Problem 28 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\frac{x \ln x}{1+\ln x}$$

Step-by-Step Solution

Answer

The derivative is \(y' = \frac{1 + 2\ln x + \ln^2 x}{(1 + \ln x)^2}\)."

Step 1: Identify the Function Structure

Identify the structure of the function. The given function is \(y = \frac{x \ln x}{1 + \ln x}\). This is a quotient of two functions of \(x\).

Step 2: Recall the Quotient Rule

The Quotient Rule states that the derivative of \(\frac{u}{v}\) is \(\frac{u'v - uv'}{v^2}\), where \(u\) and \(v\) are functions of \(x\). Identify \(u(x) = x \ln x\) and \(v(x) = 1 + \ln x\).

Step 3: Differentiate the Numerator

Find the derivative of the numerator \(u(x) = x \ln x\). Use the product rule: \(u' = (x)'(\ln x) + (x)((\ln x)') = \ln x + 1\).

Step 4: Differentiate the Denominator

Find the derivative of the denominator \(v(x) = 1 + \ln x\). The derivative is \(v' = 0 + \frac{1}{x} = \frac{1}{x}\).

Step 5: Apply the Quotient Rule

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the Quotient Rule formula: \(y' = \frac{(\ln x + 1)(1 + \ln x) - \left(x \ln x \right)\left(\frac{1}{x}\right)}{(1 + \ln x)^2}\).

Step 6: Simplify the Expression

Simplify the expression for the derivative: \(y' = \frac{\ln x + \ln^2 x + 1 + \ln x - \ln x}{(1 + \ln x)^2} = \frac{1 + 2\ln x + \ln^2 x}{(1 + \ln x)^2}\).

Key Concepts

Understanding Derivatives

In calculus, a derivative represents the rate at which a function changes. It gives the slope of the function at any given point. When you differentiate a function, you are essentially finding a formula that tells you the gradient of the function as it changes with respect to a variable, such as x.

Here's how to think about it:

• The derivative of a function at a point is the slope of the tangent line to the function's graph at that point.
• If the derivative is positive, the function is increasing.
• If the derivative is negative, the function is decreasing.

For the function given in the exercise, you need to determine how it changes as x varies. Calculating the derivative will tell us how y changes in relation to x.

Applying the Quotient Rule

The quotient rule is a technique used to find the derivative of a function that is the division of two other functions. For a function in the form of \( \frac{u(x)}{v(x)} \), the quotient rule provides a method to differentiate it.

The rule is as follows:

• If \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \).
• Here,\( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \) with respect to x.
• It's essentially saying "the derivative of the top function times the bottom, minus the derivative of the bottom function times the top, divided by the bottom function squared."

In our exercise, \( u(x) = x \ln x \) and \( v(x) = 1 + \ln x \). By applying the quotient rule, you can find the derivative of the entire function as shown in the solution steps.

Using the Product Rule

In the numerator of the function \( y = \frac{x \ln x}{1 + \ln x} \), we encounter a scenario where we must use the product rule. The product rule is used when differentiating a product of two functions.

Here's the rule:

• If \( z = f(x) \cdot g(x) \), then \( z' = f'(x)g(x) + f(x)g'(x) \).
• This means you take the derivative of the first function and multiply by the second function, plus the first function times the derivative of the second function.

For \( u(x) = x \ln x \), apply the product rule to find: \( u' = (x)'(\ln x) + (x)((\ln x)') = \ln x + 1 \). Knowing this allows us to take another step in differentiating the entire function correctly.

Grasping Differentiation

Differentiation is the process of finding a derivative, which is key in understanding how functions change. It helps in comprehending how variables relate, particularly in calculus.

Key points about differentiation:

• Differentiation applies to virtually any function you come across.
• It is a fundamental tool in calculus for finding rates of change.
• Through differentiation, you obtain a new function, the derivative, that provides insights into the behavior of the original function.

In the given exercise, differentiation was used extensively to navigate through more complex functions involving both the product and quotient rule. The goal is to simplify and find how y changes concerning x effectively.