Slope of a secant line Slope of a tangent line:
 If , we can find the slope of a secant line between the points where x = 2 and x = 5 very easily:
This is the average rate of change of over the interval and is the slope of the secant line connecting two points (2, 4) and (5, 25).
 Suppose we want the exact (or instantaneous) rate of change at the one point (2, 4). We cannot calculate this slope using standard methods because we need two points to calculate slope. But suppose that when and a is “very close” to 2, we calculate the slope of a secant line. We get:
To calculate the slope at exactly , imagine “ a” getting “closer” to the number 2. We have:
This means that the instantaneous rate of change of at is 4.
We call this the derivative of at and we write this as (read: “f prime of 2 equals 4”).
To calculate the derivative of at any point x (not just ), we use the same approach as in the work done above. We have:
Therefore we say that the derivative of is .
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The derivative of :
 We can use this method to calculate the derivative of other functions, for example .
This process can become difficult since we are relying upon “nice” factorization. Notice what we have thus far:
A pattern is developing here. We would next have
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We generalize this into what is called the power rule.
Let's Practice:
 Find .
We have by the power rule .
 Find .
We have by the power rule .
 Find .
We rewrite as .
Therefore, .
 Find .
We have for any constant such as 4.
 Find .
We have .
 What is an equation of the line tangent to at the point where ?
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We get the “yvalue” (the function value) for from
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