 So let's put a few ideas together. When finding the derivative, we can differentiate the algebraic expression, use values from a table, or use the graph, because remember the derivative corresponds to the slope of the line tangent to the graph. So let's have a couple of graphs, f of x and g of x, and let's find the derivative of f of g of x at x equals 3. So first, we can apply our chain rule, and we want to find the derivative at x equals 3, so equals means replaceable. Every place we see an x will replace it with a 3, and so it seems the first thing we need to worry about are the values of g of 3 and g prime of 3. So on the graph of y equals g of x, g of 3 is the y value when x equals 3. And so now we notice that y equals g of x goes through the point 3, 1, and so g of 3 is equal to 1. And on the graph of y equals g of x, g prime of 3 will be the slope of the line tangent to the graph at x equals 3. Now since the graph through x equals 3 is a straight line, the tangent line will be the line itself. And so g prime of 3 will be the slope of the line, and we see that slope is negative 2. On the graph of y equals f of x, f of 1 will be the y value when x equals 1. But we don't actually need this value, so we won't worry about it. However we do need the value of f prime of 1, so on the graph of y equals f of x, f prime of 1 will be the slope of the line tangent to the graph at x equals 1. Now since the graph through x equals 1 is a straight line, the tangent line will be the line itself. And so f prime of 1 will be the slope of the line, and so f prime of 1 will be equal to 3. And so the derivative will be negative 6. Or let's try something else, let's find derivative of g of f of x at x equals 6. So we'll apply the chain rule. We want to let x equals 6, so we need to find f of 6. So on the graph of y equals f of x, f of 6 is the y value when x is equal to 6. And so we see the graph of y equals f of x goes through 6, 5. And so f of 6 is equal to 5. To find g prime of 5, we want to find the slope of the line tangent to the graph of y equals g of x at x equals 5. So if we look, the line tangent to y equals g of x at x equals 5 has slope, um, has slope, um, and I'm not entirely, it's positive something, but what? Now if we absolutely had to get that slope, we could draw the tangent line and estimate the slope, but that would give us at best an approximation to the actual derivative. So let's procrastinate a little bit and see what else we can do. So on the graph of y equals f of x, f prime of 6 is the slope of the line tangent to the graph at x equals 6. Now that's around here someplace, and since the graph is a horizontal line at this point, f prime of 6 is going to be 0, and since it will be multiplied by 0 anyway, the actual value of g prime of 5 doesn't matter. We know it's positive something, but when we multiply positive something by 0, we get 0. And if we don't have the ability to get the actual value of the derivative, we might at least be able to say something about the sign. So maybe I want to find the sign of d of f of x at x equals negative 1. Now in the previous few problems, we started out by finding the chain rule, and so this time we should apply the chain rule. Now remember, we only care about the sign of the derivative at x equals negative 1, and so we might begin by noting here that f prime of negative 1 is going to be negative. Meanwhile, if we want to find the sign of g prime of f of negative 1, it would actually help if we knew the value of f of negative 1. And so we see that on the graph, f of negative 1 is equal to 1. And so now we want to say something about g prime of 1. And we see from the graph the sign of g prime of 1 is positive. And since f prime of 1 is negative and g prime of 1 is positive, their product will be negative.