Okay, so let's start off with differentiation, which is all we'll do this whole two-hour lecture. So as I was saying a couple minutes ago, is that the differentiation basically describes the rate of change of one variable with respect to the other. Okay, the method that you actually derive an equation gets you the gradient of the function, okay?
Obviously, of the original curve, okay? The derivative is the, oh my gosh, this is so hard to explain. The gradient function of the original curve is known as the derivative.
which is the gradient function. Yeah. Okay. I'm going to give you a bit more less wordy stuff in a sec. Okay.
The derivative function measures the rate of change of the gradient at any x value. So yeah. Okay.
I'm not going to bother with any of this, but I'm going to go into this. So the actual notation to write your first derivative is f dash x y dash or dy dx. Okay so let me get my pen out let's take a look so I'm sure most of you guys would be quite familiar with the notation of like f of X is equal to blah blah blah So the way that I can write the first derivative is f dash x is equal to blah blah blah. Okay, so the first derivative is just one dash. Okay, then we have these notations.
Pretty much, you know how we'll write y is equal to blah blah blah. You can write y dash is equal to blah blah blah. Or if you want to be super fancy. dy dx is equal to blah blah blah. So basically if I'm going to actually describe dy dx to you, if you're going to say it in words it will be the rate of change of y with respect to x.
So the y is at the top so the rate of change of y with respect to x. This is some more high level stuff that you can go a lot more into, probably not today, but maybe not even in year 11. Definitely when you get a little further in year 12 though. Okay, next slide. Okay, so as you know, and you've learned already, functions can be either continuous or discontinuous.
Okay, continuous functions have, you know, no discontinuities. So they have... no holes in the graph, no asymptotes, and they have no jumps. It's important to actually note whether or not there will be any discontinuities, because the function does not exist at the point where there is no derivative at the point.
Sorry, there's no derivative where the function doesn't exist. Okay, let's go into differentiation by first principles. It's a little bit of a complex formula.
I will explain it to you guys. But this little rule will get you the first derivative of any basic function. Okay, so it's f dash of x is equal to the limit of h approaches 0 of f of x plus h minus f of x. all on h.
Okay, I think we've got an example in the next slide. Okay, here we go. Find the derivative of f of x is equal to x squared plus 4x plus 4. Okay, so I'm going to write the formula up for you. So f of x is equal to the limit of h approaches 0 of f of x plus h minus f of x All on H. I'm gonna give you a second to try this on your own.
I don't expect a lot of you guys to actually know how to do this so don't fret. Okay. Just for the interest of time, I'm going to get you to pause the video if you're not yet done, work through it, and then come back and take a look at my working, okay?
So let's change my pen color because I don't like red. Let's go to orange. Okay, so I'm going to sub everything in.
So f of x is... equal to the limit of h approaches zero. For those of you wondering why it's the limit of h approaches zero and why you can't just make h equal to zero, it's because of the denominator down here.
If h was equal to zero, then anything on zero is undefined if you chuck it into your calculator, so it just won't work. So that's why it's the limit as h approaches zero. And I'll explain why we even bother doing that.
in just a sec. So let's first use this example to explain it. So let's do f of x plus h.
So that will be x plus h squared plus 4 x plus h plus 4. So that's f of x plus h minus f of x. So minus x squared. plus 4x plus 4. Okay, and that's all on h.
So let's expand things up. So that's x squared plus 2x h plus h squared plus 4x plus 4 h plus 4 minus x squared minus 4x minus 4. all, oh that was an ugly line, all on h. I'm going to change my pen color and we're just going to take out any, cancel anything out that can be cancelled.
So 4x can be cancelled, so can x2 and so can 4. Okay so let's rewrite this to make it look a little bit prettier. So it's equal to of h approaches zero of h squared plus 2x h plus 4h on h. Now what I'm going to do is I'm going to put h out the front as a common factor so lim h approaches to zero of h h plus 2x plus 4 all on h.
Now what I can do is I can cancel these h's out. Okay so it's equal to the lim h approaches to 0 of h plus 2x plus 4. But now I don't have the problem of the limit of h approaches zero because if I make h equal to zero nothing's going to um you know explode my my like function won't be undefined anymore because I don't have h in my denominator so let's just make it equal to 2x plus 4 because I can get the h to equal to zero so my final answer is therefore f dash of x is equal to 2x plus 4. Okay, hopefully this process made sense. I think we have another example in the next slide. No, we don't.
Never mind. But I'm going to get you guys to take a look at this original function. Here it is. And now let's take a look at the derivative function.
It looks Find the similar? Maybe, maybe not. But I want you to take a look. We have a 4 and we have a 2x, and I'm just going to leave you with that in the moment. So I'm going to actually explain the general rule for taking the derivatives.
And that is when f of x is equal to x to the power of n, where n is any number, then the derivative is f dash of x is equal to n times x to the power of n minus 1. Okay, so let's do our example of y is equal to x squared. Then y dash will be equal to, so n is equal to 2. So I can bring my power out the front. So n times x to the power of 2 minus 1. So it'll just be equal to 2x. Okay, but yeah, you're gonna bring out the power to the front and then reduce the power up the top by one, okay?
And then the derivative of a sum, so f of x is equal to gx plus hx. Well, basically, all you need to do if they're being added together is you're just going to individually differentiate everything. Going back to our original example of y is equal to x squared plus 4x plus 4. Okay, I'm going to give you a quick second to try and differentiate that on your own, just using that rule that I just introduced you to.
Okay, let's oops so annoying Cool. Okay, so Let's do the x squared first. So it will be bring the power out the font. So to times x to the power of 1 plus we have an invisible 1 over here so it'll be 1 times 4 times x to the power of 0. Okay and then for just the 4 we have an invisible x to the 0 already so what I'm going to do is I'm going to have it like 0 times 4 to the x to the negative 1 so my overall differential will just be 2x plus 4. So that is another method of doing the differential. Okay, and you can see it's a lot less working and a lot easier as well, less time consuming.
Okay, then we have the derivative of a multiple which I've kind of already showed you in the derivative of 4x. If we have a some constant out the front then the constant itself won't be differentiated, the coefficient. So the example of y is equal to 4x, the 4 itself will just stay the same.
So it'll be 4 times the derivative of x, which will just be 1. So it'll just be 4 times 1, which is 4. Okay so the actual coefficient stays the same. Okie dokes. Okay so here are a couple of examples. So y is equal to x cubed plus 3x squared plus 5. Take a look at the power and the coefficient that's brought out the front. Okay, so the 3 over here, oh, yeah, there's a bad color.
Let's pick a different color. Let's go purple. 3 over here is brought out the front, and this reduces to 2. Okay, the 3 out the front over here is just a coefficient.
We're going to keep it out the front. So it'll be 3 times 2 times x to the power of 1. Okay and then 5 has an invisible x to the power of 0 so it's being times by 0 which is why the 5 just disappears. Okay? Okay so then we have the example of this one which I find a little bit annoying sometimes when we have the a half out the front because you still need to think of it as though it's any other number. So the a half is still brought out the front and then a half minus one gets you negative a half.
So it's one on two times x to the negative a half aka one on two root x. Okay um the x squared we've done a couple times and then the two just disappears so yes you do you just vanish. Okay so let's do these two examples. So differentiate the following, y is equal to x squared plus 7x plus 5. And then we have a little bit of a tricky one. We have f of x is equal to x to the negative 4 plus 2x to the power of a half minus 3x.
Okay. And also, yeah, just a quick note. You only need to differentiate by first principles when specifically asked.
So the question will be like, differentiate this using first principles. That's when you use it. It's something that can be assessed in the HSE. It'll only be one question if it is asked. But obviously you want to get as many marks as you can, and it is a big part of year 11 prelims.
I would recommend doing a little practice on differentiation by first principles as well. But yeah, otherwise just use the basic rule that I just went through. Okay, I'll give you like 30 seconds more to do this question.
Okay, let's get started. So for the first question, I'll just rewrite the actual question. So y is equal. to x squared plus 7x plus 5. Okay, so I'm going to separate them out, we'll do x squared, we'll do 7x, we'll do 5. So y dash will be equal to, we're going to bring the power of 2 down the front, so it'll be 2 times x to the power of 2 minus 1, so just 1. So plus 7, the coefficient stays out the front, times, we have an invisible 1 over here, so we're going to bring the 1 out the front, 1 times x to the 0. Okay, then we have our invisible x to the 0 over here, so it'll be plus 0. So our y dash will be equal to 2x plus 7, and that's the final answer.
Okay, so that's for A. Let's do B. For b, f is equal to x to the negative 4 plus 4x to the 1 half minus 3x.
So f dashed of x is equal to, let me make that a little bit smaller, this comes here, this comes here, this comes here. Okay. So I'm going to bring negative 4 out the front so the negative also is included in my power. So minus 4 times x, guess what the power is going to be this time? If you guess negative 3, unfortunately that would be wrong because negative 4 minus 1 is equal to negative 5, okay?
Plus my coefficient of 4 out the front. times by my power of a half times x to the a half minus one, which is minus a half. And then we're going to bring the minus three times one times x to the zero.
So it'll be minus four x to the minus five plus a half on two. Well, sorry, four on two, which will be 2x to the negative 1 half minus 3. And that is my final answer. Okay.
Hopefully, this process made a little bit more sense now that we've gone through a couple of questions. Okay. Now let's go into finding the gradient.
So we're actually going to start applying differentiation now. Okay. So Once we've actually learned how to differentiate, we can now differentiate to find the gradient at a certain point on the curve. The derivative, as we were saying before, is the gradient function because we can sub in any x-coordinate and we'll get our y-value for that specific x. Subbing in the x-coordinate into the derivative, so y-or f-x, will result in the gradient at that specific x-coordinate.
Okay, I'm going to show you how to do a because it's going to be a bit tricky if I just like say do it and then I'm going to give you a sec to do b by yourself. Okay, so a y is equal to x squared plus x minus 6. Let's do y dash. Y dash will be equal to So x squared, we're going to bring the 2 out the front, so 2x to the 1, so just 2x, plus x to the power of 1, which bring the power down and then subtract 1 by 0, which will just give me 2, plus 1. And then the minus 6 disappears because we're Thanos.
Okay, so at x is equal to 2, y dash is equal to 2 times 2 plus 1. which is 5. Therefore, the gradient at x is equal to 2 is 5. It's quite simple, right? Okay, I'm going to give you a minute or so to do b by yourself. Okay, let's go through this second question. So find the gradient on the curve y is equal to x squared minus 5x plus 4, where the gradient of minus 3. This is tricky because it's actually the inverse of what we just did. Okay, so b.
y is equal to x squared minus 5x plus 4. I'm going to make my y dash function. So x squared becomes 2x minus 5x becomes minus 5. 4 disappears into the ethers. Now, what I'm going to do is I'm going to say x is equal to what? We'll make y'equal to minus 3. right and now I need to solve for that so 2x minus 5 is equal to minus 3 so let's do our reverse algebra let's change the color to yellow again I'm going to plus 5 plus 5 so 2x is equal to 8 x is equal to 4 now I'm just going to recheck that because just in case I made a mistake. So let's sub x is equal to 4 into y dash.
And I can just right check. y dash is equal to 2, 4, minus 5. And you can see that I did indeed make a mistake. So let's redo that. So minus 3 plus 5 gets me to 2. So x will be equal to 1. Now let's check that again.
Check sub x is equal to 1 into y dash. So y dash will be equal to 2 times 1 minus 5, which is indeed negative 3. Therefore, gradient, or if you want it to be a bit fancy, you can write So the gradient of the tangent is equal to negative 3 at x is equal to 1. Okay? So that is how you can approach a question like this. I have another question for you guys to do real quick.
Okay? So down below, tell me what the answer is. If y is equal to 3x to the power of 4 minus 2x to the negative 1 half, plus 3x plus 1, then what will y dash be?
I'm going to give you a few seconds to do that. Okay, hopefully everyone's had a chance to answer the question below. If not, I'll give you a sec to pause the video and have a go. But I'm going to get started.
So y is equal to 3x to the power 4 minus 2x to the negative 1 half plus 3x plus 1. So y dash will be equal to 3 times 4x cubed minus 2 times minus a half x to the minus, what's minus a half minus 1? That's equal to negative 3 on 2 plus 3 times x to the 0, so just 3 times 1 plus 0. That's going to get me to 12x cubed minus minus, right? So that's going to get me to plus x to the negative 3 on 2 plus 3, okay? So that answers my question to C.
So anyone that got C, good job. It was definitely a bit tricky, especially with the double negatives.