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Calculating derivatives is a fundamental part of calculus, which helps us understand how a function changes through its rate of change. In simple terms, the derivative of a function represents the slope of the tangent line to the function at any point. For any given function, the derivative calculation involves some specific rules and steps that simplify the process.

When tasked with finding the derivative of a polynomial function, you encounter a set of rules, such as the power rule or the product rule, which are essential tools in differentiation. By understanding and applying these rules, one can efficiently determine the derivatives of various functions, whether it’s a simple monomial or a complicated product of multiple terms. This makes derivative calculations not only manageable but straightforward once these rules are mastered.

The power rule is one of the most straight-forward rules in calculus differentiation and is crucial for finding derivatives of functions in the form of a single term raised to a power. The rule states that if you have a function: - \( f(x) = x^n \), then the derivative of this function is - \( f'(x) = n imes x^{n-1} \).

The power rule simplifies the differentiation process significantly, as instead of expanding power expressions, we just multiply the power by the coefficient and decrease the power by one. For instance, when you have a function like \( r(x) = 100x^{2.1} \), applying the power rule directly gives:

• Multiply the exponent with the coefficient: \( 2.1 \times 100 = 210 \)
• Decrease the exponent by 1: \( 2.1 - 1 = 1.1 \)

Hence, the derivative becomes \( r'(x) = 210x^{1.1} \). The power rule's simplicity helps in quickly working through derivatives and is applicable whenever you encounter polynomials or similar function forms.

The product rule is another critical component of calculus differentiation when dealing with products of two functions. It is necessary when you have two differentiable functions multiplied together. The product rule states:

• Given two functions \( u(x) \) and \( v(x) \), the derivative of their product \( uv \) is: \((uv)' = u'v + uv'\)

In the context of our function \( r(x) = 100x^{2.1} \), if you consider \( u(x) = 100 \) (a constant) and \( v(x) = x^{2.1} \), application of the product rule involves:

• Finding \( u'(x) \), the derivative of 100, which is 0, since it's a constant.
• Calculating \( v'(x) \) using the power rule: \( 2.1x^{1.1} \)
• Combining these using the product rule: \( uv' = 100 \times 2.1x^{1.1} = 210x^{1.1} \)

Thus, applying the product rule also yields \( r'(x) = 210x^{1.1} \). This not only reaffirms our results from the power rule but also illustrates that the product rule’s utility shines when separating out terms into different functions, particularly when you're faced with intricate combinations of differentiable expressions.