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*) = *x*^{2}. Calculate *f'*(*x*).
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We use the limit definition, putting in *x* instead of *a*: Therefore *f'*(*x*) = 2*x*. 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*.
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Let *f*(*x*) = *x*^{3}. 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} = *x*^{3} + 3*xh*^{2} + 3*x*^{2}*h* + *h*^{3}. Therefore Once we have a formula for the derivative of a function, we can calculate the value of the derivative anywhere. | |

Let *f*(*x*) = *x*^{2}. We found that *f'*(*x*) = 2*x*. Find *f*(3).
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Since *f*'(x) = 2*x*, we have
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Let *f*(*x*) = *x*^{3}. Use the formula we found for *f'*(*x*) to evaluate

Let *f*(*x*) = *x*^{3}. Use the formula we found for *f'*(*x*) to evaluate

Let *f*(*x*) = *x*^{3}. Use the formula we found for *f'*(*x*) to evaluate