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In calculus, the concept of derivatives plays an integral role in understanding how functions change. Derivatives represent the rate of change of a function with respect to its variables.
A simple way to think of a derivative is as a way of identifying how a function is changing at any particular point. If you visualize a graph of a function, the derivative at any point is equivalent to the slope of the tangent to the curve at that point.
For the function given as the average state cigarette tax, denoted by \( T(t) \), the derivative \( T'(t) \) provides insight into how the tax changes over the years. In mathematical terms, it allows us to examine the pace at which this tax increases between 2001 to 2007.

An increasing function is a function that consistently increases as the input values increase. This concept can be best understood when you compare it to climbing up a hill: as you move ahead, you're always going up.
For a function to be classified as increasing, its derivative must be positive over the entire interval in question. This positivity signals that as the input \( t \) (in the given example, representing the year) grows, the output \( T(t) \) (the tax amount) also grows.

• In the provided exercise, \( T'(t) \) was calculated and found to be positive across the entire interval from 1 to 7.
• Thus, the function is increasing during the period of 2001 to 2007, showing us that the average state cigarette tax per pack was indeed rising consistently each year.

The rate of change in mathematics describes how a quantity changes concerning another quantity. It's essentially the core idea behind the concept of a derivative in calculus.
This rate is particularly crucial in understanding real-world scenarios where you want to know not just if something is increasing or decreasing, but precisely how fast that change is happening.

• In the context of the provided exercise, the derivative \( T'(t) = 0.1849t^{-0.57} \) expresses the rate of change of the cigarette tax over the years.
• The rate was found to decrease over time, due to the negative exponent in the derivative. This indicates that even though the tax was increasing, the speed at which it increased was slowing down with each passing year.

The power rule of differentiation is one of the simplest and most used rules in calculus. This rule allows us to find the derivative of functions of the form \( x^n \). According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \).
This rule was applied in the exercise to find the derivative of the function \( T(t) = 0.43t^{0.43} \).
By using the power rule, we calculated the derivative as \( T'(t) = 0.43(0.43)t^{-0.57} \), simplifying the process of finding the rate at which the tax was changing.
The power rule is extremely handy and allows for straightforward differentiation of polynomial expressions, saving time and preventing errors when calculating complex derivatives.