Problem 27 - Calculus Single Variable

Question

In each of Exercises \(27-49,\) differentiate the given expression with respect to \(x\). $$ \ln (2 x) $$

Step-by-Step Solution

Answer

The derivative of \( \ln(2x) \) with respect to \( x \) is \( \frac{1}{x} \).

Step 1: Understand the Problem

We need to differentiate the function \( \ln(2x) \) with respect to \( x \). The given expression is a natural logarithm.

Step 2: Apply the Chain Rule

The logarithmic function \( \ln(u) \) differentiates to \( \frac{1}{u} \cdot \frac{du}{dx} \), where \( u = 2x \) in this case. First, we'll differentiate \( u = 2x \) with respect to \( x \).

Step 3: Differentiate the Inner Function

The derivative of \( u = 2x \) with respect to \( x \) is \( \frac{du}{dx} = 2 \).

Step 4: Differentiating Using the Chain Rule

Now that we have \( \frac{du}{dx} \), differentiate \( \ln(2x) = \ln(u) \). According to the chain rule, the derivative is \( \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{2x} \cdot 2 \).

Step 5: Simplify the Expression

Simplify the derivative expression: \( \frac{1}{2x} \cdot 2 = \frac{2}{2x} = \frac{1}{x} \).

Key Concepts

Chain Rule in Calculus Differentiation

The chain rule is a fundamental tool in calculus used to differentiate complex functions. It comes into play when you have a function nested inside another function, often represented as a composite function. Imagine it as peeling an onion layer by layer; you work from the outermost layer inward. The chain rule states: if you have a function \( f(g(x)) \), the derivative is given by \( f'(g(x)) \, \cdot \, g'(x) \).

• Outer function: In our context, the outer function is the natural logarithm \( \ln(u) \).
• Inner function: The inner function for the exercise \( \ln(2x) \) is \( u = 2x \).

To use the chain rule, find the derivative of the outer function, \( \frac{1}{u} \), and multiply it by the derivative of the inner function, \( \frac{du}{dx} \). This process is how we tackle the differentiation of nested functions and simplifies working with more intricate expressions.

Understanding the Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a special type of logarithm with base \( e \) (approximately 2.71828). It has unique properties that make it particularly useful in calculus and real-world applications, especially for exponential growth or decay processes.

• Basic property: \( \ln(e^x) = x \), which reflects its base.
• Derivative of \( \ln(x) \): The derivative is \( \frac{1}{x} \). This means, for any function \( \ln(u) \), the derivative is \( \frac{1}{u} \), a key detail when applying the chain rule.

In the exercise, knowing that \( \ln(2x) \) implies taking the derivative of a logarithm function, we use this property to transform differentiation into a series of simpler steps, fitting the pattern of \( \ln(u) \) and applying the chain rule effectively.

Derivative Simplification Techniques

Simplifying derivatives is often the final touch to a differentiation problem. It makes the resulting expression more readable and easier to use in further calculations. After applying the chain rule or any other differentiation technique, simplifying the expression involves basic algebraic manipulation.

• In this problem, we initially found the derivative \( \frac{2}{2x} \).
• The expression \( \frac{2}{2x} \) simplifies directly to \( \frac{1}{x} \).

The simplification process typically involves cancelling and combining like terms or numbers. This step ensures the solution is presented in its simplest form, thus streamlining the application of the derivative in broader mathematical or applied contexts. Remember, students often find algebraic simplification as crucial as the differentiation itself.