technique in calculus for finding the derivative of a quotient of two functions. Imagine you have two functions, say \( u(x) \) and \( v(x) \), and you form a ratio \( \frac{u}{v} \). When you're tasked with differentiating this quotient, the Quotient Rule comes into play. The rule states:

• The derivative \( y' \) of \( \frac{u}{v} \) is given by \( \frac{u'v - uv'}{v^2} \).
• This formula requires you to know both \( u' \) (the derivative of \( u \)) and \( v' \) (the derivative of \( v \)).

What's happening here is that you're accounting for how both the numerator and the denominator change with respect to \( x \). This is crucial because the behavior of the overall fraction isn't just about how the numerator changes—it also depends on how the denominator is changing. This gives an accurate reflection of the curve’s slope at any given point where both functions are defined and differentiable.

Logarithmic Function

A logarithmic function, such as \( \ln t \), is a mathematical operation that's the inverse of exponentiation. Specifically, it's answering the question: "To what power must we raise \( e \) (Euler's number, approximately 2.71828) to get \( t \)?" Logs are extremely useful for solving equations where the variable appears as an exponent, among many other applications.

• The derivative of \( \ln t \) with respect to \( t \) is \( \frac{1}{t} \).
• This takes advantage of the fact that the slope of \( \ln t \) is steeper when \( t \) is small and flattens out as \( t \) increases.

Understanding the nature of the logarithmic function helps in simplifying derivatives and in interpreting data scenarios where you see exponential growth or decay—common in physics and economics.

Simplification

Simplification is the process of making a mathematical expression easier to understand or solve. After applying the Quotient Rule, you often end up with a complex expression that can benefit from simplification. In our problem, after applying the quotient rule:\[y' = \frac{(\frac{1}{t})t - (\ln t)(1)}{t^2}\]

• You can recognize that \((\frac{1}{t})t = 1\), which simplifies the expression considerably.
• Now, you are left with the expression \(1 - \ln t\) as the numerator over \(t^2\) as the denominator.

By reducing the expression to \( \frac{1 - \ln t}{t^2} \), you make it more manageable and easier to use for further calculations or interpretations. Simplifying expressions not only aids in solving problems faster but also helps in gaining deeper insights into the behavior of functions.