Earlier, we saw that the tangent line on a graph at a point \( (x_0, f(x_0)) \) was the line passing through that point with the slope:

$${\lim_{ d \rightarrow 0 }} {{ f( x_0 + d ) - f( x_0 ) } \over d } $$

as long as that limit actually exists.

We can just call that the tangent line at \( x = x_0 \).

That particular limit is so common it has a name. We call it "f prime of \( x_0 \)":

$$ f'(x_0) = {\lim_{ d \rightarrow 0 }} {{ f( x_0 + d ) - f( x_0 ) } \over d } $$

If we drop the subscript, we get a very special function. It is called the derivative function, or just the dirivative:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{ f( x + d ) - f( x ) } \over d } $$

If a function has a tangent line at some point, the derivative of that function is the slope of the tangent at that point.

In our example of John shooting the arrow, we found that the derivative times the horizontal velocity of the arrow gave us the vertical velocity.

We have talked about limits where the denominator goes to zero, but we have not yet shown how to find that limit. That means we don't (yet) actually know what the slope of the line at any point is.

Here is where all those little rules about doing arithmetic on limits come in handy. Let's find the derivative of a function similar to (but simpler than) the function of the arrow:

$$ f(x) = x^2 + 1 $$

The derivative is:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{ f( x + d ) - f( x ) } \over d } $$

Let's plug in \( x^2 + 1 \) everywhere we see an x:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{ f( [x^2 + 1] + d ) - f( [x^2 + 1] ) } \over d } $$

Now simplify using the definition of a function:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{ [{(x + h)^2  + 1 ]} - [x^2 + 1] } \over d } $$

Then we expand the first bracketed expression:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{{     x^2 + 2xd + d^2 + 1    } - x^2 - 1 } \over d } $$

Some of the terms now cancel out:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {{    2xd + d^2     } \over d } $$

Now we can factor out that troublesome denominator:

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {    2x + d       } $$

And, since \( d \) is going to zero, we can remove it (since adding zero does nothing):

$$ f'(x) = {\lim_{ d \rightarrow 0 }} {    2x        } $$

And we finally get to use what we learned about limits. We can move the constant through the limit sign:

$$ f'(x) = 2 {\lim_{ d \rightarrow 0 }} x $$

And the limit of a constant function at any point is the constant:

$$ f'(x) = 2 x  $$

We have found that the derivative of \( x^2 + 1 \) is \( 2 x \).

That means the slope of the tangent of that curve at any point x is just 2x.

For example, the slope of the tangent at \( x = 4 \) is 8.

Even though we can't divide by zero, we were able to find the formula for the tangent line.

That's what calculus is for.