 Hello and welcome to the screencast where we're going to continue to think about derivatives of a function at a point. But this time we're going to be using no formulas at all and just using graphs to help us estimate these derivative values. So here I have a function f. It acts kind of crazy. And the important thing here to notice is that I don't have a formula for f in front of me. So if I wanted to calculate say f prime of one, I would not go back to the limit definition necessarily. That would only get so far because I don't have any formula to put into the limit definition. So rather than using the limit definition and a lot of algebra like we saw in the previous screencast, we're going to have to rely upon our graphical concept of the derivative to calculate derivatives in this case. That concept is this, that the derivative of f at the point x equals a tells me the slope of the line that is tangent to the graph of f at x equals a. So for example, if I wanted to calculate f prime of zero, that would be saying if I went up to x equals zero, that's right here on the y-axis, it looks like around y equals four, but x is equal to zero. And drew a line that touched this graph right at the point zero comma four, right where the graph goes through x equals zero, and that line follows the overall direction of the curve at that point. Then its slope should be the value of the derivative of f at zero. Let's do that specifically. The line tangent to the graph of f at x equals zero would touch right here and would look something about like this maybe. It certainly touches the graph there and follows the overall trajectory of the curve at that point. You could quibble about the accuracy of this blue line that I just drew. We would get better accuracy if I could somehow zoom in on this picture and draw a better line. But with this picture and this graph, it seems like that line there is a reasonable approximation to the tangent line, to the graph at x equals zero. The tangent line will look very, very different at different points, but at this point x equals zero, that appears to be pretty accurate. Now the derivative f prime of zero is just going to be the slope of this line. And I can estimate that slope just by looking at the grid that you see underneath the graph here. Without that grid, it's going to be pretty hard to estimate the derivative value. Now importantly here, possibly even oversimplifying the problem is that the grid is one to one. Every square you see here in the grid is the same units wide as they are tall. Namely, it's a half a unit wide and a half a unit tall. So we can go through and estimate the slope just by doing a quick estimate of rise over run. It appears that if I start with my graph here and go over by one grid, it looks like I go up roughly two thirds of a grid. Now since the grid squares are in a one to one ratio, they're not say ten units wide by one unit tall. The scaling is not distorted. Then we can say that the slope of this blue line is therefore rise over run. That's two thirds divided by one or a slope of two thirds. And therefore we could estimate the derivative f prime of zero to be equal to two thirds. Let's calculate the derivative at some different points here. Let's say x equals two. Let's go over here to x equals two and go up to the graph. And what I want to do is draw a tangent line or at least a reasonable estimate to a tangent line right there. I think I'm going to actually leave this one going only forward. I could pick it up and move it backwards but it kind of loses some resolution. So I'm going to pretend this line is only moving forward here. Now that's a reasonable estimate to the tangent line. If I were to move this back then it seems like it's touching the graph at this one point and following the overall direction of the curve there. So what's its slope? The slope of this blue line that I just calculated is going to be equal to the derivative of f at two. What is that slope? Well we're going to go back to rise over run again. It looks like if I go over by one I'm going up by one, two, and roughly a half units. So the slope again because the grid squares are in a one to one ratio to each other. They're not distorted. That would give me a slope or a derivative value there, g prime of two of approximately two and a half. Let's do one more example. I think that'll be good enough for us. Let's go over to x equals three and calculate the derivative at three. So we're going to go up to the graph that's at around just south of four point five on the graph. What's the slope right there? So I'm going to take my drawing tool here and if I were doing this on paper I'd just have a pencil and a straight edge and draw what I think is a reasonable approximation to the tangent line to the graph at that point. That's kind of hard to draw but I think this is about right. Let's move the tangent line up a little bit and yeah that seems like a pretty good tangent line here. So what's the slope of this tangent line? The slope of that tangent line is going to be equal to g prime of three. So what is it? Well here's the point of tangency. Here's the point where the line is actually touching the graph. And it appears using the grid that if I go over by one unit I'm going to go down by two units. So that's a negative slope. In fact it's in slope of approximately negative two. So therefore we can estimate g prime of three to be equal to negative two. So we can go through and estimate the derivative of this function at really any point we wish because I have a nicely drawn graph and because I have a nice grid underneath it. And with those two ingredients we can use the graphical concept of the derivative as the slope of the tangent line to estimate our derivative values. Thanks for watching.