Problem 19 - Calculus Single Variable

Question

Calculate the derivative with respect to \(x\) of the given expression. \(1 /(4+\ln (x))\)

Step-by-Step Solution

Answer

The derivative is \( -\frac{1}{x(4 + \ln(x))^2} \).

Step 1: Identify the function to differentiate

The given function is \( f(x) = \frac{1}{4 + \ln(x)} \). This function is a composite function where the inner function is \( 4 + \ln(x) \) and the outer function is the reciprocal function, \( \frac{1}{u} \).

Step 2: Apply the chain rule

To differentiate a composite function like this, we will use the chain rule. The chain rule states that the derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). For our function, let \( u = 4 + \ln(x) \). Then, \( f(u) = \frac{1}{u} \).

Step 3: Differentiate the outer function

Differentiate the outer function with respect to \( u \). The derivative of \( \frac{1}{u} \) with respect to \( u \) is \( -\frac{1}{u^2} \).

Step 4: Differentiate the inner function

Differentiate the inner function \( u = 4 + \ln(x) \) with respect to \( x \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), so \( u' = \frac{1}{x} \).

Step 5: Apply the chain rule

Combine the results from previous steps using the chain rule. Substitute \( f'(u) = -\frac{1}{u^2} \) and \( u' = \frac{1}{x} \) to find \( \frac{d}{dx} \left( \frac{1}{4+\ln(x)} \right) \). This gives you \( -\frac{1}{(4 + \ln(x))^2} \cdot \frac{1}{x} \).

Step 6: Simplify the result

The derivative of the original function is \( -\frac{1}{x(4 + \ln(x))^2} \).

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. It gives us a way to understand how a small change in input affects the output. In more simple terms, it's like knowing how fast your car is going at a particular moment.
When we calculate the derivative of the function given, we are essentially trying to find out how the function's value changes as we make small changes to the variable, here represented as \(x\).
The notation for a derivative is often shown as \(\frac{dy}{dx}\), where \(y\) is the function and \(x\) is the variable. This notation tells us how \(y\) changes with respect to \(x\).

• The value of the derivative can highlight moments when a function is increasing or decreasing.
• If the derivative is positive, the function is increasing.
• A negative derivative indicates a decreasing function.

Derivatives are foundational in finding other important features of a graph, like the slope of the tangent line at any point, and are widely used in optimization problems.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It's like peeling an onion - you deal with one layer at a time! Whenever you have one function inside another, you apply the chain rule.
For our exercise, the function \( f(x) = \frac{1}{4 + \ln(x)} \) is a composite function with an outer layer \( \frac{1}{u} \) and an inner layer \( u = 4 + \ln(x) \).
Here's how the chain rule works:

• First, differentiate the outer function with respect to the inner function.
• Then, differentiate the inner function with respect to \(x\).
• Finally, multiply the two derivatives together.

This process lets us seamlessly combine rates of changes from nested functions. In formula terms, for a function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).
This way, the chain rule efficiently tackles the complexity of layered functions, ensuring a smooth derivation process.

Composite Function

A composite function involves nesting one function inside another. Imagine a scenario where while dressing, you decide to wear a jacket over a shirt. Here, putting on a shirt is a function, and adding a jacket is another one. Composite functions work the same way!
In the exercise provided, \( f(x) = \frac{1}{4 + \ln(x)} \) is composed of two functions:

• The inner function is \( g(x) = 4 + \ln(x) \)
• The outer function is \( h(u) = \frac{1}{u} \)

When \( g(x) \) is plugged into \( h(u) \), we get the composite function \( h(g(x)) \).
Understanding composite functions is crucial because it helps us break down complicated expressions into manageable parts. This simplifies calculations, making it easier to analyze and differentiate the function.