Problem 296 - Calculus Volume 1

Question

For the following exercises, find the derivative \(\frac{d y}{d x}\) \(y=\ln (2 x+1)\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2}{2x + 1} \)

Step 1: Identify the Function

We are given the function \( y = \ln(2x + 1) \). Our goal is to find \( \frac{d y}{d x} \), the derivative of \( y \) with respect to \( x \).

Step 2: Use the Chain Rule

To find the derivative of \( y = \ln(2x + 1) \), we apply the chain rule. The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \), and we multiply by \( \frac{d u}{d x} \) given \( u = 2x + 1 \).

Step 3: Derivative of the Inner Function

In \( y = \ln(2x + 1) \), the inner function \( u = 2x + 1 \) has the derivative \( \frac{d u}{d x} = 2 \).

Step 4: Apply the Chain Rule

Using the chain rule, the derivative \( \frac{d y}{d x} \) is \( \frac{1}{2x + 1} \times 2 \).

Step 5: Simplify the Expression

Simplify the expression \( \frac{2}{2x + 1} \). This is the final form of the derivative.

Key Concepts

Derivative Calculation

In calculus, the calculation of derivatives is key to understanding how functions change. The derivative of a function represents the rate of change or the slope of the function at any point. For a given function such as \( y = \ln(2x + 1) \), our main task is to find \( \frac{dy}{dx} \), which tells how \( y \) changes as \( x \) changes.

We start by identifying the function and differentiating it step by step. Remember, the goal is to apply rules of differentiation to each component of the function. This approach ensures that we correctly capture how each part of a composite function contributes to the overall rate of change.

Understanding derivatives will empower you to solve various real-world problems where change is continuous.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. A composite function is formed when one function is nested inside another. In our problem \( y = \ln(2x + 1) \), \( 2x + 1 \) is inside the natural logarithm function. To differentiate such functions, we use the chain rule.

The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Here's how you apply it:

• First, differentiate the outer function using the inner function as the variable.
• Next, multiply by the derivative of the inner function itself.

This step-by-step application allows us to handle complex functions in a manageable way. In our example, we start with the derivative of the outer function \( \ln(u) \), which is \( \frac{1}{u} \), and multiply by the derivative of the inner function \( u = 2x + 1 \), which is \( 2 \). This gives \( \frac{1}{2x + 1} \times 2 \), simplified to \( \frac{2}{2x + 1} \).

Using the chain rule effectively can simplify even the most intricate functions.

Natural Logarithm Differentiation

Differentiating natural logarithms is a common task in calculus with its straightforward rule: the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). This makes it a linear characteristic when approaching problems involving logarithmic functions.

The natural logarithm \( \ln(x) \) emerges frequently in mathematics due to its properties and relationship with exponential functions. When differentiating \( \ln(2x + 1) \), consider the inner function \( u = 2x + 1 \). The differentiation process involves:

• Finding the derivative of \( \ln(u) \), which is \( \frac{1}{u} \).
• Multiplying this by the derivative of \( u \), if \( u \) itself is a function of \( x \).

In our example, \( u \) has a simple derivative, which is 2, leading us to the derivative expression \( \frac{2}{2x + 1} \).

This simplicity is what makes logarithmic differentiation a powerful technique to integrate into your calculus toolkit.