Problem 21 - Thomas Calculus

Question

In Exercises \(7-38,\) 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 of \( y = \frac{\ln t}{t} \) is \( y' = \frac{1 - \ln t}{t^2} \).

Step 1: Identify the appropriate derivative rule

The function given is \( y = \frac{\ln t}{t} \). This is a quotient of two functions, so we will use the Quotient Rule. The Quotient Rule states that if you have a function \( \frac{u}{v} \), its derivative \( y' \) is given by \( y' = \frac{v u' - u v'}{v^2} \). In our case, \( u = \ln t \) and \( v = t \).

Step 2: Find the derivative of the numerator

Differentiate the numerator \( u = \ln t \). The derivative of \( \ln t \) with respect to \( t \) is \( \frac{1}{t} \). So, \( u' = \frac{1}{t} \).

Step 3: Find the derivative of the denominator

Differentiate the denominator \( v = t \). The derivative of \( t \) with respect to \( t \) is \( 1 \). So, \( v' = 1 \).

Step 4: Apply the Quotient Rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the Quotient Rule formula: \[ y' = \frac{v u' - u v'}{v^2} = \frac{t \left( \frac{1}{t} \right) - (\ln t) (1)}{t^2} \]Simplify the expression to find the derivative.

Step 5: Simplify the expression

Carry out the operations: \[ \frac{t \cdot \frac{1}{t} - \ln t}{t^2} = \frac{1 - \ln t}{t^2}\] The simplification results in the derivative \( y' = \frac{1 - \ln t}{t^2} \).

Key Concepts

Quotient Rule

In calculus, the Quotient Rule is a crucial technique used in the differentiation of functions that are expressed as a ratio of two functions. If you have a function represented as \( \frac{u}{v} \), the derivative, denoted \( y' \), is obtained using the Quotient Rule:

• \( y' = \frac{v u' - u v'}{v^2} \)

The terms \( u \) and \( v \) represent the numerator and denominator of your function, respectively. The derivatives of these functions, \( u' \) and \( v' \), are essential for applying this rule. Essentially, it breaks down the differentiation into manageable steps, allowing you to handle complex expressions more easily.
This rule simplifies the process of finding derivatives when functions involve division, which otherwise can be quite cumbersome.

natural logarithm

The natural logarithm, denoted as \( \ln \), is a special type of logarithm that has the mathematical constant \( e \) as its base. The constant \( e \) is approximately equal to 2.71828 and is an irrational number that appears frequently in mathematical contexts.

• It is the inverse operation of exponentiation with base \( e \).
• For example, if \( e^x = a \), then \( \ln a = x \).

In the context of differentiation, the derivative of \( \ln x \) is particularly simple: \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).
This characteristic makes it relatively straightforward to deal with natural logarithms when differentiating, especially when incorporated with other operations, such as the Quotient Rule.

differentiation

Differentiation is a fundamental concept in calculus, focused on finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. It is the foundation for determining tangent lines to curves, and it aids in understanding how variables change in relation to one another.

• Derivatives are used to find slopes of tangent lines to a curve.
• They are also crucial for solving problems related to rates of change in physics and engineering.

By applying derivative rules like the Quotient Rule or working with specific function derivatives, like the natural logarithm, one can systematically solve a wide range of calculus problems.
Understanding differentiation involves not just calculating derivatives but also interpreting them in various practical contexts.

calculus exercises

Working through calculus exercises is an effective way to master concepts like the Quotient Rule and differentiation. These exercises involve applying theoretical principles to specific problems, which can include functions represented as ratios, products, or other more complex expressions.

• Helps reinforce the application of differentiation rules.
• Aids in developing problem-solving skills through practical examples.

Regular practice with varying function types solidifies understanding and builds confidence in handling derivatives. For instance, exercises that involve the natural logarithm and the Quotient Rule together provide valuable experience in combining different techniques.
Such practice prepares you for more advanced mathematical concepts and real-world applications, offering a solid foundation in calculus.