91Ó°ÊÓ

In calculus, the first derivative of a function gives us the rate at which the function values are changing. For the given function, \( y = x \ln x + 3x - 2 \), we apply differentiation to find the first derivative. We start by using the product rule, which is essential when a term involves a product of two functions, like \( x \ln x \). The product rule states that if \( f(x) \) and \( g(x) \) are functions, their derivative \(( f \cdot g )' = f' \cdot g + f \cdot g' \). Applying this, we find that the derivative of \( x \ln x \) is \( \ln x + 1 \). For the other terms, the derivative of \( 3x \) is simply \( 3 \), and \( -2 \) is a constant whose derivative is \( 0 \). Combining these results, the first derivative of our function is: \( y' = \ln x + 4 \).

The second derivative provides information on how the rate of change itself is changing. It addresses the concavity or convexity of a function's graph. For the first derivative \( y' = \ln x + 4 \), we find the second derivative by differentiating it again. The derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of a constant \( 4 \) is \( 0 \). Therefore, our second derivative is \( y'' = \frac{1}{x} \). Understanding the second derivative can help in assessing acceleration in physics, or concavity in curve sketching in calculus.

Verifying a solution to a differential equation involves substituting the proposed solution back into the original equation and ensuring both sides are equal. For our equation \( y'' - \frac{1}{x} = 0 \), we substitute \( y'' = \frac{1}{x} \). Doing so reorganizes the equation to \( \frac{1}{x} - \frac{1}{x} = 0 \), which simplifies to \( 0 = 0 \). Since this holds true, the proposed function \( y = x \ln x + 3x - 2 \) is indeed a valid solution to the given differential equation. Verification is crucial in confirming that a derived solution is correct and satisfies the conditions of the original problem.

The product rule is a fundamental principle in differentiation. It allows us to find the derivative of a product of two functions. Mathematically, it is represented as \( ( f \cdot g )' = f' \cdot g + f \cdot g' \). This rule is used specifically when a term in the function is a multiplication of two distinct functions. In our worked example, for \( y = x \ln x + 3x - 2 \), the term \( x \ln x \) necessitated the use of the product rule. By applying it, we determined \( ( x \ln x )' = \ln x + 1 \). Understanding this rule is essential for students as it frequently occurs in calculus problems, making the differentiation of product functions straightforward and systematic.