A LEVEL MATHS: Differentiation Exam Practice LESSON 12

Hey guys, just to let you know that you can catch these lessons live every Thursday from 6pm until 7pm, UK or BST time. And I'll just allow you to ask any questions that you want and get some answers to them. Without further ado, let's get on with the lesson. Hey guys, what's up? Welcome back to another A-Level Maths Revision Guide. Revision Guide? Revision Lesson? So today we're going to continue with what we did last week, but instead we're going to go on to the proper differentiation. So in general, if you have something that looks like this, then when you differentiate it, all you need to do is, actually let's change that square, let's generalise it to an n, is you bring the power forward and then you have x to the power of n minus 1. So for example, if we do a couple as an example, if we have 3x squared, then dy by dx, you're bringing the power forward, so it's 3 to x, so it's going to be 6x. So again, when we did the first principle stuff, you should know that these are some of the results we got. Likewise, if you have something else, let's say you have y equals 4 over x, you would first need to change this using your rules of indices to 4x minus 1, and then when you differentiate, you're going to again do the same thing, bring the minus 1 forward, x to the minus 2, so it's minus 4x minus 2, or minus 4 over x squared. Now this seems quite basic, but there are a few other things that we have to know. First of all is something called stationary or turning points. So with most coordinates on a graph, they're actually kind of boring. So if I were to draw a generic curve, let's make it a cubic, most of these coordinates are boring. However, some interesting stuff happens here and here. You can also have here. And what happens is that the graph here goes from a positive number to a negative, a negative to a positive, and then here it's also a positive to a positive. You can also have a negative to a negative, which would just be the same as this graph but flipped around. And we call those stationary or turning points, because the gradient, which is in this case represented by dy by dx, is equal to zero, it becomes flat. So we also need to be able to evaluate stationary points. Now here's where things can get a bit interesting. If we're looking at stationary points, we can work out where the stationary points are, but we can't tell what they are, if they're positive, negative or otherwise. So how can you tell? Well, one way that you can tell is you can draw the graph and have a look. So if you get coordinates like negative four, and let's say this was plus one, well, you can see that the coordinate minus four corresponds to a maximum point. So this is a maxima, because it goes from a positive to negative. And you can see the one at one is a minima. This one here is called a point of inflection. So how can we tell which one it is mathematically? Because sometimes we don't want to draw the graph, or we don't know what it looks like. Well, that means you're going to take the second order derivative. So this is really just fancy talk for differentiating again. If you differentiate it again, and it is greater than zero, i.e. the rate of change is positive, then you're looking at a minima. And this is because the change in the gradient is positive. In other words, it's going up, just like this curve here. If you find that the change in the gradient is negative, then that means you're going from a positive to a negative, so that means you must have a maxima. And if you get that it's equal to zero, this is where we have our point of inflection. Because as you can see, it goes to zero for a bit, and then it just continues on. So overall, it's just not changing. So those are the things we need. And this does look kind of scary, the notation. But again, all you're doing is differentiating the same thing again. So for example, if we were to take this and differentiate again, well then we just do the same rules. So that's enough talking, let's get into actually solving some equations. So we're going to start off relatively easy. So first of all, we have this equation, and it asks me to find dy by dx. So let's do it. So once again, we're going to bring the powers forward. So we have 2, 3, x to the 2, plus 5, 2, x to the 1. So again, we don't really write the 1. Now what happens when there is an x, but it's brought to the forward? Well, this is basically 7x to the 1. And now it becomes 7, 1, x to the 0. But anything to the power of 0 is just going to equal 1. So we can kind of ignore it. Now with constants, they just go to 0. So if we simplify this, we have 6x squared, plus 10x, minus 7. And now they're saying when x is equal... So for part b, it just says find the gradient of the curve. Well, this thing here is an equation for the gradient of the curve. So find the gradient when x equals 2 just means we're subbing in x equals 2 into this expression. So 6, 2 squared, plus 10, 2, minus 7. 2 squared is 4, times 6 is 24, plus 20, minus 7. So that's 44 minus 7, which is 37. That's it. Not too shabby, hopefully. Here are some other things that I can get a bit interesting. So we have y equals 3x plus 1 over x. So I'm actually going to rewrite this again. x to the minus 1. And we need to first find dy by dx. So again, this power just disappears, right? Because it's 3x to the 0. So if we write x to the 0, that becomes 1. Plus, bring the minus 1 forward, x to the minus 2. So in other words, dy by dx equals 3 minus x to the minus 2. Now it says find the x-coordinates of the points where the gradient is 0. Now where the gradient is 0 is our turning points, but we don't need to differentiate again. When you want to find the coordinates of the turning points, you only take the first differentials of this thing. So all I'm doing is I'm solving 3 minus x minus 2 equals 0. Because again, turning points are when the gradient equals 0. The gradient is dy by dx. It's not d squared y by dx squared. So it's not the second differential, it's the first. So that's a key mistake that certain pupils make, is they don't know whether to make y equals 0, dy by dx is 0, or d squared y by dx squared is 0. So talking about the gradient, we're always going to take the first differential as 0. Now if we were to solve this, we can get plus x to the power of minus 2 to both sides, so we get 3 equals x to the power of minus 2. And then what we can do is times both sides by x squared in order to get rid of the minus thing. So we have 3x squared equals 1. If you're unsure about that algebra, we have x squared currently. And if I times that by x squared, I get 3x squared equals x squared over x squared. So all we need to do now is divide both sides by 3, so we get x squared equals third, and then x is just going to equal plus or minus the square root of 1 over 3. And those are our two x coordinates. And this should make sense because if we have a graph like this, it will have two turning points. The other kind of notation that they can give you is f of x. It's fairly common. So in this case, what we're going to do is once again bring those to the top. Now what they mean by f prime x, that's what that little apostrophe means, is the exact same as just doing dy by dx. So in this case, bringing the powers to the front, we have 3, 3 over 2, x to the half, plus 3 times minus 2, x to the minus 3, minus 6. So simplifying that now, that's going to be 9 over 2, x to the power of a half, minus 6, x to the minus 3, and then just minus 6. So hopefully that's all okay. I'll give you a second just to process that. Again, just let me know if it's okay or not. It's not too bad yet, but we're going to come on to some exam questions in a second. I just want to make sure that this is all okay. And then we'll do some more exam questions. So again, let's write this out in terms of indices. So when you have a square root of a number, that just means the number to the power of half, plus, this might be a bit tricky, but a half, x to the minus 1, plus 10. So then if for part A we're finding dy by dx, bring the half to the front, so we have 4 times a half, plus a half, times minus 1, x to the minus 2. So that would be 4 times a half is 2, so we have 2. I forgot to put the x to the power of minus a half, so 2x to the minus a half, minus a half, x to the minus 2, for 3 marks. And then for part B we're just taking the second order differential, so doing the exact same thing again. So that's 2, minus a half, x to the minus 3 over 2, minus a half, x to the minus 3, so that becomes minus 1, x to the minus 2. Hang on a minute, I didn't actually bring the power forward. That would be times minus 2, so it would be plus x to the minus 3. And that would be it. So we're close to the exam questions, which is where things will get a bit more interesting, but again you need to make sure you're okay with the kind of basic principles first before we go on to doing the exam questions. So hopefully that's all okay. And again, just stop me in a second if it doesn't make sense. The next kind of style of question they can give you is something that looks something like this. So this is where some students can get a bit confused. So what I'd recommend doing is if we have a divide by x, is all we do is we divide everything on top by x. So that would be 2x to the power of 1, minus 5, and then plus 3 over x, which is the same as just having 2x minus 5 plus 3x to the minus 1. Then all we need to do for part A is differentiate, which is equal to 2, minus 5 goes to 0 because it doesn't change, so there's no rate of change, and that becomes minus 3x to the minus 2. And then lastly for part B, all we need to do is find the gradient when x equals 3. So when x equals 3, dy by dx equals 2 minus 3, which is 2. So in other words, that's 2 minus 3 over 9, that's 2 minus a third, which would be 5. Perfect. So again, I don't think this is too bad once you've practised it a little bit. Cool. Again, so we have just a few more styles of questions before we go on to the exam questions. This one can be fairly interesting, so part A once again. If we're just differentiating, bring the powers forward and subtract 1 from the powers, so we can start doing it without even thinking really. And now, here comes the interesting bit. It now asks us to find the range of values of x for which y is increasing. So for y to increase, what does that mean? It means the gradient is positive, because otherwise the y would be decreasing or staying the same. So all we're trying to look for is dy by dx is greater than 0. So this is a quadratic inequalities question in disguise. So we have 3x squared minus 8x minus 3 is greater than 0. So you can use a quadratic formula or you can factorise whatever one you prefer, so times to minus 9, add to minus 8, so minus 9 and plus 1. So we have 3x squared minus 9x plus x minus 3 is greater than 0. Again, if you need help with the factorising, just let me know. 3x x minus 3 and plus 1 x minus 3. So we have 3x plus 1 x minus 3 is greater than 0. So if we think about this, we have either x is equal to minus a third and x is equal to 3, but those aren't our values, because we want a range of values, because we need to make sure we're looking at the inequality part. So again, even though this is a gradient question, it's differentiation, there is actually quadratic inequalities here. So if we were to draw out this graph, we have two crossing points here and here, minus a third plus 3. We're going to have a graph that looks something similar to this. And we want the values for which it's positive. Because again, for y to increase, which is what the question says, the gradient has to be positive. So we're looking for these values here and these values here. And that occurs when x is less than minus a third, not less than equal to, or x is greater than 3. That would be your final answer. So again, it's quadratic inequalities, which is interesting, because it's only three marks. I'd say it could be a four mark question, it should be, but yeah, some of these older questions are a bit brutal. So again, they can combine topics like this. This is why your understanding needs to be very strong. If anyone needs any help with that, just let me know. Perfect. Okay, now here's the question that I was kind of warning you about earlier. We have, find the coordinates of the curves at local maximum. Again, even students that are really good at differentiation will mess up on this kind of question. And we'll go through it and I'll point out where the common mistakes are. First of all, dy by dx, as usual, that would be 6x squared, because again, 2 times 3, and then x, plus 18x minus 24. That bit should be pretty fine. Now, what we're looking for is this is a local maximum, so that is a turning point. Now, if it's a turning point, then we're looking for the values where this is equal to 0. And what I can do is I can use the quadratic formula, or factorization to work it out. I'd recommend quadratic formula. So if we just use the quadratic formula, I'll show you the two values of x you're going to get. So we have minus b, so minus 18. And then again, we have plus minus, so let's start with plus, b squared, so that's 18 squared, b squared minus 4, a is 2, c is, sorry, a is not 2, I made a mistake there, a is 6, and c is minus 24, over 2 times 6. So 1, well that's anticlimactic, we get 1, or, this equals, minus 4. So now the question becomes, okay, which one of these is going to be a maxima, and which one is going to be a minima? Well, there's two ways to do it. One way, you can sketch the graph, but that's a bit painful sometimes, it's a bit long and tedious. The other way to do it is to find d squared y by dx squared, which would be 12x plus 18. And now we're looking for the value of x that would give us that this is equal to less than 0. So when x equals 1, d squared y by dx squared equals 12, I mean, to be honest, you can probably tell just from looking which value it's going to be, just based on the equation, that is going to be 30, which is greater than 0, so it's a minima. Which means the other one has to be the maxima, and that makes sense, right? Minus 4 times 12 is minus 48, plus 18 is negative. So this one is our maxima. So now, how do we find the y-coordinate? Well, it's simple. We sub it into the very original equation, and that's where students mess up. It's when you're subbing into three different equations. We're going to sub into this equation, we've kind of subbed into this, we've used it to work out two values of x, and we've subbed into this third equation. Students can get very confused with that. d squared y by dx squared only gives you, only tells you, if it's a max, min, or point of inflection. You have to have a value of x. You can't solve this and get a value of x. It doesn't make sense. Because it's not giving you a gradient or a coordinate. So be really, really careful with that. So, all we need to do now is say when x equals minus 4, y equals, and I'm going to use a calculator for this because I'm going to be lazy, 2 minus 4 cubed. Again, if that doesn't make sense, please let me know. But just be really, really careful. A lot of students will find this topic relatively easy and rush through it and as a result make a few mistakes. Which, again, these questions can give you, this question here would be 5, 6 marks. So it is worth doing. And this would be, again, a mock question, most likely. So come up in your mocks, which determines what kind of universities you can apply for. So, it is quite important. And, minus 24. 125. So my coordinates are minus 4 for the maximum. Okay, so I'm hoping that that's all right. Okay. And here we come to the first actual exam question. So the previous questions are exam questions, but they're kind of old spec. But I picked them mostly for how they test the different aspects of what they can ask you. So here is an exam question from far more recently. 2019 paper, actually. So, covers equation this. Find the equation of the tangent of the curve at the point 2, 13 by answering the form y equals mx plus c, where m and c are about integers to be found. So, if we want y equals mx plus c, there are two things we need. One thing is the y-intercept and the other is the gradient. Now the y-intercept is very difficult to work out without the equation of the line, and without knowing what it looks like. So we have to work out the gradient. Now tangents, right, have the same gradient as a curve at this point. Gradients equal at this point. Actually, you know what, let's not be lazy and write that. Write it out properly. The gradients are going to be equal at this point. So what I can do is I can differentiate the curve, sub in the x-value, get my gradient, and sub it into the equation. Five marks, not too bad. So first of all, dy by dx equals 6x squared minus 4. Now all we have to do is say that when x is equal to 2, dy by dx equals 6 2 squared minus 4. That's going to be 2 squared is 4, times 6 is 24 minus 4, so it's 20. So now my equation for our tangent so far is 20x plus c. If you have a look, we have a set of coordinates, so we can sub those in. So now we have 13 equals 20 times x, which is 2, plus c. So 13 equals 40 plus c. So c is equal to negative 27. So now, and again, I can't stress this enough, you have to actually write out what the equation looks like. So y is equal to 20x minus 27. That's your five marks. And it only took me about two minutes. If that. That small amount of work for five marks. So again, just think about what you need and what you can get. Combine those two things and you should be able to work out pretty much anything in A-level maths. I gave myself way more room than I needed. OK. This kind of question we've come across before. An example question like it. And let's go through it again. So first of all, we need to write y in terms of indices. So we want to change this to x to the minus 1. Now for part A, we're going to just simply differentiate. Easy, right? Three marks for that. Not too bad. Now for this, hence find the exact range of values for x, which is increasing. All we will need to do, again, as I showed you before, dy by dx has to be greater than zero. So we have 6x minus 24. I'm going to write this in the other way, over x squared. Has to be greater than zero. So if we add to both sides, we have 6x greater than 24 over x squared. Times both sides by x squared, we have 6x cubed. It's greater than 24. Divide by 6. 6x cubed is greater than 4. So x is just greater than the cube root of 4. And that's our answer. Not too bad. I think this is the last one. And that was the last question. So that's all of the AS differentiation, but I cannot stress this enough. These were all from AS papers. So those mocks that determine what grades you're going to get in your predicted grades, which are the grades you use to apply to university, that's where these questions are from. Now this doesn't mean you can neglect this, because next year, in A2, differentiation gets put on steroids. There are so, so, so, so much more. And it gets pretty difficult. So you need to make sure that you're OK with this in order to move forward. In your textbooks, you may find some more difficult questions. Just so you're aware. If you would like me to answer any questions, feel free to email me. My email is in the description, as it always is. Mitchell.educatio at gmail.com Let me just type that out for you. There is no capital. It puts capital there because it's annoying. And again, on Reddit, you can kind of see me posting in some subreddits. It's just there. But just so you guys are aware, there is a class on Saturday in the morning, which is going to be... Sunday morning, sorry. So from 10 until 11.30 that I'm going to run online and also in person in a school in Hounslow where we do A-level maths. So just so you're aware, if you are interested in signing up, again, if you don't know where Hounslow is, then don't bother with what I just said. But if you are, you can email me at this email address just to book your place. And again, there's going to be an online session on Saturdays as well if you're interested in that. Other than that, I think we can finish for now. So I hope you guys have a really good evening.