 So here's a few more quick examples of some basic derivatives. So let's find the derivative of the function 3x squared minus 5 over x cubed. So again, what we have here is, well, let's see, I take x squared times 3, I take x cubed it, divide it into 5. The last thing I do is I subtract these two terms. So the function is a difference, it is an addition or subtraction, and that's good because I know the rule that the derivative of a sum or difference is the same as the sum or difference of the derivatives. So I can take this function here, break it into two parts 3x squared and 5x cubed, 5 over x cubed, and then I can differentiate each part separately. So here goes nothing. The derivative of the difference is the same as the difference of the two derivatives. And let's take a look at each part separately. This is the derivative of 3 times x squared. So this is the derivative 3, 3, 3, never changes its value, it's always equal to 3. So this is the derivative of constant times function. And I know that derivative of constant times function is constant times the derivative of the function. So I know what I'm going to be able to do with this. This is 5 over, oh wait, I only know how to differentiate x to the n. Well, that's okay. I know that if I have an x to the third in the denominator, it's the same as an x to the power minus 3. So I'll make that substitution there. And let's see. This is, again, reading the expression 5 times x to the power minus 3, and 5 also is a constant. So this is derivative of constant times function, which means that I can rewrite that as constant times the derivative of the function. So those constants, 3 and 5, I can pull those out front, and then I only have to worry about the derivative of x squared and the derivative of x to the power minus 3. Well, both of those are of the form x to the n. So if I want to find derivative of x to the n, it's going to be nx to the power n minus 1. So that's going to be x to the 2 is going to differentiate as 2x to the power 1. And the derivative of x to the power minus 3 is going to differentiate as minus 3 x to the power minus 3 minus 1. And finally, we do a little bit of algebra to simplify this. This is 6x minus 15x to the power of minus 4. And again, it's convenient, not absolutely necessary. Either of these two lines is a reasonable form to leave the derivative in, but because the dialect we were asked the question uses only positive whole number exponents, we want to write our final answer also only using positive whole number exponents. So our properly written final answer is 6x plus 15 over x to the fourth. If we run out of time, that's a good answer. If we don't even get that far, this is not a bad starting point. Well, nobody said variables have to be called x. I could call my variable something else. How about r? So here I want to find the derivative of some function where f of r is 2 pi r squared plus 800 pi over r squared. So again, I'm looking for the derivative of this function. And let's see if I know what r is, what do I have to do? Square it times pi times 2, square it, divide it into 800 pi, add. The last thing I do is I add, so this is the derivative of a sum. And I know that the derivative of a sum I can find by finding the sum of the derivatives. And again, this 800 pi over r squared can also be rewritten 800 pi times r to power negative 2. Now, let's see what we have. We want to find the derivative of 2 times pi times r squared. The thing to realize here is that 2 pi is a constant. 2 is a constant, pi is a constant, it's somewhere between 3 and 35. And so 2 times pi is just a constant number. So this is derivative of constant times function. And I can rewrite that as constant times the derivative of the function. Likewise, 800 times pi, well pi is a real number somewhere between 1 and 10. And so 800 times pi is also going to be just a real number. So this constant times r to power minus 2, the derivative can be found by multiplying constant by the derivative. So those constants factor out. And now I have to find derivative of r to power 2, well that's an x to the n type of expression. So the derivative is going to be 2r, because that's the name of our variable, to the power 2 minus 1. And the derivative of this minus 2, r to power minus 2 minus 1, which is minus 3. And a little bit of algebraic simplification gets our answer into a final nice form.