Problem 14 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ g(r)=r^{2} \ln (2 r+1) $$

Step-by-Step Solution

Answer

The derivative is \( g'(r) = 2r \ln(2r + 1) + \frac{2r^2}{2r + 1} \).

Step 1: Identify the Differentiation Rule

To differentiate the function \( g(r) = r^2 \ln(2r + 1) \), we need to use the product rule, since it is a product of two functions: \( u = r^2 \) and \( v = \ln(2r + 1) \). Recall that the product rule for differentiation is \( (uv)' = u'v + uv' \).

Step 2: Differentiate Each Function

Differentiate \( u = r^2 \) with respect to \( r \). This gives \( u' = 2r \).Next, differentiate \( v = \ln(2r + 1) \). To differentiate this, use the chain rule: \( v'(r) = \frac{1}{2r + 1} \cdot \frac{d}{dr}(2r + 1) = \frac{2}{2r + 1} \).

Step 3: Apply the Product Rule

Using the results from the previous steps, apply the product rule:\[ g'(r) = u'v + uv' = (2r) \ln(2r + 1) + r^2 \left(\frac{2}{2r + 1}\right) \].

Step 4: Simplify the Expression

First, write the expression from the product rule:\[ g'(r) = 2r \ln(2r + 1) + \frac{2r^2}{2r + 1} \].You can simplify the second term further if desired, but \( g'(r) = 2r \ln(2r + 1) + \frac{2r^2}{2r + 1} \) is already in a simplified form.

Key Concepts

Product Rule Differentiation

Differentiating functions often involves dealing with products of two or more components. This is where the product rule becomes essential. It's a key tool in calculus used to differentiate functions that are multiplied together. Suppose you have two functions, say \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product can be found using the formula:

• \( (uv)' = u'v + uv' \)

This means you first differentiate \( u \) while keeping \( v \) constant, and then, differentiate \( v \) while keeping \( u \) constant. In the provided exercise, \( u(r) = r^2 \) and \( v(r) = \ln(2r+1) \). Each function needs to be differentiated separately and placed into the product rule formula.
Remember, it's the summation of the product of the derivative of one function with the original of the other.
Always ensure to apply the rule correctly by maintaining respect for the order of operations during calculations.

Chain Rule Application

The chain rule is an essential differentiation technique used when differentiating composite functions. A composite function is a function inside another function, like \( f(g(x)) \). The chain rule states:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

This implies you first differentiate the outer function while keeping the inner function unchanged, and then multiply it by the derivative of the inner function. In our context, \( v'(r) = \ln(2r+1) \), required the chain rule.
Consider \( 2r+1 \) in the natural log function as the inner function, and ln() as the outer function. The derivative of ln(x) is \( 1/x \), applied to \( 2r+1 \) gives \( \frac{1}{2r+1} \), but the chain rule directs us to further multiply it by the derivative of \( 2r+1 \), which is 2.
This results in \( \frac{2}{2r+1} \). The chain rule helps navigate through such complicated layers, making cumbersome differentiations manageable.

Differentiation Techniques Overview

Differentiation is at the heart of calculus and involves various techniques that cater to different function types. These techniques provide a framework to find derivatives systematically. Here you'll find a summarized overview of some techniques you might encounter:

• Power Rule: Used for functions of the form \( x^n \). The derivative is \( nx^{n-1} \).
• Product Rule: Applied when differentiating products of functions, as discussed. \((uv)' = u'v + uv'\) is the method to remember.
• Quotient Rule: Used for dividing one function by another, stating \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
• Chain Rule: For nested functions. Essential for differentiating composite functions effectively.
• Exponential and Logarithmic Differentiation: Specific rules cater to these, such as differentiating \( a^x \) or \( ln(x) \).

Each technique is suited to particular function structures. Familiarity with them enhances your problem-solving skills and makes complex calculus problems more approachable.
Practice differentiating with these techniques reinforces understanding and sharpens analytical thinking in solving calculus problems.