 Since the derivative gives us the slope of the line tangent to a graph, we can also use the graph to find the derivatives. For example, given the graphs of y equals f of x and y equals g of x, we can try to find the sine of the derivative of fg at 0. So remember this notation, this is the derivative of the function f of x times g of x evaluated at 0. Since we want to find the derivative of a product, let's go ahead and apply the product rule. And so our derivative is going to be f of 0 times g prime of 0 plus g of 0 times f prime of 0. Now keep in mind we're only trying to find the sine and not the actual value. So we can keep track of the sines of these four components of the derivative of the product. So since we have the graph of y equals f of x, it appears that f of 0, the y value when x is equal to 0, is equal to 0. And likewise, since we have the graph of y equals g of x, then g of 0, the y value when x is equal to 0, is going to be positive. We also need to know g prime of 0 and f prime of 0. f prime of 0 is going to be the slope of the line tangent to the graph of y equals f of x at the point where x equals 0. So we might imagine what that tangent line looks like. And it appears that that tangent line will slope downwards, so f prime of 0 is going to be negative. Similarly, we want to find something about g prime of 0, that's the slope of the line tangent to the graph of y equals g of x where x equals 0. So let's sketch that tangent line, and it appears that this tangent line has a positive slope, so that tells us g prime of 0 is going to be positive. So now let's consider our formula for the derivative of the product. f of 0 is 0 and g prime of 0 is positive, so this first term will be the product of 0 and a positive number, so it's going to be 0. Meanwhile, in our second term, g of 0 is positive, f prime of 0 is negative. So this second term is going to be the product of a positive and a negative number, so that'll be negative. And so our derivative is going to be the sum of 0 and a negative number, so it's going to be negative. A little bit later on, we'll be interested in knowing where our derivative is equal to 0. So given this graph, suppose somebody claims that the derivative of g over f of x equals 0 has a solution at x equals 4. Is this a plausible claim? So again, let's consider what that derivative is going to look like, so we'll use the quotient rule. And again, let's consider the signs of these different components of the derivative when x is equal to 4. At x equals 4, the denominator is going to be f of 4. And since we have the graph of y equals f of x, then f of 4 is going to be the y value on the graph when x equals 4. So f of 4 is the y value on the graph of y equals f of x, where x equals 4. So if we go to this point, we see that f of 4 is going to be negative because the y value is negative. And so when we square it, we're going to get a positive number. Most importantly, that denominator is not equal to 0. How about g of 4? g of 4 is going to be the y value on the graph of y equals g of x when x equals 4. And so the sign of g of 4 is positive. f prime of 4 and g prime of 4 are going to be the slopes of the line tangent to the graphs at x equals 4. If we sketch those tangent lines, we see that f prime of 4 seems to be positive and g prime of 4 seems to be negative. So now let's consider the derivative of the quotient. At x equals 4, the denominator is not equal to 0, but since it's squared, will actually be a positive number. And while the numerator will consist of two terms, f of 4 times g prime of 4 minus g of 4 times f prime of 4. So f of 4 times g prime of 4, that's going to be a negative number times a negative number. g of 4 times f prime of 4 is going to be a positive number times a positive number. And so our numerator is going to be a positive number minus a positive number. But that could be zero, so the claim that it is zero at x equals four is plausible.