91Ó°ÊÓ

When you encounter a function that is the product of two other functions, like \( x \ln x \), you'll want to employ the product rule to compute its derivative. The formula for the product rule is \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). This rule essentially tells us to take the derivative of the first function and multiply it by the second, then add the first function multiplied by the derivative of the second.
Here's how it applies: Suppose \( u(x) = x \) and \( v(x) = \ln x \). The derivative of \( u(x) = x \) is \( u'(x) = 1 \). The derivative of \( v(x) = \ln x \) is \( v'(x) = \frac{1}{x} \). Using the product rule gives us the derivative as \( 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \).
This method allows you to handle more complex functions that aren't just simple algebraic ones, by breaking them into components that are easier to work with.

Logarithmic differentiation is a technique that can be used when you're faced with a function involving logarithms. It simplifies the process of differentiation, particularly when the function involves a product or quotient of several expressions.
Here's a simple trick: if \( y = \ln x \), then the derivative \( \frac{dy}{dx} = \frac{1}{x} \). In our exercise, this principle helps us differentiate \( x \ln x \), as it's a product of \( x \) and \( \ln x \). So, when you know the derivative of \( \ln x \) is \( \frac{1}{x} \), you can easily substitute this value wherever needed.
While the problem at hand didn’t require other forms of logarithmic differentiation, it's a crucial technique when handling functions where \( \ln x \) is taken to a power or multiplied by other complicated expressions. It gives you a powerful tool to simplify and handle potentially complex mathematical scenarios with more ease.

Once you've calculated the derivative using rules like the product rule, the next step is often simplification. Simplifying will help to make the final derivative expression cleaner and easier to understand or use in further calculations.
In the exercise, after applying the product rule and calculating the derivatives separately for \( x \ln x \) and \( -x \), the results are combined. This gives us: \( f'(x) = (\ln x + 1) - 1 \). As you can see, \( +1 - 1 \) cancels itself out, simplifying the expression to \( f'(x) = \ln x \).
In general, always look for opportunities to combine like terms and to eliminate constants or coefficients that add up to zero. This step, though seemingly minor, can make a big difference in the readability and functionality of your result. Plus, a simplified expression often means fewer errors and a greater understanding when you use the derivative for further analysis.