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Calculating the derivative of a function is a fundamental skill in calculus, especially when you're seeking to understand behaviors like extrema. The derivative tells us the rate of change at any given point, analogous to a slope of a tangent line. In the problem with the function \(k(x)=\frac{2x}{5}-(x-1)^{2/5}\), we aim to find \(k'(x)\) by taking the derivative of each component separately.

Start by identifying simpler sub-problems. For instance, \(\frac{2x}{5}\) differentiates to \(\frac{2}{5}\), because the derivative of \(x\) is 1. The expression \(-(x-1)^{2/5}\) requires more work due to the exponent and subtractive operation—it’s here that the chain rule becomes essential.

Once you've broken down the derivatives of each term, combine them to find \(k'(x)\). Such breakdown and recombination is standard when handling complex derivatives, helping you see and trace the effects of each part of the original function.

The chain rule is a crucial technique in calculus for differentiating composite functions. It essentially helps us tackle problems where a function is nested within another. Think of it as peeling layers of an onion, where you differentiate each layer separately, then multiply the results.

In our problem, we apply the chain rule to differentiate \(-(x-1)^{2/5}\). We break this down by letting \(u(x) = x-1\), making our inner function, \(f(u) = u^{2/5}\). By the chain rule, the derivative \(k_2'(x) = f'(u) \cdot u'(x)\).

The inner derivative, \(u'(x) = 1\), is straightforward as the derivative of \(x\) is 1, and constants like \(-1\) disappear. For the outer function, \(f'(u) = \frac{2}{5}u^{-3/5}\), apply the power rule to \(u^{2/5}\). Multiply these derivatives according to the chain rule: \(-\frac{2}{5}(x-1)^{-3/5}\). This helps simplify the process of differentiating complex expressions.

The first derivative test is a method that tells us whether a critical point is a minimum, maximum, or neither. By examining the sign change of the derivative, we determine whether the function is increasing or decreasing.

In our example involving \(k(x)\), the derivative \(k'(x)\) is \(\frac{2}{5} - \frac{2}{5}(x-1)^{-3/5}\). We set this equal to zero and solve for zero and undefined points to find critical points, like \(x=2\). However, just solving for zeros isn't enough.

After identifying critical points, we inspect the behavior around them:

• If \(k'(x)\) changes from positive to negative at a critical point, there's a relative maximum, indicating a peak.
• If it changes from negative to positive, there's a relative minimum, indicating a valley.

For \(x=2\), \(k'(x)\) goes from positive to negative, confirming a relative (and absolute) maximum at this point.

Critical points play a central role in understanding a function's extrema. These are points where the function's derivative equals zero or is undefined. They are potential candidates for local maxima, minima, or points of inflection.

In our case, we found the critical point \(x=2\) by setting the derivative \(k'(x)\) to zero. Critical points are not always extrema. Therefore, after finding \(x=2\), we use the first derivative test to confirm its nature as an extremum.

After pinpointing the location of critical points:

• Check the value of the original function at these points to understand their effect on the function's graph.
• Evaluate their counterparts at the end points of the domain, as extrema can also occur here.

Consequently, by confirming that \(x=2\) is indeed a maximum, and calculating \(k(2)=\frac{4}{5}\), we establish the precise extremum of the function within its domain \([0, +\infty)\).