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When dealing with functions in calculus, you often encounter scenarios where you need to differentiate the product of two functions. This is precisely where the product rule comes into play.

The product rule states that to differentiate a product of two functions, say, h(x) that is equal to f(x) multiplied by g(x), the derivative h'(x) is the sum of f(x) times the derivative of g(x) and g(x) times the derivative of f(x). In mathematical terms, that is:

\[h'(x) = f'(x)g(x) + f(x)g'(x)\].

Applying this rule is crucial in finding the derivative of complex expressions like the one in our example, where we have a logarithmic function of x multiplied by x itself. By understanding and accurately applying the product rule, we can seamlessly determine the derivative of the function and consequently find the critical points necessary for further analysis of the function's behavior.

Logarithmic functions can be intimidating at first glance, but their derivatives follow a consistent rule that, once mastered, simplifies the process of finding rates of change and slopes of tangent lines for these types of functions.

The general formula for the derivative of the natural logarithm function \(\ln(x)\) is: \[\frac{d}{dx}\ln(x) = \frac{1}{x}\].

When applying this to our original function \(f(x) = x \ln x\), it's part of what makes the product rule so useful. By incorporating the derivative of the logarithmic function into the product rule's formula, the resulting expression helps us further understand how \(f(x)\) behaves, and the process seamlessly connects to finding critical points that are often associated with local maxima and minima.

After finding the critical points of a function, the next question often involves classification—are these points local maxima, local minima, or points of inflection? Here, the second derivative test offers a straightforward method for this classification.

The second derivative test states that if the second derivative of a function at a critical point is positive, \(f''(x) > 0\), the function has a local minimum at that point. If the second derivative is negative, \(f''(x) < 0\), there's a local maximum. And if the second derivative is zero, the test is inconclusive.

In our exercise, the second derivative of \(f(x)\) at \(x = \frac{1}{e}\) is positive since \(f''(\frac{1}{e}) = e\), which is greater than zero. Therefore, according to the second derivative test, \(x = \frac{1}{e}\) is a local minimum. This result highlights the utility of the second derivative test in confirming what the graph of \(f(x)\) is doing at critical points.