Problem 35 - College Algebra and Calculus: An Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=x\left(1-\frac{2}{x+1}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \( f(x) = x - \frac{2x}{x+1} \) is \( f'(x) = 1 - \frac{2}{(x + 1)^2} \)

Step 1: Rewrite the expression

First, it's easier to manage the equation if we simplify it. We can distribute \( x \) in the equation, giving us \( f(x) = x - \frac{2x}{x+1} \).

Step 2: Differentiate using the product rule and the quotient rule

Differentiate using the product rule and quotient rule. The derivative of \( x \) is 1. To find the derivative of \( \frac{-2x}{x+1} \), we use the quotient rule which states that the derivative of \( \frac{u}{v} \) is \( \frac{u'v - uv'}{v^2} \). Here, \( u = -2x \), \( v = x + 1 \). Thus, \( u' = -2 \) and \( v' = 1 \). Substituting these values gives us \( \frac{-2(x + 1) - (-2x)}{(x + 1)^2} \).

Step 3: Simplify the expression

Simplify the expression obtained in the previous step: \( 1 + \frac{-2x - 2 + 2x}{(x+1)^2} = 1 - \frac{2}{(x + 1)^2} \)

Key Concepts

Understanding the Product Rule

When dealing with derivatives of functions that involve products of two or more terms, the product rule becomes very handy. The product rule states: if you have two differentiable functions, say \( u(x) \) and \( v(x) \), their derivative is given by the formula \[ u'(x)v(x) + u(x)v'(x) \].
Here's how you can easily remember and apply it:

• Differentiation is simply the process of finding the function's derivative.
• Think of the product rule as taking the derivative of each function, one at a time, while keeping the other function constant.
• Apply this rule only when you multiply functions together, not when you have functions added or subtracted.

In the context of our exercise, the product rule helps manage situations where products appear after simplifying expressions. However, in this particular exercise, we directly use the product rule as a compliment to managing other rules like the quotient rule.

Mastering the Quotient Rule

The quotient rule is vital when you deal with division of functions. If \( u(x) \) and \( v(x) \) are differentiable functions, the quotient rule states that the derivative of \( \frac{u(x)}{v(x)} \) is given by:\[\frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\]Let's break it down:

• The numerator involves multiplying the derivative of the top with the bottom function, minus the product of the top function with the derivative of the bottom.
• The whole expression is divided by the square of the bottom function.
• This rule is crucial when you have a fraction where both the top and bottom parts are functions of \( x \).

In our exercise, applying the quotient rule helps in differentiating the term \( \frac{-2x}{x+1} \). Make sure to find the derivatives of both the numerator and the denominator separately before applying the rule. This will give you a structured approach to simplify the overall expression in the end.

Differentiation Rules Overview

Differentiation rules are fundamental techniques in calculus used to find how a function changes at any point. These rules provide a framework for calculating derivatives efficiently:

• The Power Rule: For any function \( f(x) = x^n \), the derivative \( f'(x) = nx^{n-1} \).
• The Constant Rule: The derivative of a constant is always zero. If \( c \) is a constant, then \( f(x) = c \), and \( f'(x) = 0 \).
• The Sum/Difference Rule: The derivative of a sum/difference of functions is the sum/difference of their derivatives. If \( f(x) = u(x) \pm v(x) \), then \( f'(x) = u'(x) \pm v'(x) \).

Understanding these rules is essential for tackling derivative problems efficiently. Each rule provides unique insights and methods needed to approach different algebraic expressions. In our exercise, utilizing the product and quotient rules within these broader differentiation rules allows for accurate computation of derivatives in more complex algebraic forms.