Problem 53 - Calculus Single Variable

Question

Let \(A\) and \(B\) be constants. Calculate the derivative of \(A \cdot x \ln (x)+B \cdot x\) with respect to \(x\). Show that there are values of \(A\) and \(B\) such that $$ \int \ln (x) d x=A \cdot x \ln (x)+B \cdot x+C $$ where \(C\) is an arbitrary constant.

Step-by-Step Solution

Answer

The derivative is \( A \ln(x) + A + B \). For \( A = 1 \) and \( B = -1 \), the integral holds: \( \int \ln(x) dx = x \ln(x) - x + C \).

Step 1: Differentiation of Each Term

To find the derivative of the function \( f(x) = A \cdot x \ln(x) + B \cdot x \), we differentiate each term separately with respect to \( x \).

Step 2: Differentiate the First Term

The first term is \( A \cdot x \ln(x) \). Use the product rule, which states that if \( u = x \) and \( v = \ln(x) \), then \( \frac{d}{dx}[uv] = u'v + uv' \). Here, \( u' = 1 \) and \( v' = \frac{1}{x} \). So,\[\frac{d}{dx}[A \cdot x \ln(x)] = A(1 \cdot \ln(x) + x \cdot \frac{1}{x}) = A(\ln(x) + 1)\].

Step 3: Differentiate the Second Term

Now, consider the second term, \( B \cdot x \). Its derivative with respect to \( x \) is simply \( B \), because \( B \) is a constant. Thus:\[\frac{d}{dx}[B \cdot x] = B\].

Step 4: Add the Derivatives Together

Combine the results from the previous steps to find \( f'(x) \):\[f'(x) = A(\ln(x) + 1) + B = A \ln(x) + A + B\].

Step 5: Integration to Find Constants

The problem states that the integral \( \int \ln(x) \, dx \) can be expressed as \( A \cdot x \ln(x) + B \cdot x + C \). By integrating \( \ln(x) \) using integration by parts with \( u = \ln(x) \), \( dv = dx \), \( du = \frac{1}{x}dx \), and \( v = x \), we get:\[\int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - x + C\].Matching terms gives \( A = 1 \) and \( B = -1 \).

Step 6: Verify

Plug \( A = 1 \) and \( B = -1 \) into the derived formula and verify:\[\int \ln(x) \, dx = x \ln(x) - x + C\]is consistent with the expression \( 1 \cdot x \ln(x) - 1 \cdot x + C \) from the problem statement.

Key Concepts

Product Rule

The product rule is a fundamental technique used in calculus to find the derivative of the product of two functions. When you have two functions multiplied together, like in the term \(A \cdot x \ln(x)\), you can't just differentiate each factor alone. Instead, the product rule comes into play. It states: If \(u(x)\) and \(v(x)\) are functions of \(x\), then the derivative of their product \(u(x) \cdot v(x)\) is \(u'v + uv'\).
The notation \(u'\) and \(v'\) represent the derivatives of \(u\) and \(v\) with respect to \(x\). In our example, where \(u = x\) and \(v = \ln(x)\), we find \(u' = 1\) and \(v' = \frac{1}{x}\). Applying the product rule correctly gives us the derivative of the first term as \(A(\ln(x) + 1)\).
The product rule is pivotal for any differentiation task involving multiplication of functions, ensuring accuracy in derivative calculation.

Integration by Parts

Integration by parts is a powerful method in integration similar to the product rule in differentiation. It's based on the integration of products of functions and is used when a direct integration is complex. The integration by parts formula is derived from the product rule:

\[\int u \, dv = uv - \int v \, du\]
This comes in handy for integrals like \(\int \ln(x) \, dx\). By selecting \(u = \ln(x)\) and \(dv = dx\), we set up the integration by parts framework. Differentiating and integrating these parts gives \(du = \frac{1}{x}dx\) and \(v = x\).
Substituting into the formula, we have \(uv - \int v \, du = x \ln(x) - \int x \cdot \frac{1}{x} \, dx\), which simplifies to \(x \ln(x) - x\). This process unveils the structure for finding constants as shown in the exercise, linking integration and differentiation smoothly.

Constant Multiplication

Constant multiplication refers to finding derivatives or integrals when constants multiply functions. Constants do not change during differentiation or integration, which simplifies calculations significantly. For a function like \(B \cdot x\), where \(B\) is a constant, the differentiation path is straightforward.
The derivative of \(B \cdot x\) with respect to \(x\) is simply \(B\). Direct differentiation over constants follows the rule: the constant remains and the derivative of \(x\) is 1. This highlights why knowing constant multiplication's role is vital in calculus, helping to easily manage terms involving constants.
When integrating, constants simply multiply the integral of the function. If \(B\) was involved in an integral, it would simply "factor out," making integral evaluation more direct.

Differentiation

Differentiation involves finding how a function changes at any point along its curve. It helps us understand the rate of change or the slope of the function. Calculus defines this through derivatives, integral in comprehending behavior of functions.
In this problem, we differentiated \(f(x) = A \cdot x \ln(x) + B \cdot x\). Separating the function into terms, we apply both the product rule and constant rules effectively, leading us to \(f'(x) = A(\ln(x) + 1) + B\).
Understanding differentiation fully aids in grasping the dynamic features of functions like linearity and curvature, also reinforcing fundamental calculus strategies.

Integral Evaluation

Integral evaluation is the process of finding the integral of a function, representing the accumulated sum or area under a curve. In this exercise, we evaluate \(\int \ln(x) \, dx\) using integration by parts.
The integral ultimately becomes \(x \ln(x) - x + C\), revealing the values of \(A\) and \(B\) as 1 and -1, respectively, making sense of our derived expression from the integration task.
Integral evaluation is crucial in physics, engineering, and economics, where understanding accumulation is key. Through problem-solving, integrating functions helps solidify calculus applications, joining with differentiation knowledge to complete the calculus toolkit.