INSTRUCTOR: Section 3.3, differentiation rules. These notes are from OpenStax calculus volume 1. At the bottom of every page, there is a link to the textbook, somewhere around here. And I highly suggest you go there. Because these notes and these videos are intended to go along with the textbook. 

So they go very well together. So please do that at some point, if you haven't already. All right. 

So our first differentiation rule-- actually, before we get there, I just want to say I'm sorry. Because you have to be exposed to where those rules are coming from. It's really an important process of becoming a mathematician. And that's what you are. You're studying math. 

So the how-to, the limiting process-- I think I said several times, the limiting process is the most important, most valuable part of all of calculus. So we use limits for everything in the last couple sections. 

Well, in this section, we are going to not prove things. We are going to use some rules. And these are really observations. If you use the limit definition over and over again, you'll notice some patterns, if you haven't already. 

And so these rules are the patterns. So you're going to go, why didn't you talk about this before? Why didn't you show me this first? But the fact is, you have to know where it comes from to appreciate where you're going, and the limiting process that is calculus. 

So our first rule is constants. What f of x equals c, where C is a constant. Then, f prime of x is 0. So think of it this way-- lines, no xs, just constant horizontal lines, the slope is always 0. That's what this is. 

It is a horizontal line. And so therefore, its slope is 0. So find the derivative f of x equals 8. Well, the slope of that is f, prime of x equals 0. And if you do apply the limit definition, you will get 0. 

We're not going to dwell on that one too much longer. The power rule is a really valuable rule. Let f of x equal x to the n, where n is an integer. Then, f prime of x equals n times x to the n minus 1. Or said another way, ddx of x to the n equals nx to the minus 1. 

So let's look at example 2. Find ddx, the derivative of the function x-cubed. By this rule here, our n is 3. So our n is 3. So our derivative is 3 x to the 3 minus 1 or 2. So our derivative function is 3 x-squared. 

That's all there is to that one. Example 3, differentiate x to the 10. Well, the derivative of that is going to be 10x to the n minus 1. So that'll be 10x to the 9. Go back in the last section and look at those quadratic functions and see if you can make sense of those a lot quicker, now. 

The derivative of x to the 7 is 7x to the 6. You reduce the exponent by 1. And we multiply by the old exponent. 

Example 5, find the derivative of g of x equals 3x-squared. So g prime of x, that is going to be if we multiply by 2, that'll be 6x to the 1st. And compare it to the derivative of f of x equals x-squared. So f prime of x is bring down the 2, 2x. 

All right. Well, to compare this, we want to notice this. The derivative is 3 times 2x. Well, what do you notice about the original function? It had a 3. So there's the thing. 

Since the limit can be applied over sums and indifferences, and actually constant multiples as long as those limits exist, the derivative holds, as well. So that is, if we have differentiable functions f and g, if we have a sum or a difference, we can take the derivative of them separately just as we could the limits, and add or subtract those, depending on what the original operation was. 

If we have a constant times our function, then that is the constant times the derivative of that function. All right. So let's look at number 6, here. 

f prime of x-- the derivative of that is-- we're going to take the derivative separately. The derivative of this portion right here is 10x to the 4. Our power-- we ignore the 2. We say 5 times 2, and then make that x to the 4. 

The derivative of just 7 is 0. So we would say plus 0, or not write it at all, in that case. And there is our derivative function. 

Number 7, the derivative of f of x equals 2x-cubed minus 6x-squared plus 3. The derivative of this one is 6x-squared decreasing the exponent by 1. The derivative of this is minus 12x to the 1st, or just x, and then plus 0. And that is our function. 

So for our next example, find the equation of the line tangent to the graph. f of x equals x-squared minus 4x plus 6 at x equals 1. Now, in the last couple of sections, that would have been a big ordeal. But let's go ahead and find f prime of 1. 

f prime of 1 would be the derivative function evaluated at 1. So I'm going to go ahead and write that like this-- this that is dydx evaluated at x equals 1. So dydx would be 2x minus 4 evaluated at x equals 1, which would be negative 2. 

Now, I wrote that out in the longest way I could have possible. But that is some of the notation to keep in mind. We're taking that function evaluating it at x equals 1-- the vertical bar means that. So that's the slope. 

Now, we have a point, 1 comma. Evaluate that. That's 1 minus 4 negative 3. So that's positive 3. 

All right. So if we plug that in to our point slope form, that would be y minus 3 equals our slope, x minus 1. y minus 3 is negative 2x plus 2. And therefore, y equals negative 2x plus 5. 

Things go a lot quicker. So first, find our derivative and evaluate it at that point. Then, onto our point slope, which leads us to our slope intercept form. Next, we have the product rule. 

This is if we have two functions that are multiplied. So let f of x and g of x be differentiable functions. Then, the derivative of f of x times g of x-- this right here-- is the derivative of f times g plus f times the derivative of g. 

Or you can put it in his prime notation. fg prime is f prime g plus fg prime. So let's go ahead and evaluate that with this first one. For j equals x, j of x equals f of x times g of x. Use the product rule to find j prime of 2. 

Well, by this definition, since j is f-- so let's go ahead and say, we'll add that here. j prime of 2 is f of 2 times g of 2 prime. By definition, that is going to be the derivative of f. So that's a f prime of 2 times g of 2 plus f of 2 times g prime of 2. 

Now, I'm really abusing notation with that first bit. That is not quite accurate, the way I wrote that. So bear with me. 

But I'm applying that rule to that. And that would be f prime of 2 is negative 4 times g of 2 is 1 plus f of 2 is 3. g prime of 2 is 6. 

So that is 14. All right. So j prime of 2 is 14. 

Now, I'm going to write this another way, that rule. A lot of times you might see it written like this-- uv prime equals u prime v plus v prime u. That's just another way that I often see it written, with us and vs. 

All right. So in this case, I'm going to go ahead and use that u and v. We have a product of two functions. So f prime of x is u prime. That will be 2x times v, 3 x-cubed minus 5x plus v prime. 

That would be 9x-squared minus 5. I'm using those other rules. v prime times u, which is x-squared plus 2. Now, perhaps we should clean that up just a little bit. 

That'll be 6 x to the 4 minus 10x-squared plus 9. x to the 4. 9x next to the 4. 

And then, we've got plus 18x-squared minus 5x-squared minus 10. So f prime of x was going to be 15x to the 4th. Negative 10, so that is 8 plus 3 x-squared minus 10. And there is our derivative function based on that product rule. 

Number 11. Let's go ahead and call these u and v. So our f prime of x is derivative of the first one. That'll be 10x to the 4th times our v 4x-squared plus x plus v prime. 

That would be 8. 8x plus 1 times u, 2x to the 5th. So this would be 40x to the 6. All right. 40x to the 6th plus 10x to the 5th. 

Distributing that, 16x to the 6 plus 2x to the 5th. So that f prime is 56x to the 6 plus 12x to the 5th. So remember, the derivative of the first, and then you add the derivative of the second. 

All right. One of the last things we're going to have, here, is the quotient rule. In fact, I believe is going to be the last rule that we are going to have. Yes, it is. 

The quotient rule. So rather than having a product of two functions, f of x times g of x, we have f of x divided by g of x. Now, this rule right here, f prime of x times g of x minus g prime of x times f of x divided by f of x-squared. That is our rule. 

Now, the way I remember this, the way I actually learned it, is Lo-De-Hi MINUS Hi-De-Low OVER Lo-Lo. If that helps, go for it. Hold on to that. Lo-De-Hi MINUS Hi-De-Low OVER Lo-Lo. Yodel to yourself. 

I'm going to go ahead and write this in u v notation, like I did earlier. u divided by v prime is Lo-De-HI. So v du, derivative of u, minus Hi-De-Lo over Lo-Lo, v-squared. It's a common way to see that. 

So in our case, this would be u and this would be v. So 5x-squared divided by 4x plus 3. So for our derivative, k prime of x. Lo-De-Hi would be 4x plus 3. Derivative of the numerator, there, is 10x, so times 10x, minus Hi-De-Lo, so 5x-squared. 

The derivative of the denominator is 4 over the denominator squared. So 4x plus 3 quantity in parentheses squared. All right. So that would be 40x-squared plus 30x minus 20x-squared. 

Good. Divided by 4x plus 3-squared. Going to combine like terms and factor the numerator. If I combine like terms and factor out a 10x, that'll be a 20x-squared. 

So 2x plus 3 over 4x plus 3-squared. I'm going to leave that. And the reason I'm factoring that is because, later on in the section and certainly in another chapter or so, being able to find the solutions, the places where that derivative is zero and where it's undefined are going to be very, very vital. So factor your numerator as much as you can in these cases. 

Number 13, find the derivative of this function. All right. Well, if here is our u, and this is are v, then our derivative k prime of x Lo-De-Hi. So that is 4x minus 3. Derivative of that is 3 minus Hi-De-Lo 3x plus 1 De-Lo is 4 over 4x minus 3-squared. 

So multiplying that out, I get a 12x minus 9 minus 12x minus 4 over 4x minus 3-squared. So that will be negative 13. Negative 13 over 4x minus 3-squared. 

All right. Next, we want to try something a little different. f of x equals x to the negative 4. I want to point out, this can be written as 1 over x to the 4. 

So let's take the derivative of that. If we're going to do that, f prime of x-- so that would be my u and my v, Lo-De-Hi. So that would be x to the 4th times 0. Minus Hi-De-Lo. There we go. Minus Hi-De-Lo, so that would be 1 times 4x-cubed over Lo-Lo. 

If we square that, we get an x to the 8. All right. So this is going to be negative 4 x-cubed over x to the 8, or negative 4 over x to the 5th. 

Now, suppose that rather, I did this. It'd be really nice if the power rule worked. That would made this a little easier. That would have been negative 4x to the negative 5, subtracting 1, which is negative 4 over x to the 5th. And lo and behold, the power rule works. 

The power rule works for all integers, and in fact, for all real numbers. So if you have square roots, it comes in very handy with things like that. So you can use the power rule on something with a negative exponent. 

Well, let's try a couple that are also negative exponents. Let's write these first as 6x to the negative 2. Well, in that case, that is negative 12x to the negative 3, or negative 12 over x-cubed. That is our derivative. 

So if you see something that is rational like this, it's a really good idea to write them without a rational component to them. Make them where they are x to the something. So that would be f prime of x is negative 7x to the negative 8, or negative 7 over x to the positive 8. Moving on. 

For k of x equals 3 H of x plus x-squared g of x, find k prime of x. This will require us to apply a lot of our rules. So for the first, I have a constant multiple of that. So I'm going to ignore my constant and just say 3 times h prime of x, which I don't know what h prime is. 

I don't know what h is. So just 3 times h prime of x plus. Now, for the second component, I am going to have to use my product rule. 

My product rule would say the derivative of the first. So here's my u and v. So I take the derivative of the first. That's 2x times g of x plus v prime u, so g prime of x times x-squared. 

And that would be k prime of x. So if you don't know what the functions are, we can still arbitrarily write them as derivative-- g primes and h primes. 

For k of x equals f of x times g of x times h of x, express k prime in terms of f, g, h, and their derivatives. All right. So let's go ahead and think of this as our u, and these last two is our v function. 

In that case, we would have derivative of the first, f prime of x times our v, so g of x, h of x plus derivative of the second. That would b g prime of x-- no, I can't do that. It's a product. 

g of x, h of x prime times just the first one, the u, so f of x. Well, I'm going to have to use the product rule again. Let's leave that part. 

That would be derivative of the first times the second plus the second, so the derivative of the second, so h prime of x, g of x times f of x, there. All right. Oh, do not do that. h of x. 

So if I distribute the f of x, what I get is this. f prime-- I don't know why I keep on doing that-- f prime of x, g of x, h of x plus f of x, g prime of x, h of x plus f of x, g of x, h prime of x. This is the general power rule for three functions. The derivative of the first, leave the other ones alone. 

The derivative of the second, leave the other two alone. The derivative of the third, leave the others alone. This would work for any number of functions. 

So remember that, in case it comes up. So these rules can actually apply to much bigger things. Find the derivative of h of x is 2x-cubed k of x over 3x plus 2. 

So we will use our quotient rule, here. Here is our here u. Here is our v. So h prime of x is Lo-De-Hi. So 3x plus 2, the derivative of u would be a product rule. 

So that would be, let's see, 6x-squared k of x plus k prime of x, 2x-cubed times 2x-cubed. All right. that's Lo-De-Hi minus Hi-De-Lo. 

2x-cubed k of x, De-Lo would be 3, all over Lo-Lo, 3x plus 2-squared. Now, this may not get much prettier, here. You know what? 

I'm just going to leave it this way. I'm sure it will simplify a little bit. I'm not sure it'll be worth it, especially with the space I've got, here. 

Number 20. Find the derivative of h of x is x-cubed times f of x. This'll be our product rule. So the derivative of the first, 3x-squared times f of x. There's our u and our v. Plus derivative of the second, f prime of x times x-cubed. For the sake of this big idea, here, I want to just leave that the way it is. 

All right. Number 21. Determine the values of x for which this function has a horizontal tangent line. A horizontal tangent line means we have something like this, where our slope is 0. So let's first find the derivative of this. 

The derivative of that is 3x-squared minus 14x plus 8. We want to know where that is 0. Let's see. Does that factor? 

That would be 24. 7 and 4 will not work. 3 and 8. I'll just go and use the quadratic formula. It'll work. 

So if I want to solve this, the values of x, for where this is 0, negative b plus or minus the square root of b-squared minus 4ac. Believe that's 96. 4a c, so that's 12. Yeah. 96 Over 2a, so that's 6. 

So that's 14, or that's 100. Square root of that's 10, so plus or minus 10 over 6. 

So that would be 24. That'd be a value of 4. So this would have factored. And, oh-- not quite. 4 and 2/3. 

So x equals 4 and x equals 2/3 would cause our tangent line to be 0. All right. Moving on. 

Given the position function, what is the initial velocity? Now remember, initial velocity means v of 0. And that means s prime of 0. 

So let's go ahead and find s prime. s prime of t. Well, if we have our u and our v, Lo-De-Hi. t-squared plus 1 times 1 minus Hi-De-Lo, t times 2t over Lo-Lo, t-squared plus 1-squared. 

So that would be 1 over t-squared plus 1-squared. Well, if we evaluate that-- so s prime of 0, which is s prime of t evaluated at t equals 0, plugging that in, we get 1 over 1. 

Well, that's just 1. All right. So s prime of 0, or v of 0, is 1. All right. 

Our last question, I believe-- yes, our last one in the section-- determine the values of x, where this has a horizontal tangent line. So follow the same process we did in the last one. 

First, take the derivative, so 3x-squared minus 14x plus 8. And it looks like we got this one free, because that is the same function. All right. 

So that brings us to the end of our function on our derivative rules. These are going to be very handy, and make your life a whole lot easier, as opposed to that limit definition. But remember, calculus is all about those limits. That's what it comes down to. That's it.