So you might be asking how is calculus, how is differentiation actually helpful in a business setting? Let's look at a situation. Suppose that we have a graph of profit for an organization. On this axis is dollars and this axis is quantity. We're going to draw a curve. This curve right here, represents the total amount of profit that a firm is experiencing. So how do I get this profit curve? Well, there's some revenue. So every time I make an additional unit of quantity, I'm taking on some revenue. That is, I will sell each one of these products and in order for me to actually sell the product, I have to produce that product. In order to produce that product, I'm going to be taking on additional costs. So the difference between my total revenue my total costs at each unit of quantity, at each unit of output will reveal some level of profit. So then, I get this curve that represents the amount of profit that I get at each level of quantity. So now, there's a really nice mathematical characteristic. The mathematical characteristic that I'm talking about is the fact that somewhere up here, at the top of this profit curve, you will reach a maximum and at that point, the slope of that curve is zero. That is, if I were to draw this, that's a tangent line and it's a tangent line at the very top of that curve. That tangent has a slope of zero. So I can say, the derivative, the change in my profit with respect to the change in my quantity outputs at that point is equal to zero, has a zero slope. So now, we have a way to identify the quantity associated with that height of that profit curve. We'll call that quantity Q star. Q star is the quantity where the derivative of profit with respect to quantity is equal to zero. That is, that identifies the maximum of my profit. So what I really want to do is if I have my profit function, I could write it just like this. Let's say my profit is sum function of quantity. Then, what I want to do is I want to say, well, take the derivative of this function, dY dQ and set it equal to zero. So the way that we write that would be f prime of Q and set it equal to zero. Then, we would solve for that derivative. This is the calculus that you're likely to do in your studies within business. It's pretty straightforward. So I'm going to then let's go to the next tape and let's talk about some of the different rules and some of the ways you actually go about calculating the derivatives. Then, we'll go through a couple of examples with respect to this problem. Where we're saying "Okay. What is my function for my profit and how do I calculate the derivative in order to identify the profit maximizing quantity?" Let's look at some more helpful rules for more complex functions. That is, how to take the derivative of something that's different than just Y is equal to 3X squared. There's a two rules here, the sum and difference rule. The sum and the difference rules is really saying that, the derivative of a sum or a difference of two functions is the sum or difference of the derivatives of those two functions. Namely, let's say that Y is a function of X plus a function of a different function of X. So say Y is f of X and it's also g of X. That's because let's say Y is slightly more complex. In this case, you might say that Y is equal to f of X plus g of X. That is, there's two different function of X. When we sum those two, then we get a value of of Y. That is to say that Y has two different components, and X enters into Y through two different functions, f of X and g of X. For example, let's suppose that Y is equal to 3X squared plus 5X cubed. Then, we know that dY dX is the derivative of the first function plus the derivative of the second function. So we use our power rule here and remember for 3X squared, we're going to take that two and put it down in front of the three and then subtract one from my exponent. So we end up with 2 times 3 times X to the power of 2 minus 1, which gives me 6X. My second function is 5X cubed. Again, we're going to take that three and put it down in front of my function and then subtract a one from my power. So I end up with 3 times 5 to the power of X to the power of 3 minus 1, so we end up getting 15X squared. So the sum of two functions, the derivative is the sum of the derivatives. Let's suppose that we had a difference. So instead of looking at this, we're looking at something like this. Suppose that Y is equal to 4X squared minus 2X. Now, instead of a sum we have a difference. Well, the derivative of Y with respect to X is now the difference of the derivatives of the first part and the second part. So the first part, the derivative here is remember using the power rule, the two comes down in front of my four. So I end up with 2 times 4 to the power of 2 minus 1, so I end up with 8X. Then down here, I'm going to subtract, so it's 2X to the power of 1. So I'm bringing that one in front of the two and multiplying. Then, I have to subtract a one from this. Now, you might say, that's really weird, X to the power of 0, what happens? Well, anything to the power of zero becomes a one. So this becomes 2 times 1 or 2. So dY dX in this example, is equal to 8X minus 2. When Y is a function of two functions, f of X and g of X and their summed, then the derivative is the sum of the derivatives. If it is equal to the difference f of X minus g of X, then dY dX becomes the difference of the derivatives.