Derivative Of Ln(2x/(x+1)) A Step-by-Step Guide

Finding the derivative of a function is a fundamental concept in calculus, and it allows us to understand the rate at which a function's output changes with respect to its input. In this article, we will delve into the process of finding the derivative of the function $f(x) = \ln\left(\frac{2x}{x+1}\right)$. This function involves a natural logarithm and a rational expression, making it an excellent example to illustrate the application of various differentiation rules. We'll explore these rules step-by-step, ensuring a clear and comprehensive understanding of the solution. This exploration will not only provide the answer to the specific derivative but also reinforce the broader concepts and techniques essential for calculus.

Understanding the Function

Before we dive into the differentiation process, let's take a moment to understand the function $f(x) = \ln\left(\frac{2x}{x+1}\right)$. This function is a composite function, which means it's a function within a function. Specifically, it's the natural logarithm function applied to the rational function $\frac{2x}{x+1}$. The natural logarithm, denoted as $\ln$, is the logarithm to the base e, where e is an irrational number approximately equal to 2.71828. Logarithmic functions are particularly important in calculus due to their unique properties and applications in various fields, such as physics, engineering, and economics. Understanding the domain of the function is crucial as logarithms are only defined for positive arguments. Therefore, $\frac{2x}{x+1}$ must be greater than zero. This condition will influence the steps we take and the potential simplification we can apply.

Differentiation Rules

To find the derivative of $f(x)$, we'll primarily use two important rules: the chain rule and the quotient rule. The chain rule is used when differentiating composite functions, and it states that the derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$. In our case, $f$ is the natural logarithm and $g(x)$ is the rational function $\frac{2x}{x+1}$. The quotient rule is used to differentiate functions that are the ratio of two other functions. It states that if $h(x) = \frac{u(x)}{v(x)}$, then $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$. These rules are the bedrock of differential calculus and understanding how to apply them correctly is essential for mastering the subject. Now, let's apply these rules to our function.

Applying the Chain Rule

The first step in finding the derivative of $f(x) = \ln\left(\frac{2x}{x+1}\right)$ is to apply the chain rule. The chain rule, as mentioned earlier, is crucial for differentiating composite functions. In this case, we have the natural logarithm function as the outer function and the rational function $\frac{2x}{x+1}$ as the inner function. The derivative of the natural logarithm function, $\ln(u)$, with respect to $u$ is $\frac{1}{u}$. Therefore, the first part of applying the chain rule involves finding the derivative of the outer function, which is $\frac{1}{\frac{2x}{x+1}}$. This simplifies to $\frac{x+1}{2x}$. Next, we need to find the derivative of the inner function, $\frac{2x}{x+1}$, which requires the application of the quotient rule. This step is critical as it combines two fundamental rules of differentiation, the chain rule and the quotient rule, to solve a more complex problem.

Applying the Quotient Rule

Now, let's apply the quotient rule to find the derivative of the inner function, $\frac{2x}{x+1}$. The quotient rule states that if we have a function $h(x) = \frac{u(x)}{v(x)}$, its derivative $h'(x)$ is given by $\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$. In our case, $u(x) = 2x$ and $v(x) = x+1$. The derivative of $u(x)$, $u'(x)$, is 2, and the derivative of $v(x)$, $v'(x)$, is 1. Plugging these values into the quotient rule formula, we get: $\frac{2(x+1) - 2x(1)}{(x+1)^2}$. This expression simplifies to $\frac{2x + 2 - 2x}{(x+1)^2}$, which further simplifies to $\frac{2}{(x+1)^2}$. This result is the derivative of the inner function, and it's a key component in completing the chain rule application.

Combining the Results

Now that we have the derivative of the outer function (from the chain rule) and the derivative of the inner function (from the quotient rule), we can combine these results. The chain rule dictates that we multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function. In our case, this means multiplying $\frac{x+1}{2x}$ by $\frac{2}{(x+1)^2}$. This gives us: $\frac{x+1}{2x} \cdot \frac{2}{(x+1)^2}$. Simplifying this expression involves canceling out common factors. We can cancel a factor of 2 and a factor of $(x+1)$, leading to a simplified expression. This simplification is crucial as it reduces the complexity of the derivative, making it easier to analyze and use in further calculations.

Simplifying the Derivative

Let's simplify the expression we obtained after applying the chain and quotient rules: $\frac{x+1}{2x} \cdot \frac{2}{(x+1)^2}$. As we identified earlier, we can cancel out the common factor of 2. Additionally, we can cancel one factor of $(x+1)$ from the numerator and the denominator. This leaves us with: $\frac{1}{x} \cdot \frac{1}{x+1}$. Multiplying these two fractions, we get: $\frac{1}{x(x+1)}$. This is a simplified form of the derivative, but we can further expand the denominator to obtain: $\frac{1}{x^2 + x}$. This final form of the derivative is often preferred as it clearly shows the relationship between the derivative and the original function. This simplified form is not only easier to work with but also provides more insight into the behavior of the function's rate of change.

Final Answer

After applying the chain rule, the quotient rule, and simplifying the resulting expression, we have found the derivative of $f(x) = \ln\left(\frac{2x}{x+1}\right)$. The derivative, denoted as $f'(x)$, is: $f'(x) = \frac{1}{x(x+1)} = \frac{1}{x^2 + x}$ This final answer represents the instantaneous rate of change of the function $f(x)$ with respect to $x$. Understanding the derivative allows us to analyze the function's increasing and decreasing intervals, find critical points, and understand its concavity. This result is not just a numerical answer; it's a powerful tool for understanding the behavior of the original function.

In conclusion, finding the derivative of $f(x) = \ln\left(\frac{2x}{x+1}\right)$ involved a multi-step process that included the application of the chain rule and the quotient rule, followed by simplification. This process highlights the interconnectedness of different differentiation rules and the importance of algebraic manipulation in calculus. The final derivative, $f'(x) = \frac{1}{x(x+1)}$, provides valuable information about the rate of change of the original function, making it a crucial result for further analysis and applications.