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When dealing with the differentiation of a product of two functions, it’s essential to utilize the product rule. Imagine you have two functions, say \( u(x) \) and \( v(x) \), and you need the derivative of their product. The product rule allows you to find this by freeing you from dealing with their complexity as a singular block. According to the rule, the derivative of their product is \( u'v + uv' \). This means you differentiate one function and hold the other steady, then swap and add the results together. For our specific problem, with \( f(x) = x \sqrt{x-x^2} \), you let \( u(x) = x \) and \( v(x) = \sqrt{x-x^2} \). Applying the product rule simplifies the differentiation process as each part is dealt with separately, easing calculation.

The chain rule is the go-to method when working with composite functions, or functions within functions. When you see something like \( \, \sqrt{x-x^2} \, \), the chain rule helps develop the derivative by simplifying it layer by layer. Think of peeling an onion; each function has its skin. Begin with the outermost function, in our case the square root, then dive into the inner layer, here \( x-x^2 \). To achieve this, differentiate the outer function first, keeping the inner function intact. The derivative of the square root is noted as \( \, \frac{1}{2}x^{-1/2} \, \). Next, move inward and find the derivative of the inside function, which in our case is \( 1-2x \, \). Combine these products following the chain rule, adjusting for each layer peeled. This step-by-step method unveils complexities smoothly, easing derivative calculations for intricate compositions.

Critical points of a function are essentially its 'highlight reel'. They are where the function changes, peaks, or bottoms out, hinting where it stops going up and starts going down (or vice versa). These occur when the first derivative \( f'(x) \) is zero or undefined. In our exercise, critical points are found by solving \( \, x \cdot \frac{1}{2}(x-x^2)^{-1/2}(1-2x) + \sqrt{x-x^2} = 0 \, \). By solving this equation, you pinpoint where the function could potentially have local peaks or valleys. Remember, not all critical points lead to maxima or minima; some might merely be inflection points, indicating neither a peak nor valley.

To determine where a function is increasing or decreasing, the first derivative test is your best friend. After finding critical points, consider what happens between these points by examining the sign of the first derivative (whether \( f'(x) > 0 \) or \( f'(x) < 0 \)). Here's the process: split the number line by critical points, then test the sign of \( f'(x) \) within each resulting interval. If \( f'(x) \) switches sign from positive to negative, the function transitions from increasing to decreasing, indicative of a relative maximum. Conversely, a switch from negative to positive represents a relative minimum. This test provides a clearer image of the function's behavior over its domain, highlighting intervals of ascent and descent.

Relative extrema are key plot points of a function that stand out locally; these are the hills and valleys in our function’s path. They tell us the peak highs and lows within certain sections. Determining these involves noting where the first derivative equals zero (critical points), then using the first derivative test to see if they are indeed maxima or minima.

• A relative maximum occurs at a point where function changes from increasing to decreasing.
• A relative minimum appears where it shifts from decreasing to increasing.

In our exercise, relative extrema guide us to identify such crucial transition points on the graph of \( f(x) \). These points help inform practical conclusions about the function's performance and behavior within its designated interval.