Problem 35 - Brief Applied Calculus

Question

Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(x)=\frac{3 x+1}{2+x} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{5}{(2+x)^2} \).

Step 1: Identify the functions (u and v)

For the function \( f(x) = \frac{3x+1}{2+x} \), we assign the numerator \( u(x) = 3x + 1 \) and the denominator \( v(x) = 2 + x \).

Step 2: Differentiate u(x)

Find the derivative of the numerator function: \( u'(x) = \frac{d}{dx}(3x + 1) = 3 \).

Step 3: Differentiate v(x)

Find the derivative of the denominator function: \( v'(x) = \frac{d}{dx}(2 + x) = 1 \).

Step 4: Apply the Quotient Rule

The quotient rule states that \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the formula: \( \left( \frac{3x+1}{2+x} \right)' = \frac{(3)(2+x) - (3x+1)(1)}{(2+x)^2} \).

Step 5: Simplify the derivative expression

Expand and combine like terms in the numerator: \( 6 + 3x - 3x - 1 = 5 \). Now the expression is \( \frac{5}{(2+x)^2} \).

Step 6: Final Simplified Derivative

Thus, the derivative of \( f(x) = \frac{3x+1}{2+x} \) is \( f'(x) = \frac{5}{(2+x)^2} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. It is represented as the derivative of the function. This concept is crucial for understanding how variables interact with each other. To differentiate a function means to calculate its derivative, which provides insights into the function's behavior, such as its slope and rate of change.

When dealing with differentiation, one of the main tools is applying the rules of differentiation, such as the power rule, product rule, and in this case, the quotient rule. Each rule helps simplify the process of finding the derivative under different circumstances, like when dealing with polynomial terms, multiplying functions, or dividing functions. These rules ensure efficiency and accuracy, making it easier to obtain derivatives for various types of functions.

In our exercise, the key rule applied is the quotient rule because our function is a division between two expressions, specifically a numerator and a denominator.

Derivative Calculation

Calculating derivatives involves following specific mathematical techniques to find how a function behaves as its input changes. With the example function provided, \( f(x) = \frac{3x+1}{2+x} \), the task is to determine the derivative using the quotient rule.

The first step is to identify the components of the function: the numerator \( u(x) = 3x+1 \) and the denominator \( v(x) = 2+x \). Next, we compute the derivatives of these parts:

• The derivative of \( u(x) \), noted as \( u'(x) \), is computed as \( 3 \).
• Similarly, the derivative of \( v(x) \), noted as \( v'(x) \), is \( 1 \).

Having these results, the quotient rule can be applied, which is
\(\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). By substituting the values, the expression becomes:
\(\frac{(3)(2+x) - (3x+1)(1)}{(2+x)^2} \).

The quotient rule's main advantage is handling fractions of two differentiable functions, which simplifies the complexity of the derivative calculation.

Simplifying Expressions

Once a derivative expression is calculated, it's often necessary to simplify it to make it more understandable or easier to use in further calculations. Simplifying involves rewriting the expression in its most basic form without changing its value.

In our example, after applying the quotient rule, the expression becomes:
\(\frac{(3)(2+x) - (3x+1)}{(2+x)^2} \). Simplification then follows by distributing and combining like terms in the expression, resulting in:

• First, expand the terms in the numerator: \( 6 + 3x \) and \( -3x - 1 \).
• Combine these to simplify to \( 5 \).

This results in the simplified final derivative expression of \( f'(x) = \frac{5}{(2+x)^2} \).

Simplifying helps clarify the result and potentially reveals additional properties of the function, making it easier to interpret and use in other mathematical analyses or applications.