Problem 67 - Calculus and its applications

Question

Find the first through the fourth derivatives. Be sure to simplify each derivative before continuing. $$ f(x)=\frac{x-1}{x+2} $$

Step-by-Step Solution

Answer

The first four derivatives are: \( f'(x) = \frac{3}{(x+2)^2} \), \( f''(x) = \frac{-6}{(x+2)^3} \), \( f'''(x) = \frac{18}{(x+2)^4} \), and \( f^{(4)}(x) = \frac{-72}{(x+2)^5} \).

Step 1: Identify the function

The function given is a rational function: \( f(x) = \frac{x-1}{x+2} \). We will find its derivatives using the quotient rule.

Step 2: Apply the Quotient Rule for the first derivative

The quotient rule states that if \( f(x) = \frac{u}{v} \), then the derivative \( f'(x) \) is \( \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u = x-1 \) and \( v = x+2 \). So, \( u' = 1 \) and \( v' = 1 \). Substitute these into the formula:\[f'(x) = \frac{(x+2)(1) - (x-1)(1)}{(x+2)^2} = \frac{x + 2 - x + 1}{(x+2)^2} = \frac{3}{(x+2)^2}\]

Step 3: Differentiate again for the second derivative

To find \( f''(x) \), apply the quotient rule again to the first derivative \( f'(x) = \frac{3}{(x+2)^2} \). Let \( u = 3 \) and \( v = (x+2)^2 \). Then, \( u' = 0 \) and \( v' = 2(x+2) \). The second derivative is:\[f''(x) = \frac{(x+2)^2(0) - 3\cdot2(x+2)}{((x+2)^2)^2} = \frac{-6(x+2)}{(x+2)^4} = \frac{-6}{(x+2)^3}\]

Step 4: Find the third derivative

Differentiate \( f''(x) = \frac{-6}{(x+2)^3} \) using the quotient rule again. Here \( u = -6 \), \( u' = 0 \), \( v = (x+2)^3 \), and \( v' = 3(x+2)^2 \). Apply the quotient rule:\[f'''(x) = \frac{(x+2)^3(0) - (-6)\cdot3(x+2)^2}{((x+2)^3)^2} = \frac{18(x+2)^2}{(x+2)^6} = \frac{18}{(x+2)^4}\]

Step 5: Calculate the fourth derivative

For the fourth derivative, differentiate \( f'''(x) = \frac{18}{(x+2)^4} \). Use the quotient rule again, with \( u = 18 \), \( u' = 0 \), \( v = (x+2)^4 \), and \( v' = 4(x+2)^3 \):\[f^{(4)}(x) = \frac{(x+2)^4(0) - 18\cdot4(x+2)^3}{((x+2)^4)^2} = \frac{-72(x+2)^3}{(x+2)^8} = \frac{-72}{(x+2)^5}\]

Key Concepts

Understanding the Quotient Rule

When differentiating a rational function, like \( f(x) = \frac{x-1}{x+2} \), we often use the quotient rule. This rule is crucial when dealing with functions written as a ratio of two other functions. The quotient rule helps us find the derivative of a function that is the quotient of two differentiable functions.

According to the quotient rule, if you have a function \( f(x) = \frac{u}{v} \), its derivative \( f'(x) \) is given by \( \frac{v \cdot u' - u \cdot v'}{v^2} \). Here's the breakdown:

• \( u \) and \( v \) are functions of \( x \) that you're dividing.
• \( u' \) represents the derivative of \( u \), and \( v' \) is the derivative of \( v \).
• It's important to first find these derivatives \( u' \) and \( v' \) separately, then substitute them into the formula.

The quotient rule is particularly handy because it lets us work with these types of fractions easily, making complex derivative calculations more manageable.

Exploring Higher-Order Derivatives

Higher-order derivatives refer to the derivatives of derivatives, going beyond just the first derivative. In our example, we go all the way to the fourth derivative. Here's the step-by-step:

Start with the first derivative by applying differentiation rules, like the quotient rule. Once you have \( f'(x) \), repeat the process for each subsequent derivative:

• Second Derivative \( f''(x): \) Apply the derivative rules to \( f'(x) \).
• Third Derivative \( f'''(x): \) Continue the differentiation process on \( f''(x) \).
• Fourth Derivative \( f^{(4)}(x): \) Lastly, differentiate \( f'''(x) \).

Each step involves simplifying the expressions, which is essential. This makes it easier to see patterns or recurring terms across derivatives.

Using higher-order derivatives can help analyze functions, particularly in understanding concavity, inflection points, or in more advanced calculus like Taylor series.

Rational Functions and Their Role

A rational function is a fraction where both the numerator and the denominator are polynomials. In our example, \( f(x) = \frac{x-1}{x+2} \) illustrates a standard rational function, where both the top \( (x-1) \) and bottom \( (x+2) \) are polynomials of degree one.

Rational functions are interesting in calculus because they exhibit behaviors both similar to and distinct from simpler polynomial functions. Studying their derivatives can:

• Reveal vertical or horizontal asymptotes, as the function values behave differently at particular points.
• Help in determining limits, as the presence of \( x \) as a variable in both polynomials can lead to interesting limit-related observations.
• Illustrate continuous or discontinuous behaviors based on the zeros of the denominator.

In practical situations, understanding rational functions and their higher-order derivatives assists in modeling real-world phenomena, like rates of change or understanding complex systems in physics or engineering.