Problem 17 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ g(p)=p \ln (2 p+1) $$

Step-by-Step Solution

Answer

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

Step 1: Differentiate using the Product Rule

The function \( g(p) = p \ln(2p + 1) \) is a product of two functions: \( p \) and \( \ln(2p + 1) \). To find the derivative \( g'(p) \), we apply the product rule: \((uv)' = u'v + uv'\), where \( u = p \) and \( v = \ln(2p + 1) \).

Step 2: Differentiate the First Function

Differentiate \( u = p \). The derivative \( u' = 1 \).

Step 3: Differentiate the Second Function Using the Chain Rule

To differentiate \( v = \ln(2p + 1) \), use the chain rule. The derivative of \( \ln(x) \) is \( \frac{1}{x} \), so \( v' = \frac{1}{2p + 1} \cdot \frac{d}{dp}(2p + 1) = \frac{2}{2p + 1} \).

Step 4: Apply the Product Rule

Substitute \( u, u', v, \) and \( v' \) into the product rule formula: \( g'(p) = u'v + uv' = 1 \cdot \ln(2p + 1) + p \cdot \frac{2}{2p + 1} \).

Step 5: Simplify the Derivative Expression

Simplify \( g'(p) = \ln(2p + 1) + \frac{2p}{2p + 1} \) to get the final expression for the derivative.

Key Concepts

Product Rule

In calculus, the product rule is a fundamental tool used to find the derivative of a product of two functions. The rule is articulated as follows: if you have two differentiable functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) can be found using the formula:

• \((uv)' = u'v + uv'\)

To understand this better, consider our example function \( g(p) = p \ln(2p + 1) \). Here, \( u = p \) and \( v = \ln(2p + 1) \). We need to find the derivative of \( g(p) \) using the product rule.

• First, differentiate \( u = p \) to get \( u' = 1 \).
• Then, differentiate \( v = \ln(2p + 1) \), which is handled using the chain rule (more on that soon!).
• Finally, combine these derivatives using the product rule formula: \( g'(p) = 1 \cdot \ln(2p + 1) + p \cdot v' \).

Chain Rule

The chain rule is another important technique in calculus for finding the derivative of a composite function. In simple terms, it states that if you have a function \( y = f(g(x)) \), then the derivative \( dy/dx \) is found by multiplying the derivative of the outer function \( f \) with respect to the inner function \( g \), by the derivative of the inner function \( g \) with respect to \( x \). The formula is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In our problem, when differentiating \( v = \ln(2p + 1) \), we notice it's a composite function.

• The outer function is \( \ln(x) \) and the inner function is \( 2p + 1 \).
• Applying the chain rule, the derivative of \( \ln(x) \) is \( 1/x \), so we have \( v' = \frac{1}{2p + 1} \cdot \frac{d}{dp}(2p + 1) = \frac{2}{2p + 1} \).

Seamlessly, the chain rule fits with the product rule to derive the complete derivative of complex functions like this one.

Logarithmic Differentiation

Logarithmic differentiation is a powerful method in calculus, especially useful when differentiating functions involving products, quotients, and powers. The technique leverages the properties of logarithms to simplify complex derivatives.
However, in this particular problem, while we are dealing with a logarithmic expression \( \ln(2p + 1) \), we primarily apply the chain and product rules. Logarithmic differentiation shines more brightly in cases like finding the derivative of functions of the form \( f(x) = x^x \) or products with several terms where direct application of rules would be cumbersome.

• Using logarithmic properties can transform multiplication into addition and division into subtraction, simplifying the differentiation process.
• Understanding when and how to apply logarithmic differentiation can save time and effort in finding derivatives of complex expressions.

Even though not directly used here, knowing about logarithmic differentiation provides a broader strategy toolbox for tackling diverse calculus problems.