Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Differentiation rules derivative of a constant function. If F has the constant value F of X = C, then the derivative of C = 0. The proof F prime of X equal the limit as X + H -, F of X all over H Well, if this is a constant function, then no matter where we're at, every X value is going to have AY value of C. So if we're here at X and this is X + H, our F of X + H and our F of X are both going to be C. So we get C -, C / H zero over H. And that makes sense because a slope of a horizontal line is always 0. Power rule for positive integers. If N is a positive integer, then the derivative of X to the north is really n * X to the north -1 so S. So the original function F of X equal X to the north. Power F prime of X equals by definition. I'm going to use the alternative definition here. The limit as Z approaches X of F of Z -, F of X / Z -, X. When we plug that in, we'd have Z to the north minus X to the north divided by Z -, X. Now, this is something that you probably have not seen factor before, but this does factor and it factors into the C -, X * Z to the north -1. If you think about the Z * Z to the n - 1, that would give us the Z to the n + Z to the n - 2 * X + Z to the n - 3 * X ^2 plus... What will happen is as we distribute this out, we're going to have one term become positive and the following term being negative, and they're going to cancel each other. So really all we get left with is the very first term and the very last term when we distribute, which is that Z to the north minus X to the north. So the Z -, X is now can cancel and we get limit as X as Z approaches X of Z to the north -1 + Z to the north minus two X + Z to the north -3 X squared. Plus since Z is approaching X, we're going to replace Z with XS. So X to the north -1 + X to the north minus two X + X to the north -3 X squared plus... So we have X to the north -1 + X to the north -1 + X to the north -1 + X to the north -1. And the question becomes how many of these do we have? And we have N of them, so we get n * X to the north -1 the constant multiple rule. If U is a differential function of X&C is a constant, then the derivative of the constant times the function is really equal to the constant times the derivative of the function. Proof. Let H limit as H approaches zero of F of X + H -, F of X / H So the original function is going to be the C * U of X. So C * U of X + H -, C * U of X all over H. There's AC in common, so we can factor it out. Then we have the UU times X + H -, U of X all over H. This constant has nothing to do with the limit as H approaches 0, so we can pull the constant outside. So the constant times limit as H approaches zero of U of X + H -, U of X all over H is just the constant times the derivative of the function derivative sum rule. So if I have the derivative of two functions that I'm going to add together, I'm going to have the limit as H approaches zero of UX plus H plus VX plus H -, U of X + v of X all over Hu of X + H -, U of X + v of X + H -, v of X all over H. If I split this up into two different functions, we have EU of X + H - U of X / H and the V of X + H - v of X / H all of the limit as H approaches 0. By definition, this first one is just the derivative of U in terms of X, and the second one is just the derivative of V in terms of X. We have a derivative product rule. If U&V are differentiable at X, then so is their product UV and the derivative of the product is equal to EU times the derivative of V in terms of X + v times the derivative of U in terms of X. In prime notation, UV prime equals U prime v + v prime U. So the derivative. Here it's a little longer. We have the derivative of the product, so we're going to have the limit as H goes to 0 of U of X + H * v of X + H -, U of X * v of X all over H. Now I need to figure out how to make this so that I can get some derivative formulas out. So we are truly just going to choose to add some terms and the terms that I'm going to add is the negative UX plus HV of X and positive. And the reason I'm doing this is if I look at my grouping, then I can group this U of X plus HV of X + H -, U of X plus HV of X and realize they both had a U of X + H in common and pull it out. That leaves me AV of X + H -, v of X all over H If I split this up into the limit, I have the limit of the first one times the limit of the second. The limit of the second one, by definition, is the derivative of V in terms of X. On this first one, when H went to zero, we just get U of X. We're going to do the same concept for the 2nd portion of this limit. So we see here that they each had AV of X in common. Pulling it out, we have U of X + H -, U of X all over H. If I split that up as a multiplication of those two limits, H goes to zero. There wasn't any in this V of X, so it's just V of X. And the limit is H goes to 0 of U of X + H -. U of X all over H is the definition of DUDX. Our next one is the derivative quotient rule. Same concept. So now we're going to have U / v, and what that equals is V times DU DX minus U times DV DX all over v ^2. If we do the proof, the limit as H approaches zero of U of X + H / v of X + H -, U of X / v of X all over H I'm going to cross multiply to get a common denominator. So U of X plus HV of X -, v of X plus Hu of X all over V of X plus HV of X. Now, just like in the last one, I'm going to pull together 2 terms that are going to allow me to group. So I'm going to add and subtract U of XV of X. If I add this U of XV of X, they both have AV of X in common that I can pull out, leaving me EU of X + H -, U of X all over H. If I look at the second two terms, I'm going to pull out AU of XA negative U of X and I'm going to get left V of X + H -, V of X all over this H here and pulling this H up. The reason I'm doing that is because that's definition of the derivative as limit as H goes to 0. So I have this DUDX and the V of X over. When H went to zero, this bottom turns into VX times V of X. This next piece when H goes to zero, I get this negative U of X. This is the definition of the derivative of V in terms of X. And then H goes to zero. We get V of X * v of X in the bottom again. So V of X * d UD X -, U * d VDX all over v ^2. Simpler notation is the prime notation, which says VU prime minus UV prime over v ^2. Proof for the power rule for a negative integers. If we have DDXX to the north, where N is a negative in a negative number, what we're going to do is we're going to rewrite this to make it be positive and prove it for the positive. So let N be a negative number equaling negative M where M is a positive integer. Because we know that X to the north is really 1 / X to the M. So the derivative of X to the north equal the derivative of 1 / X to the M. Now using the quotient rule that we just developed, the derivative of the top, which is a constant, the derivative of constant is 0 times the bottom minus the derivative of the bottom X to the M power. So we're going to bring down the M and take it to 1 less power times the top all over the bottom squared. So if we simplify this up, we get negative MX to the m -, 1. If the bases are the same, we subtract the exponents so -2 M so equals negative MX to the negative m -, 1. Well, negative M back here was really N so NX to the north -1 Thank you and have a wonderful day.