The "derivative of f at a," written f ' (a), is a number that is equal to the slope of the function f at a.
For any differentiable function f there is another function, known as the derivative of f and written f ' (x). We write f '(x) to show that this is a function.
We can calculate the derivative function using the limit definition in the same way we calculated the value of the derivative at a point using the limit definition.
Practice:
Let f(x) = x2. Calculate f'(x).
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We 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. In formulas, the limit definition of the derivative function is 
This is the same as the limit definition of the derivative of 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. Show that f'(x) = m.
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This problem is nice, because it tells what the answer is supposed to be. "Show that f'(x) = m" means "Find f'(x), and we should find m as our answer." 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).
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We use the limit definition of the derivative: 
We leave it to you to check that (x + h)3 = x3 + 3xh2 + 3x2h + h3. Therefore 
Once we have a formula for the derivative of a function, we can calculate the value of the derivative anywhere. | |
Let f(x) = x2. We found that f'(x) = 2x. Find f(3).
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Since f'(x) = 2x, we have 
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Let f(x) = x3. Use the formula we found for f'(x) to evaluate
Let f(x) = x3. Use the formula we found for f'(x) to evaluate
Let f(x) = x3. Use the formula we found for f'(x) to evaluate