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We'll use the limit definition, putting in x instead of a:
Therefore f ' (x) = 2x. If we plug in 1 for x we find f ' (1) = 2, which agrees with our earlier calculation.
With formulas, the limit definition of the derivative function is
This is the same as the limit definition of the derivative at a point, but with x instead of a. When we evaluate the derivative of f at a point, we take a value and plug it in for x in the definition above.
Let f be a line. That is, f(x) = mx + b where m and b are constants. What is f ' (x)?
Let's see if we can make an educated guess before we dive into the limit.
Since derivative means "slope" and lines have a constant slope, we should expect f ' (x) to just be the slope of the line. Let's see if our suspicions are confirmed. Here we go:
We already thought of m as the "slope" of a line mx + b. It's nice that when we calculated the derivative we got m, since the derivative is also supposed to be the "slope." This means that for any line
f(x) = mx + b, the derivative function is a constant function:
f ' (x) = m for every value of x.
Let f(x) = x3. Find f ' (x).
We use the limit definition of the derivative:
But we'll leave it to you to check that
(x + h)3 = x3 + 3xh2 + 3x2h + h3.
Once we have a formula for the derivative of a function, we can calculate the value of the derivative anywhere.
Let f(x) = x2. Given that f ' (x) = 2x, what is f ' (3)?
Since f ' (x) = 2x, we have
Just plug the number into the function. That's all there is to it.