Problem 43 - Essential Calculus Early Transcendentals

Question

$$41-44=\( Find \)y^{\prime}\( and \)y^{\prime \prime}$$ $$y=x \ln x$$

Step-by-Step Solution

Answer

\( y' = \ln x + 1 \) and \( y'' = \frac{1}{x} \).

Step 1: Find the first derivative

The function given is \( y = x \ln x \). To find \( y' \), we need to use the product rule which states that \( (uv)' = u'v + uv' \). Here \( u = x \) and \( v = \ln x \). Thus, \( u' = 1 \) and \( v' = \frac{1}{x} \). Now, applying the product rule, we get: \[ y' = 1 \cdot \ln x + x \cdot \frac{1}{x} \]\[ y' = \ln x + 1 \]

Step 2: Simplify the first derivative

From the previous step, the first derivative is \( y' = \ln x + 1 \). This is already in the simplest form.

Step 3: Find the second derivative

Now, to find the second derivative \( y'' \), we differentiate \( y' = \ln x + 1 \) with respect to \( x \). The derivative of \( \ln x \) is \( \frac{1}{x} \) and the derivative of a constant (1) is 0. Thus:\[ y'' = \frac{1}{x} + 0 \]\[ y'' = \frac{1}{x} \]

Step 4: Simplify the second derivative

The second derivative \( y'' = \frac{1}{x} \) is already expressed in its simplest form.

Key Concepts

Product Rule

When it comes to calculus derivatives, understanding the product rule is essential. The product rule is a technique used to differentiate functions that are the product of two or more functions. Specifically, if you have a function defined as the product of two functions, say \( u(x) \) and \( v(x) \), then the derivative of the product, \( (uv)' \), can be calculated using the formula:

• \( (uv)' = u'v + uv' \)

In this scenario, \( u' \) and \( v' \) represent the derivatives of the functions \( u \) and \( v \) respectively. So, to apply the product rule:

• Differentiate \( u(x) \) to get \( u'(x) \)
• Differentiate \( v(x) \) to get \( v'(x) \)
• Multiply \( u'(x) \) by \( v(x) \) and \( u(x) \) by \( v'(x) \)
• Add the two results together

In the exercise, the function \( y = x \ln x \) makes use of the product rule. Here, we set \( u = x \) and \( v = \ln x \). The derivatives are straightforward: \( u' = 1 \) and \( v' = \frac{1}{x} \). Applying the formula gives us the first derivative.

First Derivative

The first derivative of a function gives us important information about the rate at which the function's value is changing at any point. It is often denoted as \( y' \) or \( f'(x) \) if the function is \( f(x) \).

In the exercise, after applying the product rule to the function \( y = x \ln x \), we found that the first derivative is given by

• \( y' = \ln x + 1 \)

This expression simplifies the rate of change of our original function. Each term in the derivative holds significance:

• \( \ln x \): shows the rate change corresponding to the logarithmic part
• \( + 1 \): a constant contribution to the rate of change

This first derivative can be used to analyze the increasing or decreasing nature of the function \( y \), and it's valuable for finding critical points, which leads us into exploring concavity with the second derivative.

Second Derivative

The second derivative, \( y'' \) or \( f''(x) \), provides insight into the curvature or concavity of the function. It tells us how the rate of change (first derivative) itself is changing. If the second derivative is positive, the function is concave up, indicating a local minimum. Conversely, if it is negative, the function is concave down, suggesting a local maximum.

In this particular exercise, we differentiate \( y' = \ln x + 1 \) to find the second derivative:

• The derivative of \( \ln x \) is \( \frac{1}{x} \)
• The derivative of the constant 1 is 0

Thus, the second derivative is

• \( y'' = \frac{1}{x} \)

This tells us that the curvature is dependent on the term \( \frac{1}{x} \). For \( x > 0 \), it means the curve is getting less steep as \( x \) increases, indicating that the function \( y \) is increasing at a decreasing rate. Understanding this helps in graphing functions and identifying points of inflection.