 While we can go back to the definition of the derivative every time we want to find a derivative, it's more convenient if we actually have some rules that make it a little bit easier to find other derivatives once we've found a couple. And these fall under the category of derivative rules, and so we're going to introduce a couple of the basic ones. In general, if I'm looking at the derivative of a function, I have the following relationships. First of all, the derivative of a constant times a function is the same as the constant times the derivative of the function. And another very, very, very, very useful rule is the following, the derivative of the sum of two functions is the same as the derivative of the one function plus the derivative of the other function. The derivative of a sum is the same as the sum of the derivative. And just a quick reminder of our place. Remember that we have two different ways of expressing the derivatives. We have what's called the differential notation, this d over dx of some function. And we have what's called the prime notation, f prime of x. And they both mean exactly the same thing. Good notation survives because it reminds us of something useful, or it helps guide our process. And we'll see that both of these notations have their place when finding derivatives. Alright, so let's take a look at a problem. For example, suppose I know the derivative of f at 3 is negative 2, and the derivative of g at 3 is equal to 5. So I've done whatever work is necessary, and I've found a couple of derivatives, but I decided, you know, I don't actually care about the functions f of x and g of x. What I really need to know something about is this function f plus g of x, the sum of f of x and g of x. Or maybe I need to know minus 7 times g of x, minus 7 times whatever the function g of x is. Well, I could try and find the derivative by hand as we did before, but now that I have this information about the derivatives of these functions, I can use my derivative properties to find these derivatives. So let's take a look at this. So for this first expression here, this is the derivative of f plus g of 3. I want to find the derivative of the sum of f and g, so I'm going to just look for the sum of the derivatives. And so that's going to be pretty straightforward. The derivative of a sum is the same as the sum of the derivatives. And if only I knew what f prime of 3 and g prime of 3 are, oh, good. I actually have those values up here. So I'll go ahead and just substitute those values in, and I have my derivative of the sum. Now let's take a look at that other expression. I want the derivative of minus 7 times g evaluated at 3. So here I have a constant multiple minus 7 multiplied by my g function. And so what do I know? Well, I know that a constant multiplied by the function differentiated is the same as the constant times the derivative of the function. So minus 7g prime of 3, well, it's going to be minus 7 times whatever the derivative is. So I know what the derivative is, and I can do the computation. Let's talk about some actual derivatives of specific functions. So one of the more important rules of differentiation is that if I have any actual constant, if I find the derivative of that constant, the derivative is going to be 0. Now don't get this confused with the derivative of a constant times a function. Here, I just have the derivative of the constant all by itself. Likewise, if I have any exponent n for any n not equal to 0, the derivative of x to the n is n times power x to the n minus 1. And again, this holds for any n not equal to 0. If n is equal to 0, well, it actually does apply in a degenerate case. x to the 0 is 1, which is a constant, and the derivative of a constant is 0. And I get the same thing here, and happens to be 0. I get 0 times, well, who cares what? 0 multiplied by something is just going to be 0. And between these two rules and the other two rules, I can now differentiate any polynomial or, more generally, anything I can write as a sum of exponential expressions. So for example, suppose I want to find the derivative of x raised to the power 5. Well, the first two or 30 times that you do problems like this, it's helpful to write down what the relevant derivative rule is. This is a function of the form x to something, and I know that the derivative of x to the something and x to power n minus 1. So I can just fill in the blanks. I'm dealing with the derivative of x to power 5. It's going to be 5 times x to power 5 minus 1, and believe it or not, the most common error occurs in the next step, where I have to simplify this expression. Let's see. Well, I can do 5 minus 1. Believe it or not, 5 minus 1 to 4, doing that simplification is the source of more errors and mistakes in this type of derivative problem than you would think. That was pretty easy. Let's try a different one. How about 5 times x to power 8? Now here, it's useful to think about what our function is. This is, reading it out, 5 times x to power 8. In other words, what this looks like is the derivative of a constant 5 multiplied by some function x to the 8th. And if I have a constant multiplied by a function, the derivative of constant multiplied by a function is just going to be constant times the derivative of the function. So what can I do here? Well, if I want to find the derivative of 5 times x to the 8th, I can move that constant out front. That's 5 times the derivative of x to the 8th. And then, I know how to differentiate that. That is an x to the n expression. So the derivative is going to be nx to power n minus 1. Exponent 8 comes down front. x to power 8 minus 1. And hardest step of the problem, simplifying these two expressions. That works out to be 40x to power 7. How about something a little bit more complicated? Well, here's a nice polynomial. g of x equals x cubed minus 4x squared plus 8. And I want to find the derivative. And remember that the prime notation and the differential notation mean exactly the same thing. I just want to find the derivative of g of x. And I can either express that g prime of x, or I can express that using differential notation derivative with respect to x. Now, the prime notation is actually very useful shorthand, because I can read it as differentiate something. So here's my problem. I want to find the derivative of g of x. Well, that's the derivative of, here's g of x, x cubed minus 4x squared plus 8. A little bit of analysis is useful here. If I take a look at what this function is, what I see is that it is a sum and a difference. And just as a note, we determine that by considering what is the last thing we do before we evaluate the expression x cubed minus 4x squared plus 8. And the last thing we have to do here is we have to add and subtract these quantities x cubed 4x squared and 8. So what that means is that I have the derivative of a sum and a difference. Well, the derivative of a sum or the derivative of the difference is just going to be the sum or the difference of the individual components. So as a first step, I can rewrite this. The derivative is just going to be the derivative of the individual pieces subtracted or added as they were in the original function. Well, now I have to do the actual differentiation. This first term, x to the 3, is just a function of the form x to the n, and I know how to differentiate that. The second term here, that is, let's read that out, 4 times x squared. Now 4 is a constant, x to the second is a function I can differentiate. So what we have here is constant times function. Well, the derivative of constant times function is the same as constant times the derivative of the function. And then finally, this 8 here is actually just a constant, and when I differentiate a constant I get 0. So let's now do the differentiation. So there's my x to the third derivative thereof. That's 3x to the n minus 1, 3x squared. Derivative of 4 times x squared is the same as 4 times the derivative of x squared, and then the derivative of 8 is going to be 0. And now I have this expression, perfectly good answer for the derivative, but let's do a little bit of algebra to clean that up. That's going to be 3x squared minus 8x. Let's try another one. How about function equal to square root of x minus 5 over x? And for this we have to play around a little bit with the rules of exponents. So the important thing to remember here, if I'm looking at a root, I'm talking about a fractional exponent. So the square root of x is the same as x to the power one-half, and if I have something in the denominator it's the same as having a negative exponent. So 5 over x is the same as 5x to the power negative 1, and then both of these terms now become something of the form x to the power n, x to the power one-half, x to the power minus 1. So if I want to find the derivative, I can convert them both into exponential form. Important thing to note here, I have not differentiated at this point. This is exactly the same as this. That's square root of x minus 5 over x. I haven't differentiated, so I'm going to keep the derivative hashtag. So let's see. This thing in here is a difference. There's a subtraction there, so I know that I can differentiate a difference by differentiating the individual pieces of that difference. And now I have the derivatives of x to the n for my first term, and the second term is 5 times x to the power minus 1. And again for my second term that's constant times functions, so the derivative is going to be constant times the derivative of the function. And both of them look like x to the power n, so I can bring that exponent down front one-half, x to the power of one-half minus 1. Again the exponent drops by 1. Here's 5, exponent comes out front, x to the power of minus 1, minus 1, and hardest part of the problem simplifying that expression. First step, one-half minus 1 is negative a half, minus 1 minus 1, negative 2, and 5 times negative 1 subtracted is the same as plus 5. Now while we could actually write the derivative f prime of x in this fashion here, there's a style issue in mathematics which might be described as follows. You want to give answers in the same dialect that you started with. So here the dialect that we started with involves square roots, involves rational expressions. Here we have to have these rational exponents. We have these negative exponents. Let's rewrite it so we only have square roots and rational expressions.