So far we have only been able to find the slope of a function that is a line by putting the equation of that line in slope-intercept form.

However, there are functions that are more complex than a line. Consider the function . The graph of can be seen below. Suppose we wanted to find the slope of the line at x=1. can not be put in the form y = mx + b. How do we find the slope?

Let's recap what we have done to estimate the slope. Our
calculuations of the three slopes looked like this:

On the first line, you might be wondering why we expressed f(3) as f(1+2) and why in the second line we expressed f(2) as f(1+1) and the same for the third line.

Consider the limit: . With the first slope, x = 1

and h = 2. In the
second
slope, x = 1 and h = 1, and with the third slope we have x = 1 and h =
0.05. We are letting h go closer and closer to 0 which is giving us a
closer and closer approximation of the slope. When we evaluate the limit
at h = 0 and x = 1 we have the slope of the curve, , at x = 1. Let's evaluate the limit and see what it
tells us.

.

Evaluating 2x at x = 1, gives us a slope value of 2 which is what we would expect considering our first three calculations since the slope was getting closer and closer to 2.

Now formally, the slope of at x = c is equal to which is known as the **derivative** of at x = c and is denoted f', read 'f prime' or
also denoted read 'the derivative of y with respect to x' where, in
this case,