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When faced with the task of differentiating composite functions like \( y = \ln^2(x) \), the chain rule becomes a vital tool. The chain rule helps in differentiating a function that is nested within another function. In this case, \( y = (\ln(x))^2 \) is a composition of the functions \( u = \ln(x) \) and \( y = u^2 \).

To apply the chain rule, we first differentiate the outer function with respect to the inner function, resulting in the derivative of \( y \) with respect to \( u \), which is \( \frac{dy}{du} = 2u \). Next, we find the derivative of the inner function \( u \) with respect to \( x \), which gives \( \frac{du}{dx} = \frac{1}{x} \).

By combining these results, the chain rule tells us to multiply them together to find the overall derivative: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2\ln(x) \cdot \frac{1}{x} = \frac{2\ln(x)}{x} \). This shows how seamlessly the chain rule simplifies the process.

Derivatives are a fundamental concept in calculus, representing the rate of change of a function. Essentially, they provide the slope of the tangent line to the function at any given point, describing how the function's output value changes with its input value.

In our example, finding the derivative of \( y = \ln^2(x) \) involved using the chain rule. This first derivative, \( \frac{dy}{dx} = \frac{2\ln(x)}{x} \), shows us how \( y \) changes with \( x \).

Derivatives are crucial for various applications, such as:

• Determining the instantaneous rate of change.
• Locating critical points where a function might have its local maximum or minimum.
• Understanding the behavior of functions, whether they are increasing or decreasing.

Mastering derivatives opens the door to profound insights into how functions behave across their domains.

Finding extrema involves identifying points where a function reaches its highest or lowest values, locally or globally. Extrema are critical in understanding the behavior of functions, especially in optimization problems.

To identify extrema, we first look at the critical points, which occur where the first derivative is zero or undefined. In our example, the first derivative \( \frac{2\ln(x)}{x} = 0 \) leads us to the critical point when \( x = 1 \).

At these points, the function could have a

• Local maximum: The function value at this point is greater than at nearby points.
• Local minimum: The function value at this point is lower than at nearby points.
• Saddle point: A point where the derivative is zero but not a maximum or minimum.

Once critical points are identified, further analysis can be conducted using other methods like the second derivative test to classify these points.

The second derivative test is a method to classify critical points into local maxima, minima, or saddle points, based on the concavity of the function.

In the exercise, after finding the critical point at \( x = 1 \), the second derivative test is used. By computing the second derivative \( \frac{d^2y}{dx^2} = \frac{2(1 - \ln(x))}{x^2} \) and evaluating it at the critical point \( x = 1 \), we get a positive value, specifically 2, indicating a local minimum.

Here's a simple guide to using the second derivative test:

• If \( \frac{d^2y}{dx^2} > 0 \) at a critical point, the function is concave up, indicating a local minimum.
• If \( \frac{d^2y}{dx^2} < 0 \), the function is concave down, suggesting a local maximum.
• If \( \frac{d^2y}{dx^2} = 0 \), the test is inconclusive, and further analysis is needed to classify the critical point.

This test is a powerful tool for quickly understanding the nature of critical points in calculus.