Problem 22 - Single Variable Calculus

Question

Find the derivative of each function. \(f(r)=\frac{1-r e^{r}}{1+e^{r}}\) \(\begin{aligned} 1+\frac{1}{x} & \text { 24. } y &=\frac{1+\frac{1}{x}}{1-\frac{1}{x}} \end{aligned}\)

Step-by-Step Solution

Answer

The short answer for the derivatives of the given functions are: For the first function: \(f'(r)=\frac{-2e^r-4e^{2r}+2re^{2r}}{(1+e^r)^2}\) For the second function: \(y'=\frac{-2}{(x-1)^2}\)

Step 1: Differentiate the numerator

We have \(u(r)=1-re^r\), so \(u'(r)=-e^r-re^r\).

Step 2: Differentiate the denominator

We have \(v(r)=1+e^r\), so \(v'(r)=e^r\).

Step 3: Apply the quotient rule

Using the quotient rule, we get: \(f'(r)=\frac{(-e^r-re^r)(1+e^r)-(1-re^r)(e^r)}{(1+e^r)^2}\) #Step 2: Simplify the derivative#

Step 4: Distribute terms

Expanding the terms in the numerator, we get: \(f'(r)=\frac{-e^r-e^{2r}-re^r-re^{2r}+re^r-re^{2r}}{(1+e^r)^2}\)

Step 5: Simplify the terms

Combining like terms in the numerator, we get: \(f'(r)=\frac{-2e^r-4e^{2r}+2re^{2r}}{(1+e^r)^2}\) So the derivative of the first function is \(f'(r)=\frac{-2e^r-4e^{2r}+2re^{2r}}{(1+e^r)^2}\). #Step 3: Differentiate the second function#

Step 6: Write the function

First, write the second function as \(y=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\) in a simplified form: \(y=\frac{x+1}{x-1}\).

Step 7: Differentiate the numerator

We have \(u(x)=x+1\), so \(u'(x)=1\).

Step 8: Differentiate the denominator

We have \(v(x)=x-1\), so \(v'(x)=1\).

Step 9: Apply the quotient rule

Using the quotient rule, we get: \(y'=\frac{(1)(x-1)-(x+1)(1)}{(x-1)^2}\) #Step 4: Simplify the derivative#

Step 10: Simplify the terms

Simplifying the terms in the numerator, we get: \(y'=\frac{x-1-x-1}{(x-1)^2}\)

Step 11: Simplify the expression

Combining like terms in the numerator and simplifying the expression, we get: \(y'=\frac{-2}{(x-1)^2}\) So the derivative of the second function is \(y'=\frac{-2}{(x-1)^2}\).

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in differentiation, especially when dealing with functions that are ratios of two other functions. It helps us find how the rate of change of a ratio changes as its variables change. When applying the quotient rule, keep in mind the formula: if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative \( f'(x) \) is given by:\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]Here, \( u(x) \) is the numerator function, and \( v(x) \) is the denominator function. The steps involve:

• Differentiating the top function \( u(x) \) to get \( u'(x) \)
• Differentiating the bottom function \( v(x) \) to get \( v'(x) \)
• Substituting into the quotient rule formula

This helps us systematically simplify and compute the derivative of a ratio of functions.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function. This is crucial in calculus as it helps describe the rate at which something changes. For quotient problems, we use the quotient rule among other techniques. We also apply basic differentiation rules like:

• Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Exponential Rule: If \( f(x) = e^x \), then \( f'(x) = e^x \).
• Constant Multiplier Rule: If \( f(x) = c \, g(x) \), then \( f'(x) = c \, g'(x) \).

In the exercise, we differentiated terms like \( re^r \) by applying the product rule—a technique used for differentiating a product of two functions. Stay consistent in applying these rules to transform and find derivatives accurately.

Function Simplification

Once you have applied the differentiation rules, it is time to simplify. Simplification is key to making an equation easier to work with and understand. During the exercise, terms were expanded and restructured to combine like terms and remove unnecessary complexity. When simplifying a derivative obtained through the quotient rule:

• Distribute terms correctly in the numerator or denominator.
• Combine like terms to reduce the number of terms.
• Look for common factors that might simplify the expression further.

The simplification process ensures that the derivative is expressed in its simplest form, making it easier to interpret and use in further analysis.

Numerator and Denominator Analysis

Analyzing the numerator and the denominator separately is crucial when using the quotient rule. This involves understanding each component of the function before applying calculus operations.

• Identify and differentiate the numerator \( u(r) \). Understand its contribution to the overall function and its derivative.
• Identify and differentiate the denominator \( v(r) \). This helps in determining the overall shape and behavior of the function.
• Keep track of changes in both parts when calculating derivatives since any change in one affects the entire equation.

In the given problem, examining each function separately allowed a clear view of how each part affects the rate of change. Careful analysis ensures accuracy and provides insights into the behavior and constraints of the function resulting from changes in its components.