Problem 20 - Thomas Calculus

Question

In Exercises \(5-36,\) 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 of \( y \) with respect to \( t \) is \( \frac{-\ln t}{t^2} \).

Step 1: Identify the function type

Firstly, observe the given function \( y = \frac{1 + \ln t}{t} \). This is a rational function where both the numerator and denominator are functions of \( t \).

Step 2: Apply the quotient rule

The quotient rule states that if you have a function \( y = \frac{u}{v} \), the derivative \( \frac{dy}{dt} \) is \( \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \). In this problem, \( u = 1 + \ln t \) and \( v = t \).

Step 3: Differentiate the numerator and the denominator

Calculate the derivative of the numerator \( u = 1 + \ln t \) with respect to \( t \). The derivative \( \frac{du}{dt} \) is \( \frac{1}{t} \). Next, the derivative of the denominator \( v = t \) is \( \frac{dv}{dt} = 1 \).

Step 4: Substitute the derivatives into the quotient rule

Using the quotient rule formula, substitute \( u \), \( v \), \( \frac{du}{dt} \), and \( \frac{dv}{dt} \) into \( \frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \). This gives \( \frac{dy}{dt} = \frac{t \cdot \frac{1}{t} - ( 1 + \ln t ) \cdot 1}{t^2} \).

Step 5: Simplify the expression

Simplify \( \frac{dy}{dt} = \frac{1 - (1 + \ln t)}{t^2} = \frac{1 - 1 - \ln t}{t^2} \). Further simplify to get \( \frac{dy}{dt} = \frac{-\ln t}{t^2} \).

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental concept in calculus, specifically when dealing with derivatives of rational functions. It is a method used to find the derivative of a division of two functions. In simple terms, when you have a function that is expressed as the division of two separate functions, you apply the Quotient Rule to differentiate it. The formula for the Quotient Rule is:\[\frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2}\]Where:

• \( u \) is the function in the numerator.
• \( v \) is the function in the denominator.
• \( \frac{du}{dt} \) and \( \frac{dv}{dt} \) are the derivatives of \( u \) and \( v \), respectively.

This rule helps us determine the rate at which the function \( y = \frac{u}{v} \) changes with respect to its variable. It's important to remember that the function should be both continuous and differentiable for the Quotient Rule to be applicable.

Rational Functions

Rational functions are functions that can be expressed as the ratio of two polynomials. In mathematical terms, a rational function is any function that can be written as \( \frac{p(t)}{q(t)} \), where \( p(t) \) and \( q(t) \) are polynomial functions and \( q(t) eq 0 \). Rational functions are prevalent in calculus because they can represent a variety of real-world situations and mathematical models.These functions have unique characteristics:

• They can have vertical asymptotes, where the function approaches infinity as the denominator approaches zero.
• They may also have horizontal asymptotes, describing the behavior of the function as \( t \) approaches infinity.
• They are used to model relationships where one quantity precisely depends on another by a factor or ratio, such as speed (distance/time) or density (mass/volume).

In our exercise, \( y = \frac{1+\ln t}{t} \) is a rational function where the numerator and denominator involve algebraic and logarithmic expressions, providing a perfect example to apply the Quotient Rule.

Logarithmic Differentiation

Logarithmic differentiation is another handy tool in calculus, especially useful when dealing with complicated functions that could be differentiated more easily once their logarithms are taken. While not directly applied in this exercise, understanding logarithmic differentiation broadens your calculus toolkit.With logarithmic differentiation, you:

• Take the natural logarithm of both sides of the equation \( y = f(t) \).
• Use properties of logarithms to simplify the differentiation process, such as converting product into addition and quotient into subtraction.
• Differentiate implicitly with respect to the variable, solving for \( \frac{dy}{dt} \).

This method is especially useful for functions that are products or quotients involving variables raised to variable powers, as it transforms a complex differentiation into a simpler algebra problem. Even though logarithmic differentiation wasn't needed for our original function, it shows how different methods can be used to tackle derivatives effectively.