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The first derivative of a function is crucial for understanding its behavior. It tells us how the function changes at any given point and, more importantly, helps in finding relative extrema, which are the local maxima and minima of the function. In simple terms, these are the "hills" and "valleys" of the graph.
For the function \( f(x) = x^3(x+1)^2 \), we start by taking the first derivative. By computing \( f'(x) \), we can identify where the slope of the function is zero, which indicates potential relative extrema points. This function is a product of two parts, \( u = x^3 \) and \( v = (x+1)^2 \), which means we'll need to use the product rule to differentiate it.
Once we apply the product rule and simplify, we end up with \( f'(x) = x^2(x+1)(5x+3) \), which gives us the critical points where the derivative equals zero.

The product rule is a technique for differentiating problems in calculus where two functions are multiplied together. If you have two differentiable functions, \( u(x) \) and \( v(x) \), their product derivative can be calculated with the formula: \((uv)' = u'v + uv'\). This rule ensures that each part of the multiplied expression is properly taken into account when differentiating.
In our exercise, where we want to differentiate \( f(x) = x^3(x+1)^2 \), the product rule becomes indispensable. Here, \( u = x^3 \) and \( v = (x+1)^2 \). By using the product rule:

• Find \( u' \) by differentiating \( u \) to get \( 3x^2 \)
• Find \( v' \) by differentiating \( v \) to get \( 2(x+1) \)

Once you have \( u' \) and \( v' \), substitute back into the product rule formula to calculate \( f'(x) \).
This calculated derivative can then be simplified to a more manageable expression.

Critical points are where the first derivative of a function equals zero or is undefined. These points give potential locations of relative maxima, minima, or saddle points (inflection points). Finding these points is essential because they help in analyzing the function's graph.
For the function \( f(x) = x^3(x+1)^2 \), we derived \( f'(x) = x^2(x+1)(5x+3) \). To find the critical points, solve the equation \( f'(x) = 0 \).
This equation is zero when any of its factors is zero:

• \( x^2 = 0 \), which simplifies to \( x = 0 \)
• \( x+1 = 0 \), leading to \( x = -1 \)
• \( 5x+3 = 0 \), simplifying to \( x = -\frac{3}{5} \)

Each of these solutions is a critical point that could potentially lead to a relative extremum. Identifying them is the first step towards finding the true relative maxima or minima, which can be further confirmed by using the second derivative test or by evaluating the function's values at these points.