Introduction to Derivatives

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?

To find the slope we can pick two points on the line. One point is (1,1), which are the coordinates of the point at which we want to find the slope of the line. Let's pick another point also on the line like (3,9). Let's plot (3,9) and draw the tangent line connecting the two points. We can see the result below. The slope of that line is 4. (If you don't remember how to calculate slope you can refresh your memory here.)

That looks pretty good as a first try. Let's see if we can improve our approximation with a point closer to (1,1) like (2,4). The slope of the line connecting (1,1) and (2,4) is 3/1 = 3. Let's connect those two points with a line and see how good our approximation is.

That's even better, but there is still room for improvement. Let's pick a point even closer to (1, 1) like (1.05,1.1025). The slope of the line connecting (1, 1) and (1.05, 1.1025) is 2.05. Let's plot the line and see how well we do:
That looks nearly perfect. However, what if we needed an even closer approximation? How would we go about doing that?

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,