Problem 9 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(f^{\prime}(x) .\) Do these problems without using the Quotient Rule. $$ f(x)=5 \ln \left(2 x^{2}+3 x\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=5 \ln \left(2 x^{2}+3 x\right)\) is \[f^{\prime}(x) = \frac{5 (4x+3)}{2x^2+3x}\].

Step 1: Identify Differentiation Method

Looking at the form of the function, \(f(x)=5 \ln \left(2 x^{2}+3 x\right)\), it can be observed that it is a composite function. The function inside the logarithm is \(g(x)=2 x^{2}+3 x\). So, the most applicable method for finding the derivative is the Chain Rule.

Step 2: Apply the Chain Rule

The Chain Rule states that the derivative of a composed function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function. Hence, the derivative of \(f(x)\) is:\[f^{\prime}(x) = 5 \cdot \frac{1}{(2x^2+3x)} \cdot (2x^2+3x)^{\prime}\].

Step 3: Compute the Derivative of the Inner Function

The inner function is \(2x^2+3x\), and its derivative is \(4x+3\). Replace \( (2x^2+3x)^{\prime}\) with \(4x + 3\) in the equation. So we get: \[f^{\prime}(x) = 5 \cdot \frac{1}{(2x^2+3x)} \cdot (4x+3)\].

Step 4: Simplify the Result

Now, the expression can be simplified further:\[f^{\prime}(x) = \frac{5 (4x+3)}{2x^2+3x}\].

Key Concepts

Composite Functions

Understanding composite functions is crucial when tackling differentiation problems. A composite function is essentially a function inside another function, often denoted as \( f(g(x)) \). To visualize this, imagine nesting dolls, where each doll is a function wrapped within another. In our exercise, we're dealing with the natural logarithm function taking another function as its input, specifically \( f(x) = 5 \ln(2x^2 + 3x) \). Here, the outer function is \(5\ln(x)\), and the inner function is \(g(x) = 2x^2 + 3x\).

When differentiating composite functions, we use the Chain Rule, which is a method for finding the derivative of a composite function. The importance of recognizing composite functions cannot be overstated as it ensures the correct application of differentiation rules and leads to accurate results.

Natural Logarithm Differentiation

The natural logarithm, denoted as \( \ln(x) \), plays a significant role in calculus due to its unique properties and relationship with exponential functions. Differentiating natural logarithms requires a specific rule: the derivative of \( \ln(u) \), where \( u \) is a function of \( x \), is \( \frac{1}{u} \cdot u' \).

By applying this rule to our exercise, the derivative of \( 5\ln(2x^2 + 3x) \), according to the Chain Rule, becomes \( 5 \cdot \frac{1}{2x^2+3x} \cdot (4x+3) \). This step is vital as it not only shows the differentiation of the logarithmic portion but also sets up the function for the subsequent derivative simplification by prepping the inner function's derivative.

Derivative Simplification

Simplification is the final touch in computing derivatives, turning complex expressions into more manageable forms. After using the Chain Rule and differentiating the natural logarithm, we arrived at \( f'(x) = 5 \cdot \frac{1}{(2x^2+3x)} \cdot (4x+3) \). Simplification helps by making it easier to evaluate the derivative at particular points or to integrate the function later on. For instance, in our exercise, we can simply multiply the constants and organize the terms to get a single fraction: \( f'(x) = \frac{20x + 15}{2x^2 + 3x} \).

At this point, you might notice the numerator can be factored out or the expression could be further simplified depending on the context, but even getting to this stage allows one to see the derivative in a clearer, more concise form. The ability to simplify correctly can be the difference between an expression that is usable and one that remains overly complex and challenging to work with.