Problem 40 - Brief Applied Calculus

Question

Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(s)=\frac{s^{3}+1}{s-1} $$

Step-by-Step Solution

Answer

\( f'(s) = \frac{2s^3 - 3s^2 - 1}{(s-1)^2} \)

Step 1: Identify the components of the Quotient Rule

The function given is a quotient of two functions. Let \( u = s^3 + 1 \) and \( v = s - 1 \). The derivative of the quotient \( \frac{u}{v} \) is given by the Quotient Rule: \( \frac{d}{ds}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).

Step 2: Differentiate the numerator and denominator

Find the derivative of \( u \) and \( v \). For \( u = s^3 + 1 \), the derivative \( u' = 3s^2 \). For \( v = s - 1 \), the derivative \( v' = 1 \).

Step 3: Apply the Quotient Rule

Substitute the derivatives \( u' \) and \( v' \) into the Quotient Rule formula: \[ f'(s) = \frac{(3s^2)(s-1) - (s^3+1)(1)}{(s-1)^2} \].

Step 4: Simplify the numerator

Expand and simplify the terms in the numerator: \((3s^2)(s-1) = 3s^3 - 3s^2\) and \(-(s^3+1) = -s^3 - 1\). So, the numerator becomes \(3s^3 - 3s^2 - s^3 - 1\), which simplifies to \(2s^3 - 3s^2 - 1\).

Step 5: Final simplified derivative

Insert the simplified numerator back into the derivative expression: \[ f'(s) = \frac{2s^3 - 3s^2 - 1}{(s-1)^2} \]. No further simplification is possible.

Key Concepts

Quotient Rule

The Quotient Rule is a handy tool in calculus for finding the derivative of a function that is the ratio of two differentiable functions. This is particularly helpful when dealing with complex expressions. The general formula for the Quotient Rule is:

• Given two functions, u(s) and v(s), the derivative of the quotient \( \frac{u}{v} \) is:
• \[ \frac{d}{ds}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]

The numerator of this formula contains two parts: the derivative of the top function \( u' \) multiplied by the bottom function \( v \), and the original top function \( u \) multiplied by the derivative of the bottom function \( v' \). The difference of these two gives the derivative's numerator, while the square of the denominator v(s) becomes the denominator in the derivative.
Always remember: the order matters! Subtract uv' from u'v. This preserves the difference as intended by the rule. The Quotient Rule helps you avoid algebraic mishaps when differentiating fractions.

Derivative

Differentiation is a fundamental concept in calculus. It relates to finding how a function changes at any given point, often described as the function's slope or rate of change. When you differentiate a function, you're looking for the derivative.

• The derivative indicates how one quantity changes in relation to another.
• Think of it as the function's instant rate of growth or decline.

In our example, the function is \( f(s) = \frac{s^3 + 1}{s - 1} \). Differentiating it involves finding the derivatives of the numerator \( u = s^3 + 1 \), which is \( u' = 3s^2 \), and the denominator \( v = s - 1 \), which is \( v' = 1 \).
By using Quotient Rule, you get:\[ f'(s) = \frac{(3s^2)(s-1) - (s^3+1)(1)}{(s-1)^2} \]. Understanding how to find derivatives accurately allows you to tackle more elaborate calculus problems effectively.

Function Simplification

Function simplification is an essential step in calculus, especially after obtaining a derivative. This process helps in making the expression easier to interpret or compute. In many math operations, simplifying is necessary to identify patterns or solve equations efficiently.
From our original exercise, simplifying the derivative involved carefully operating on the numerator:

• Initially, calculate \( (3s^2)(s-1) \) to get \( 3s^3 - 3s^2 \).
• Similarly, \( -(s^3+1) \) simplifies to \( -s^3 - 1 \).

By combining these results, you achieve a simplified numerator: \( 2s^3 - 3s^2 - 1 \). Conveniently, the denominator remains as is, \( (s-1)^2 \), because it doesn’t factor easily with the numerator.
With simplifying, you can verify accuracy and make subsequent calculations more straightforward. Remember, an unsimplified expression can lead to mistakes in interpretation or further calculations.