A Level Maths Made Easy - Differentiation AS

hey guys welcome back to the maths Made Easy Series this is the series where we go through each of the topics and we start from the basics and we work our way through to the really hard style questions now I will point out that you will have to do some of your own reading or learning in order to fully understand what we're doing here but this is aim to try and get you all the way up to that grade A or a star style of question just so you know with this differentiation we're doing things in year 12 so things like proving things from first principles uh stationary points that kind of thing also modeling with differentiation is actually a separate video because that is a huge topic with a ton of really really hard questions that'll probably be the next video or the one after that so this one is just going through the more basics of the differentiation and then right at the end there's actually a pretty tricky question that was nine marks in total which is a pretty fat number of marks as usual I'm going to put a link in the description for all of these questions plus the notes so that way you can retry them whenever you want and you can look at the notes and all that wonderful stuff without further Ado let's jump into what differentiation actually is so from first principles differentiation simply is a way of calculating a gradient now you might think well why the hell would we need that when we already have you know the good old fashioned change in y over change in X well that really only works for straight lines but if I have a curve for instance and I ask you to work up the gradient over here you have to do something like drawing a tangent but the thing is the tangent can be multiple different things like this I might say that's a good tangent but you might say no it needs to be a bit more Steep and honestly who's to tell who's right if you remember in gcsc when we did uh kind of equations of tangents and working them out working the gradients out you got a variety of answers right a range which isn't really good enough in maps differentiation is a way of eliminating that and finding an actual equation for the gradient you might think why do we get an equation well simply put can you see that the gradient will change along this entire curve here it goes from a negative value down to zero and then to a positive value it changes all the time which means it won't just be a number the gradient isn't five for instance it's different this might be minus 4 then it goes down to zero then it goes up to plus 4 or something like that and all the values in between so we actually need to get an equation so let's start off with how the gradient is actually defined in terms of differentiation the very first thing you're expected to be able to do is to prove from first principles that the derivative of kxq is 3 kx2 I decided to start with this because you don't actually need to know differentiation to know that this is true so differentiation relies on drawing tangents but making them infinitely small and precise which means we do it mathematically we do it algebraically so they are 100% accurate and here is how differentiation is labeled it is consider as the limit as H tends to zero I'll explain what the limit is in a second of f x + H minus f of x all over H so we Sub in these values we simplify it and then we make H equal to Z the common mistake I see students make is they try and make H equal to Z too quickly if I make H equal Z right now I'll get a math error because I'm dividing by zero that is not what you're supposed to do you're supposed to sub subtitute this all in first do all of the algebra simplify it and then when you can't simplify anymore then you make H equal to Z that's what this limit things mean thing means it means that if I make H Zer what happens to the expression so first of all let's define what f ofx is f ofx in this case is going to be KX cubed I would like to point out that I have picked a hard question so it's probably the hardest that you get in year 12 for proving from first principles that means f ofx plus h well all I do is I replace x with X+ H so we get K X+ H all cubed and then of course H is just H let's sub this in now I'm going to stop writing the limit although that isn't super accurate I'm going to stop writing the limit just so we can do the simplification first so we have K X Plus H cubed minus fact that should be so again technically speaking the um X cubed is the only thing being cubed the K is not being cubed in this case just so you can see that uh kind of a bit more clearly minus KX Cub over H so again that point that I just made make sure you know what's being cubed it's the X it's not the K that's why it's only the X Plus H being cubed here next thing we need to do is we need to expand the brackets and simplify I'm going to point out this is a bit annoying but again part of a Lev is doing some pretty annoying algebra so the way I would do it is I would expand two brackets first so if you remember X Plus H cubed is going to look like this so if I expand just two brackets first I will get x2+ 2 HX + h^ 2 and then I'm still timesing by X Plus H cuz I haven't done that yet minus KX cubed over H I do want to also point out that these questions are generally four marks so it's a very basic algebra so take your time with it every time I Mark year 12 assessments a good chunk of people will make mistakes in the algebra and not get the answer and then they spend time panicking wondering why they haven't got 3 kx^ S don't be one of those people take your time be slow and methodical so here we're still going to leave the K on the outside we have x * X sorry x * X which is X cubed x * H which is hx2 then we have 2 HX * X which is 2 H x^2 + 2 h^ 2 X plus h^ 2 X Plus H cubed close bracket minus KX cubed I know this seems like a lot but I promise we're going to start simplifying it now and um over H then what I'm going to do is I'm going to expand the bracket as in the K bracket but I'm also going to you know gather like terms so we have X Cub * K there's no other X cubes here so that's just KX cubed then if we gather up the HX s we have hx2 + 2 H X2 and that's it so that gives me 3 hx^ 2 * K so that's 3 k H x² then we have h 2 x h x so that's 3 H2 x * K so that's plus 3 k h^ 2 x and then lastly we have H Cub so we have KH sorry KH Cub yes I know this looks very fun I know over h Hello over cool the KX cub and minus KX Cub cancel now if I do the division what am I left with I'm left with well again if we cancel H cancels H cancels H cancels so we're left with three k x^ 2 + 3 k h x + k h squ okay now we go back to the limit so the limit as H tends to zer so let me move this over slightly again the four mark question the limb H TS to zero well if I have a look at this entire expression if I make H equals z the first thing will stay exactly the same so I'll still be left with 3 K x^2 but the K HX that would be 3 K * 0 so that goes away K * 0^ 2 goes away so I'm left with 3 kx^ 2 which is our answer now some people like to try and be dodgy and they try and expand this and then write therefore equals this you need to show the intermediate steps okay you can't just wishy-washy jump to the end okay so now we've seen what first principles are and again they can ask you to do all kinds of differentiation with that let's actually look at what differentiation is and how we use it so here we have an equation of a cubic so 2x cub + 5x2 - 7 x + 10 find Dy by DX this means here if you remember to work out the gradient it was the difference in y over the difference in X you can think of it as Dy as being difference in y over the difference in X all it means is the equation for the gradient or it means you're doing differentiation that's all it means but how do you actually differentiate something well all you do and I'm going to show you right now using part a as an example is you bring the power to the front and reduce the power by one so bring the power to the front we have 2 * 3 which is 6 x to the^ of 2 let's move on bring the power to the front so that becomes 5 * 2 so 10 x now let me show you how this technically works right if you give yourself anything for example X cubed if you differentiate it you would get 3x^2 in general this is how it works if y = to a x to the^ of n then Dy by DX is simply a n x to the nus1 that's it now you might be thinking what about this what about X well X is technically to the^ of 1 so that becomes - 7x to the^ of 0 but what is anything to the power of 0 it's one so it just becomes minus 7 if you differentiate anything that is a con not a constant but if it's just linear if it just has a power of one you just get rid of the X that's all that happens because it becomes X the^ of zero what about 10 though because 10 doesn't have a power of X well here's the way it works we look for the difference in y with respect to the difference in X so basically how much does this equation change every time I ch X so for example 2x Cub if I vary X will vary as if it's 6x2 5x^2 varies as if it's 10 x 7 x varies like 7 well does that make sense if I drew a line which is y is equal to - 7x it would be a linear it would be a constant gradient going downwards what about y = 10 that's a flat line there's no gradient constants will always go to Z Z when you differentiate them because they don't change they have no gradient the gradient of 2x Cub is 6x2 5x^2 is 10 x - 7 x is - 7 the gradient of 10 is nothing because it doesn't change it is a flat horizontal line whereas the 7x thing well if you drew the graph of Min - 7x it's a straight line going downwards right it's a diagonally downwards um graph so the gradient would just be - 7 okay what about Part B this is where differentiation gets really really really useful find the gradient of the curve when x = 2 this here is an equation for my gradient if I sub in x = 2 into the equation for Dy by DX it gives me the gradient of the curve precisely at that x coordinate how cool is that you don't need to use tangents you don't need to use any kind of approximations estimations it gives me the exact gradient so that would become six * 2^ 2 + 10 * 2 - 7 and whatever that is is my answer again all all of a levels is calculator so you can just start using a calculator for this very nice and easy assuming I can click the buttons anyway 37 so the gradient at that point is 37 again no tangents drawn no estimations no approximations that is the exact gradient at xal 2 that is the power of differentiation you don't have to estimate anything very useful so let's try it one more time but with a different equation if you notice we have this 1 /x you don't know how to differentiate 1 /x yet so we're going to rewrite this again I'm assuming you've done my indices video and I hope you have or at least you've learned it in school but you can write that as x^ minus1 where we do the same thing with Dy by DX okay so that remember that's 3x the^ of 1 bring the one forward so that's three I should write Dy by DX first got a bit too excited X to the^ of0 which is one so we don't need to write it this one's interesting though bring the power forward we get minus one then we get X and what happens when I subtract one from that I get to the minus 2 which you can either write as - 1x^2 you can write it as - x- 2 however you want but that's the answer cool now Part B again is where differentiation gets a bit more interesting now if you imagine a curve for most of a curve the gradient has some value positive negative whatever the interesting parts of a curve is when the gradient is equal to zero these are called stationary points so either you call them stationary or you call them turning points the reason for that is why if I show you if I draw a little cheeky little quadratic the point where the gradient equals to zero is that point right there why because if you look before the gradient is negative right the gradient is negative before this value and the gradient is positive afterwards which means at some point between being minus and being positive it must be zero okay and you can also do it the same way with the with a reverse quadratics so something like this starts of positive and it becomes negative which means somewhere in between it must have a gradient of zero that's the interesting point that's why they call it a turning point they's call it a stationary point because the idea there is the gradient is not moving there's no gradient at all those are the interesting points I mean how would you do that well think about it where the gradient is equal to zero find the x coordinates okay well this is my equation for the gradient right so if I'm looking for when the gradient is zero I just Sub in that the gradient equals zero and then I solve it okay so I'm going to move the the negative X over to the other side so we get 3 is equal to x- 2 it might be beneficial now to actually write this in terms of a a fraction so I'm going to times both sides by x^2 we get 3x^2 = 1 so we get x^2 = a 3 square root both sides and we get xal to squ < TK a 3 but remember when we square root now it's plus or minus because look at this it's plural it means there's more than one x value we're looking for so you will lose marks if you don't have plus or minus because that means you've only found one value you haven't found two there's two values for this one there are two turning points in this equation right here so again keep this in mind because it is a very useful and important thing and we're going to also go through how do we know if it's a kind of a this shape or this shape it's a good question Tada okay so curve has the equation y = 2x CU + 9x2 - 24x + 13 very long equation find the coordinates of the curves local maximum so in the previous question we spoke about turning points so we have three different turning points now I'm actually going to draw them without axes just to make it a little clearer we can have a point like this where we go from a negative gradient to a positive gradient that is called a local Minima minma there we are Minima that is an N believe it or not it's not two separate NS we can also have something that goes not like that at all that goes from a positive to a negative and this right here is called a Maxima if you want you can so basically the plural of Maximum is Maxima the plural of minimum is Minima that's where those fancy words I'm really not happy with how that M turned out so mean but there's one other possibility right we've gone from negative to positive positive to negative but what if it goes from negative to negative or positive to positive what if it stays the same so something that either looks like this or well if we look at the point in the middle it's negative to negative or positive to positive that is called there's two ways to call it I call it the point of inflection some people call it a saddle point which actually isn't mathematically correct a saddle point has to be three-dimensional this is only a 2d so it's a point of inflection but how can you tell right how can you tell which one's which cuz before we just made the gradient equal to zero how do I know if this is a maximum Minima point of inflection this is where things get interesting we have to differentiate again so it looks like this when we differentiate twice we sometimes call it a second order differential I just like to say double differentiating because it sounds uh cool I guess cool is that the right word probably not so how do we know if it's a minimum maximum point of inflection now here's where people get confused this second differentiation is going to tell me how now the gradient is changing I like to think of it as acceleration so in maths and physics when you have velocity it's how quickly your distance is changing your acceleration is how quickly your velocity is changing you can have a positive velocity and a negative acceleration right because you could be slowing down that's the kind of idea here can you see my gradient is going from a minus to a positive so the change in the gradient is positive because there's a positive change in the gradient it goes from minus to positive that means overall it's a positive change in the gradient I hope that makes sense so if my second order differential my double differential is bigger than zero it is a Minima Maxima it goes from a positive to a negative so it's the change is negative so that means the double differential or the second order differential is negative because it's again if I speaking in terms of velocity it's slowing down where this when it's speeding up point of inflection well it goes from negative to NE negative there's no change and here it goes positive to positive there's no change so it equals zero so what we going to do here this is a beautiful question it is a six marker but in my opinion it's a really good six marker you're going to differentiate so the first thing we're going to do is differentiate this if you are feeling a bit Brave I'd like you to pause the video now and actually have a go bring the power forward we get 6 x - 1 from the power so 6 x^2 bring the power forward that's 18 x - one from the power basically just get rid of the X it's - 24 and the constant on the end goes away because it's not changing with respect to X there's no change doesn't affect it cool now just like in this previous question to find the coordinates where the gradient is zero that will give me the x coordinates of my stationary points not necessarily the maximum the maximum will be one of them but there may also be a Minima or a point of inflection maybe there's two Maxima who knows but to get my x coordinates I'm going to make this equal to zero and solve it and here's the thing about a level maths you can solve this in whatever way you want you can try and factorize although I wouldn't waste my time because maybe you can't factorize this right maybe you can't no point wasting your time I would use the quadratic formula or the calculator so on your calculator menu down twice to the right equation Funk polinomial polinomial just means something with X with X in it basically x to the one X2 X Cub whatever the degree is just the biggest power which is squared then we just type in our values people do ask me can you get marks for this the answer is absolutely because they're not actually testing if you know how to solve a quadratic that's not the point of the question it's part of it so I have two x coordinates we have x = 1 and we have x = -4 okay now which one if any is the maximum don't know how can I check I differentiate again so we're going to do our second order differential or double differential how did you differentiate twice you do the same thing again power forward 12x and remove one from the power plus 18 and now what I'm going to do is I'm going to sub both of these values in and see which one gives me for a Maxima a negative answer so here we go d^2 y by dx2 will equal 12 you might be able to see already which one's going to give me a positive or negative but that's going to give me 12 + 18 is 30 that's bigger than zero which means you say Minima so it's not that one same thing with this one can you use your brain to kind of predict which one it is absolutely of course you can 12 * -4 + 18 - 48 + 18 which is- 30 yeah it's less than zero which is the Maxima that's the one I want am I done no read the question I would have gotten five out of six if I stopped here find the coordinate I've only found an x coordinate how do I find the y-coordinate funnily enough this is actually where students get very confused is finding the y-coordinate they're like oh do I sub it into Dy by DX no because you'll get zero right that's the point do I sub it into d y by DX s we've already done that that's not the y coordinate you sub it into the original so here's the problem with differentiation at a level especially year 12 which equation do I sub it into is the most common question I get here's how you know if it's the equation for y that is only for coordinates so if they ever ask you about finding the coordinates of subing into y Dy by DX that is when you want the gradient or turning points TP or stationary points it's the only times you're going to use that for d^2 y by dx2 you only need it if you want to verify which Turning Point a coordinate is okay so one more time if you ever want a coordinate now you might be thinking well hang on a minute it says find the coordinates or back here it say find the x coordinates I'm talking about if you want the whole coordinate you're going to be subbing back into the original if you want the gradient or the turning point or the stationary point or where the gradient is zero whatever that's Dy by DX if you want to verify so d^2 y by DX is quite rare it's only if you want to verify which Turning Point a coordinate is is it Maxima uh Minima point of inflection let's see only time use D2 y by dx2 so subbing into the original equation we can get our y value uh to this might take a bit of a bit of a Time bit of time plus 9 4^ squar I'm just put like that I can't remember if it's minus or plus- 24 plus 13 yeah good find that value and this is going to be our y value again go to menu go back to the top so 2 * 4 cubed + 9 4^ 2 - 24 4 + 13 gives me 189 that's my Y and for the very last thing I would literally write - 489 I'd even put it in a box so that way the examiner has no excuse that is a six mark question and look at it like this top thing is notes we did one line two 3 four five six lines six marks but look at the actual maths the actual maths we did differentiation you're going to find pretty easy eventually subbing in numbers into a calculator solving quadratics that's not bad the difficulty is knowing what to sub in and where the other difficulty and IID say this is probably the biggest difficulty is you guys write you're working out horribly lay it out neatly you know make it nice and logical like you're trying to explain it to someone because I promise you if it doesn't make sense to the exam and it probably didn't make sense to you you probably would have made some mistakes write it out neatly and your life will be 10 times easier this leads us to the Natural well I mean it's one of the two tricky questions I really put in here and the good news is this one is actually super super super common and once you know it once you really know it every single time it comes up so here we have a curve with an equation it says to find an equation of the tangent to the curve at the point 213 they can also ask for the normal but there's literally only one tiny difference between a normal and a tangent so the key thing here is that when you have a curve and a tangent the tangent at that point will have the exact same gradient as the curve so remember to get a tangent you need y mx plus C you need to have a coordinate to work out C and you need to have a gradient so we're going to differentiate this curve and sub in the x coordinate of two to find the gradient at that point which is the gradient of the line and then sub in the coordinate to find what C is that's it for five marks so differentiating once gives us 6 x^2 - 4 so then at x = 2 our Dy by DX will equal 6 * 2^ 2 - 4 so 2^ 2 is 4 * 6 is 24 - 4 is 20 that's it so so what's the difference between the tangent and the normal question well this is the only difference with a tangent our m is going to be the exact same value as the Dy by DX with the normal we would have to do the inverse sorry the negative reciprocal that's it so if they give you a normal instead of a tangent it's the same thing the same process you're just taking the negative reciprocal of the gradient and that's it so now we have y = 20x + C so to work out what C is because remember we do need to work it out we're just going to sub in these coordinates that's it we're done so x = 2 y = 13 based on those coordinates right 2 and 13 so we have 13 = 20 * 2 + C so that's 13 = 40 plus C so that means- 40 from both sides you get what 27 27 y so then our answer is just Y is equal to 20x - 27 for five marks I think that's really really good again the only difference if it's a normal is you take the negative reciprocal of that n value and that's it that's really as hard as it can get they can get you to do it with circles as well but remember you just differentiate it's the same type of thing right so with any curve you differentiate and Sub in whatever the coord are let's move on to the hardest question it's a pretty tricky one that came up in recent years here is the question and in my opinion it is one of the trickiest questions that's come up um but not really as well and the reason why I say that is It's Tricky for a-level students in year 12 because they don't know you're allowed to do this right and that's the biggest difference between GCC and a level is the problem solving that you have to do the math is going to be just as easy as it's been so far right right so if you found it hard it'll be hard if you found it easy it's going to be easy the difference is how we're going to use the information and remember what I always say in my catchr and gcsc they have to give you enough information to do the question and they will never give you information you don't need so in this case each of these things is useful now it tells me that g of X is a cubic the general equation for a cubic will be something with X cubed something with X2 something with X and then a constant on the end but remember we don't know what the numbers are in front of X cubed X2 X or on the end so that's my first step but now it says it's cubic in which the coefficient of x cubed is equal to the coefficient of x so in other words both of those values are the same so I'm going to say it's ax cubed plus b x^2 + ax plus d that's what that is so I've used that first piece of information the next thing is that it passes through the origin that means that when x equals 0 y should equal 0 so if I sub that in we get 0 = a 0 Cub honestly you probably worked out what this is already but I'm just being super thorough plus d so in other words all of these are zero so D is just equal to zero so now we have y = ax cubed + B x^2 plus ax that's it now we've used that last piece of information there you see what we're doing here we're just using all the information they give me in order to work things out now this third bullet point is actually where a lot of students get confused because it looks like one piece of information but it's not it's two pieces of information first of all it tells me there's a stationary point at xal 2 okay St Point means I differentiate so let's differentiate this 3 ax2 plus 2 BX plus a and it tells me there's a stationing point x = 2 what does that mean it means at x = 2 Dy by DX should equal zero oops so that means we have 3 a 2^ 2 + 2 b 2 + a should = 0 so if we expand this now we have 2^ 2 is 4 * 3 is 12 so 12 a but plus another a so it's 13 A+ 4 B = 0 and this is where people get to and they're like I don't know what to do now there's a hidden piece of information they've given you they've given you a coordinate so in other words when x = 2 y = 9 there's actually two pieces of information in that one line very devious they could have said you know has a stationary point at x = 2 and then said you know the coordinate 29 lies on the curve but instead they made it a bit harder by putting two pieces of information in the same line that's why this was a very hard question so if I sub that in now we have y is well we don't have we have N9 equals actually no let's sub it in from left to right like we're doing here so a 2 cubed plus B 2^2 if you unsure what I'm doing I'm subbing in X and Y into this equation but I'm going to put y on the other side uh just to make it a bit more but you know I'm just doing it you know equals 9 um if we simplify so 2 cubed is 8 so we have 8 a + 2 a so it's 10 a + 4 b equals 9 yeah that is correct cool equals 9 so now we have two equations two unknowns what do we do we do simultaneous equations so if I write them on top of each other I was going to drag them over but I realize you guys can't do that on on paper so let's do it like this instead I would just subtract them 13 a - 10 a is just 3 a 4 b - 4 B is 0 - 9 so a is equal to -3 sub that into which equation would be good I mean either one's probably going to be fine I would sub it into the original just so there's a little less um rearranging so 13 a that's 13 * - 3 - 39 + 4 B = 0 4 B = 39 so B is equal to 39 over 4 now remember we need to actually find the entire equation I believe the question actually says find G of X and now we're going to Sub in all of our values so G of x equals there was - 3x Cub cuz remember we worked out what a is this is our equation here plus b says plus 39 over 4 x^2 and then we have plus a x say - 3x job done for seven marks is actually not that bad um all things considered by the way they can give you up to 10 Mark questions in a level now but again I think the hardest part or at least the again it's noticing it that's the problem is definitely the third bullet point not easy Part B prove that the stationary point is a maximum well this is quite nice because we've already worked out what Dy by DX is all we need to do is do d^2 y by dx^ 2 so in fact I'm going to write it in terms of a being X sorry A and B so we're going to have 6 a x + 2 b then I'm going to sub in the values we worked out so a was -3 so 6 * - 3 is - 18x 2 b 2 * 39 over 4 will give you 39 over2 but you're more than welcome to check that out and then prove it's a maximum well the x coordinate was two so when x equal 2 D2 y by dx^ 2 again I would recommend always writing out the um substitution equals and let's work it out again I know it's negative but I like to just be really thorough in my answers oops - 18 * 2 plus 39 / 2 gives me - 33 /2 that is less than zero and remember the reasoning from before it means it's a maximum because we're going from a positive gradient to a negative gradient so overall the change in the gradient is negative changing gradient is d^2 y by dx^ that brings us to the end of the standard differentiation as I'm going to call it then in the next couple of videos we're going to be doing the modeling with differentiation which is where it gets honestly pretty spicy because there's so many different situations they can put you in and they are very tricky um but of course we're going to go through every single a level topic eventually that includes the applied stuff so stats mechanics um that should be very fun again I'll give you the link to all of the notes and stuff in the Des description it's on patreon but remember it's free so you don't have to pay for it it's just that's why I host it cuz it just happened to be the most convenient thing for me but I hope you have a wonderful day and I'll see you in the next one