[Music] hi everyone welcome to the April 2023 math Advanced um lecture for year 11 um my name is cisha and I'll be taking you through today's um series um so let's get started so I am obviously part of Aon so I'm just going to give you a little intro to AAR notes as a company so we've been around for a while and we've had our online um well actually our inperson lectures since 2016 so 2016 to 2019 we were running our lectures in UTS in the holidays um then 2020 hit and we moved to our online lectures which were a big hit um bigger than we even were in person because we were able to reach a lot more people um and this time we're trying something a little bit new we're trying out pre-recorded lectures so I am not actually talking to you live at the moment I recorded this a couple weeks back um but do not fret there's a Q&A section down the bottom I am actually there right now like actually answering your questions just typing instead of verbally um so still ask as many questions as you want um and I'll be sure to answer them all um a little bit more about atar notes is that we have a bunch of free resources so obviously um you've got the free lectures which you're at right now we've got um discussions um like our forums our monthly newsletter and something else that we're quite known for is our atal calculator as well we have a bunch of resources that's all for free for you guys um but if you were after something a little bit more we've also got our own um like paid resources as well so I as well as a NES I am one of the tutors for the a NES like tutoring system which is called T smart so I do the math Advanced for the HSC um and I do private tutoring for a lot of subjects um including year 11 maths so if you guys are interested I think there'll be something probably down below about that as well um we've got our study guides which is probably something that we're just as popular um and known for as our holiday lectures um and then we have this resource called Ed unlimited and Ed unlimited is kind of a little of a rockstar in that it actually has all of our study guides put into like a website pretty much it's Netflix but for the HSC um so that's also something that you could probably take a look at there'll probably be a few um discount codes down below as well for you to take a look at okay and um before I properly get started with this lecture I'd also like to give a massive shout out to um our sponsors um because if it weren't for them these holiday lectures wouldn't be occurring so thank you to them um also before I properly get started I'd like to before I even introduce myself give a shout out to you guys so you guys number one have taken the actual effort of getting up getting onto your computer and watching this lecture right now so that's a shout out to you in itself but also I'd like to congratulate you on completing your first term of year 11 I think that's like year 11 is definitely a big jump from year 10 and you've definit you've already conquered it so a big congratulations to you guys for that you're a third of the way there already which is kind of scary but also super cool so good job on that so I'll quickly introduce myself I graduated in 2021 with an ATA of 95.8 five um the subjects that I did include math Advanced extension one Chim physics DT and English Advanced um currently I do veterinary science at usid so I'm second year um so if you have any questions about that you can also pop that down in the questions below um I am you can tell by my degree a huge animal fanatic I do love to rescue and Foster um animals I I Foster a lot of kittens at the moment I've got a little Foster kitten I'm not sure if he'll make an appearance today just because if he does come in into my room um he'll probably be biting down the camera and everything so we'll see how it goes um another fun fact about myself is that I absolutely love rock climbing but I really cannot say that I'm very good at it because I don't have the strength for it but I still really enjoy it um okay so just an overview of what will be happening today what we'll be going through so there's going to be two like content blocks um that will kind of be smooshed together because we're like doing a whole new like method of presenting to you guys um and I know that this isn't a live stream that we had like last term in the Christmas holidays but I'm still here to respond to any of your questions down below um so ask me literally anything nothing is a stupid question and I am more than happy to answer okay um so all of this is brand new content for you guys um which you haven't encountered and it's perfectly okay to be lost um I definitely was when I was in your position um and I know the content that you'll be learning even year 12 students can are lost with this content as well so um today will be pretty much just an intro to calculus um so the actual content blocks we'll start with just some frequently asked questions because um even though the Q&A is down below I just wanted to address questions that I get asked all the time um um followed by something that I like to call the art of solving questions um pretty much just a method of maximizing your marks so you don't make silly mistakes um something that I started to use a little bit later in my HSC Journey that I wish I'd known earlier so I wanted to get you guys started nice and early um we're also going to start with differentiation and then we're going to go and break it down into all of the um like differentiation rules um most of these rules would mean nothing to you yet but don't worry in under two hours time you shall know it all okay so let's get started okay the frequently asked questions uh the first one that I get asked a lot is how much time should I spend studying math each week how much time should I be studying um for all my subjects Al together um so for this question I unfortunately can't give you a hard set answer um sorry um for me it would look a little something like I would come home after school and I would do my Math's homework like unless I didn't have matths that day the first thing that I would do when I come home is my math homework um after that I would mark my mat Mook and then see if I got anything wrong whatever I got wrong I'm not just going to be like oh I got it wrong okay next question I'm going to take a look and say why did I get it wrong what did I do to get it wrong um even if it was some dumb silly mistake I want you to kind of reflect upon why you got it wrong because just that process of being like hang on take a step back what did I do um that's going to stop you from making the same mistakes repetitively and until you do that reflective process you will keep making that m mes because you won't actually know what you did wrong right so um that is how you should be doing your maths um but in time in terms of time I didn't really study for maths during like the Monday to Friday week unless I was struggling with what I was learning in class so let's say we were learning about the quotient Rule and I was like this is absolutely disgusting i' probably do a few more questions on it so whatever my teacher sent me as homework and then and then some more after that um and then at the end of the week I would spend a good chunk of time um just going through everything I learned in this past week maybe even a little bit of the hardest stuff that I did the week before and then I'd also come back to something that I've been struggling in for a while so um that's how I studied for it in terms of time I don't know I kind of studied for advanced extension one kind of smooshed together but combined I would spend at least six hours a week excluding homework um but yeah and how much time should you be studying for all your subjects alt together obviously that depends on how you're feeling about your subjects what you're struggling with um and all of that but the advice that most teachers give is a general rule of Thum is an average of three hours a day take that with what with what you will um because everyone studies differently your methods your time um the way that you do it all different to each person so I unfortunately can't really give you a straightforward answer and I can only give you a beat around the bush kind of answer um second question is what studying or what did studying for Ma look like for you I kind of answered that a little bit in my explanation for the first question but yes um as I was saying before starting off doing my homework marking my homework reflecting upon what I got wrong um another good thing that I personally didn't utilize but one of my friends did and um she found it really helpful is like a mistakes diary so get like um a little book and anytime that you make a mistake you're going to write you know the date that you did it and what mistake it was um so that once again just that whole reflective process of what did I get wrong why did I get it wrong what should I do next time to not get it wrong you know I mixed up my question rule I mixed up my product rule because you don't know what they are yet but they're quite confusing they're not really but they're really easy to be confused with each other um so something like that so that you don't keep making the same mistakes um apart from that studying for maths as I was saying before was um end of the week I would reflect upon what I'd learned in my four lessons throughout the week and then I would go back to what I learned last week and then um I can't think of an advanced math like example but for all your extension one students um that just learned permutations and combinations um I'm going to guess that you found just as disgusting as I did which was truly atrocious um and it stayed that way for a while so literally from every holidays Easter holidays of year 11 until the um September holidays of year 12 every single holidays I would set aside like a good day to go through questions and like make myself super strong at that so that's like I can't think of an advanced example at the moment off the top of my head but anything that you guys are finding hard um and like you just can't wrap your head around end of the week that's the time to really do it um end of next week go again and then in the holidays now that you're in the holidays anything that you like take some time to just go back and look at what you've learned this term um anything that you're just a bit like oh oh I forgot about that oh I can't quite remember how to do that oh man I got this wrong um take a moment to just relearn learn that relearn that do the exercise again um and that is really going to strengthen your whole your whole game your whole Maths game okay but yeah for studying for maths um looked a lot like doing exercises and then past paper questions um when I when it came to that reflective end of the week or end of the term kind of thing okay I didn't get the best Marks in My term one assessment task should I drop to standard this one I'm going to guess a few of you guys would have already asked by this point um and the answer to that obviously isn't going to be like yes do it or no don't do it um but I personally would not drop to standard unless it's because you're not enjoying advanced math um if you haven't learned calculus yet wait until your school's started calculus for you to kind of um know whether you should drop to standard or not but um I don't know if this is a popular or unpopular opinion but in my very fine opinion apart from the fact that calculus is in mat advanc but I enjoyed and like found calculus quite easy um I think standard maass is even harder than Advanced because there's a lot of like wordy questions and applications questions that I personally suck at um so don't drop to standard because you think standard's going to be easier drop to standard because you don't like the content that you're learning here and take a look at the standard content and be like is that better or is that kind of just as terrible that's that's my little piece of advice for you um don't drop because you like got bad marks um fourth question I know all the techniques and content but I am so prone to making silly mistakes how can I fix this um so now that's what I'm going to go into a little something called framing AKA what I was talking about before when I said like the art of solving a question so um as I was saying before this technique I only started utilizing at the very end of my HSC Journey um so pretty much before trials like my HC trials um I was I was getting good marks for sure I was like maybe average 90s like low 90s and then once I started applying this I was getting um in my practice papers like 99 100% um and in my trials I think I got like 99% so it's just going to stop you from making dumb mistakes not fully reading the question not fully answering the question accidentally making a mistake somewhere in you're working so um something that I personally would advise you guys to use so the the steps that you go is number one read the question and read it again um underline highlight whatever you want um the information in the like the key variables so your X your y u what is impacted by what okay um look for hidden information in the questions um I'm trying to think of a year 11 example okay well um I can only think of a calculus example which you guys can't relate to just yet but you will once you finish the lecture um but something like what's the rate of change of this this and that the rate of change is another way of saying like looking at the first derivative so um when you differentiate AKA use calculus upon um you need to like kind of understand that the rate of change is also the first derivative um and until you make that connection you can't actually answer the question right because it's assumes that you're going to use the derivative to answer the question um and the third question is does my final or what does my final answer need to look like um you know do I need units do I need to round sigfigs um do I need to find the actual area do I need to just write that X is equal to this or do I need to like actually work everything out okay just so you don't overdo your answer and accidentally lose a mark because you didn't give your final answer correctly or just in casee you under did it and then they actually wanted you to show all you're working right um the second thing is to recall the relationships and formula um that are required so the the differential rules that we'll be learning later on for example um and then the actual methods and steps to getting to the final answer okay and you're actually going to write down these formul so um if you think that you're going to need to use the chain rule you're going to write down the um like equation for the chain rule so that like straight from the formula sheet or what ever is easiest for you to remember and then step three you're going to solve methodically down the page um and then you're going to actually check that your answer makes sense why first making sure all info makes sense and is correct then um the formula is written correctly and check each step after is actually making sense as well um then after that just make sure you go back to your step three in your first step which is what does my final answer need to look like does it look like that okay um this I know it's four steps and I explained it in quite a bit of time um but when you're actually applying framing it will really only add maximum 30 seconds to your time um like once you get to like once you've practiced it enough it's just your first nature and it won't take any extra time so just something to keep in mind and it's really going to like make your silly mistakes go extinct because you can't do silly mistakes with this kind of thing so yeah just something that I would like you guys to use if you wanted to okay now in terms of actually studying for mats um there are a few methods to study for mats obviously um one that I started to use once again a little bit later towards my HSC journey is mind maps um the reason being is because you're able to take a look at the you know big picture um through these little little dots so you connect them everything together and I'll show you a picture this is an example of a calculus line map a lot of these words might not mean too much yet but I've got my calculus in the middle over here and then it branches off to differentiation it branches off to integration applications of differentiation and then from each um bubble comes another bubble and another bubble and another bubble which all Branch off into their own Bubbles and then they can Branch together to like form a connection so um that's just how you can you know big big picture see your work um and then if you're not a big fan of the bubble mind maps you can also do a hierarchical map so um you know big picture calculus um breakdown from calculus we have differentiation breakdown from that is the product rule and then breakdown from that is the method of solving the product rule which is the cross method solution okay um once again this might all be a bit gibberish for you at the moment because you can't use my calculus examples um but we'll get started with the calculus now okay so let's start off with differentiation which was all we'll do this whole to our lecture so as I was saying couple minutes ago is that the differentiation um basically describes the rate of change of one variable in with respect to the other okay the method that you actually um derive an equation um gets you the gradient of the function okay obviously of the original curve okay the um derivative is the oh my gosh this is so hard to explain the the gradient function of the original curve is known as the derivative um which is the gradient function yeah okay I'm going to give you a bit more less wordy stuff in a sec okay okay um the derivative function measures the rate of change of the gradient at any x value so um yeah okay I'm not going to bother with any of this but I'm going to go into this so um the actual notation to write um your first derivative is f-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 ofx is equal to blah blah blah so the way that I can write the first derivative is f-x is equal to blah blah blah okay so the first derivative is just one dash okay then we have these notations pretty much um 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 um if I'm going to actually describe Dy DX to you um it if you're going to say it in words it will be the the um like 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 um and you've learned already um functions can be either continuous or discontinuous okay um continuous functions have you know no discontinuities so they have no holes in the graph no ASM tootes and they have no jumps right um so yeah it's important to actually note whether or not there will be any discontinuities um because um 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 um but this little rule will get you the first derivative of any basic function okay um so it's f- of X is equal to the limit of H approaches zero of f of x + hus F ofx all on H okay um I think we've got an example in the next slide okay here we go find the derivative of FX isal x^2 + 4x + 4 okay so I'm going to write the formula up for you so f ofx is equal to the limit of H approach Z of f of x + H minus f of x all on H I'm going to 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 um just for the interest of time I'm going to get you to pause the video If you're not yet done um work through it and then come back and take a look at my working okay so let's change my P color cuz I don't like red let's go to Orange okay so I'm going to sub everything in so f ofx is equal to the limit of H approaches zero um for those of you wondering why it's the limit of H approaches zero and you why you can't just make H equal to Z it's because of the denominator down here if H was equal to zero then um 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 um and I'll explain why we even bother doing that in just a sec so let's let's first use this example to explain it so let's do F of X+ H so that will be X+ h 2 + 4 x + H um + 4 so that's f of x + H minus f of x so minus x^2 + 4 x + 4 okay and that's all on H so let's expand things up so that's X2 + 2 x h + h 2 + 4 x + 4 H + 4 - x^ 2 - 4x - 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 um like cancel anything that can be cancelled so 4 x can be cancelled um so can x s and so can four okay so let's rewrite this to make it look a little bit prettier so it's equal to the limit of H approaches zero of h^2 + 2 x h plus 4 h on H now what I'm going to do is I'm going to put H out the front as a common factor so Li H approaches to zero of H H + 2x + 4 all on H now what I can do is I can cancel these H's out okay so it's equal to the limb H approaches to zero of h plus 2X + 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 + 4 because I can get the H to equal to zero so my final answer is therefore f- of X is equal to 2x + 4 okay hopefully this process makes sense I think we have another example in the next slide no we don't never mind um 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 kind of similar maybe maybe not um but I want you to take a look we have a four 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 derivative um and that is when f ofx is equal to X to the^ of n where n is any number um then the derivative is f- of X is equal to n * X to the^ of N - one okay so let's do our example of Y is equal to x^2 then Y dash will be equal to so n is equal to 2 so I can bring my power out the front so n * X to the^ of 2 - 1 so it'll just be equal to 2x okay um but yeah you're going to bring out the power to the front and then reduce the power up the top by one okay um and then the derivative of a sum um so f ofx is equal to GX plus HX um 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 um Y is = x^2 + 4x + 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 ah so annoying cool okay so let's do the X2 first so it will be bring the power out the front so 2 * x to the power of one plus we have an invisible one over here so it'll be 1 * 4 * x^ 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 * 4 to the X1 so my overall differential will just be 2x + 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 um if we have a some constant out the front um then the constant itself won't be differentiated the coefficient so um the example of Y is equal to 4X um the four itself will just stay the same so it'll be 4 * the derivative of x which will just be one so it will just be 4 * 1 which is four okay so the actual coefficient stays the same ory do okay so um here are a couple of examples so um Y is equal to X cubed + 3x² + 5 um take a look at the power and the coefficient that's brought out the front Okay so the three over here oh yellow is a bad color let's pick a different color let's go purple three over here is brought out the front and this reduces to two Okay the three out the front over here is just a coefficient we're going to keep it out the front so it'll be 3 * 2 * X the power of one okay and then five has an invisible x to the power of Z so it's being Times by zero which is why the five just disappears okay um 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 1 gets you negative a half so it's 1 on 2 * x to the a half AKA one on 2 rootx okay um the X squ 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 = x^2 um + 7 x + 5 and then we have a little bit of a tricky one we have f of x is equal to X4 + 2 x ^ half - 3x okay um and also yeah just a quick note um 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 um it's something that can be assessed in the HSC um it'll only be one question if it is asked but um obviously you want to get as many marks as you can and it is a big part of year 11 um prelims so I would recommend doing a will practice on differentiation by first principles as well okay 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 e okay let's get started so for the first question I'll just rewrite the actual question so y isal to X2 + 7 x + 5 okay so I'm going to separate them out we'll do X2 we'll do 7 x we'll do 5 so Y dash will be equal to we're going to bring the power of two down the front so it'll be 2 * x ^ of 2 - 1 so just 1 so plus seven the coefficient stays at the front times um we have an invisible one over here so it we're going to bring the one out the front 1 * x 0 okay then we have our invisible X to 0 over here so it'll be plus 0 so our y Das will be equal to 2x + 7 and that's the final answer okay so that's for a let's do B for B the f ofx is equal to X the 4 + 4 x the half - 3x so F Das of X is equal to a little bit smaller this comes here this comes here this comes here okay so I'm going to bring -4 out the front so the negative also is included in my power so -4 * X guess what the power is going to be this time if you guess -3 unfortunately that would be wrong because -4 - 1 is equal to5 okay plus my coefficient of four Out The Front Times by my high my power of a half time x to the a half - 1 which is minus a half and then we're going to bring the -3 * 1 * x 0 so it'll be - 4 x to the - 5 plus a half on two well sorry 4 on 2 which will be um 2 x to the half - 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 um so once we've actually learned how to differentiate we can now differentiate to find the gradient at a certain point on the curve um the derivative as we were saying before um is the gradient function okay CU we can sub in any x coordinate and we'll get our like y Das value for that specific X okay so subbing in the x coordinate into the derivative so y-h or f d of X um will result in the gradient at that specific x coordinate okay um 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 = x^2 + x - 6 let's do y Das Y dash will be equal to so X2 we're going to bring the two out the front so 2 x to the 1 so just 2x plus X The Power of One um which bring the power down and then subtract one by zero which will just give me two plus one and then the minus 6 disappears because we're Thanos okay so at X is = 2 Y dash is = to 2 * 2 + 1 which is five therefore the gradient at x = 2 is five it's quite simple right okay I'm going to give you a minute or so to do be by yourself okay let's go through this second question so find the gradient on the curve Y is = x^2 - 5x + 4 where the gradient of -3 this is tricky CU it's actually the inverse of what we just did okay so b y is = x^2 - 5x + 4 um I'm going to make my y- function so x^2 becomes 2x - 5x becomes - 5 four disappears into the ethers now what I'm going to do is I'm going to say at X is equal to what will make y Das equal to-3 right and now I need to solve for that so 2x - 5 is equal to -3 so let's do our reverse um algebra let's change the color to Yellow again I'm going to+ 5 + 5 so 2x is = 8 x is = 4 now I'm just going to recheck that because just in case I made a mistake so let's sub x isal 4 into Y dash and I can just write check um Y dash is equal to 2 4 - 5 and you can see that I did indeed make a mistake so let's redo that so - 3 + 5 gets me to 2 so X will be equal to 1 now let's check that again check um sub x = 1 into Y dash so y Das will be equal to 2 * 1 - 5 which is indeed -3 therefore gradient or if you want it to be a bit fancy you can write Mt so the gradient of the tangent um is equal to -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 um so down below tell me what the answer is if Y is equal 3x ^ 4 - 2x the a half + 3x + 1 then what will Y dash be I'm going to give you a few seconds to do that for e okay hopefully everyone's had a chance to answer the question below um if not I'll give you a sec to pause the video and have a go but I'm going to get started so why oops Y is = to 3x the 4 - 2x half + 3x + 1 so y Das will be equal to 3 * 4 x Cub - 2 * - half X to Theus so what's - a half - 1 that's equal to -3 on 2 + 3 * x 0 so just 3 * 1 + 0 that's going to get me to 12 x cubed minus minus right so that's going to get me to plus x to the -3 on 2 + 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 okay now let's get into the Like rules that I was talking about earlier so um if we're given a function y is equal to U * V where u and v are two different functions so let's say um an example would be like x - 2 2x + 3 so two different functions being times together um then we can use a rule to differentiate this and that is called the product rule so Y dash will be equal to V * u-h so the original V time the derivative of the U plus the original U * the derivative of v um or it's given this notation which is the exact same equation just written in a different method and this is the way that the formula sheet writes it um but I personally to this day use this one because it just looks a bit nicer right um but let's let's use an example because both of these let's face look absolutely disgusting oh okay we don't have an example let me let me write an example then my question to you guys is y is equal to let's let's use my example from before x + 1 2x + 3 that's my example I'm going to give you actually no I'm going to actually show you how to do it because it's a bit of a tricky rule if I do say so myself I'm going to Define my U and my V so U will just be my first function CU I'm not fancy like that so it's equal to x + 1 U Dash will be the derivative of U right so it will just be 1 V will be equal to 2x + 3 v-h will be equal to just two now obviously for this kind of example you can use foil and just expand um but there will be many examples that you'll path where expansion won't really help and you'll still have two um two separate functions being times together so um it's really important that you actually learn how to use this rule so let's follow the y- rule so it'll be V * u-h so this times that so it'll be 1 * 2x + 3 plus u * V Das so like that so it'll be 2 * x + 1 which gets me to 2x + 3 + 2x + 2 so my final answer is 4x + 5 that is my final D derivative okay awesome okay now let's go into the chain Ru um statistically the chain rule is the most um like the pretty much the bane of most HSC students existence it's the one that people most get confused on and I'm not sure why but as people head towards the HSC they kind of just forget this rule exists so um my mission today is to make it as you know nice for you guys to understand as I possibly can okay um but basically it's when we have a whole function to the power of n so instead of it just being like x to the power of n it's going to be some disgusting function that we'll call F ofx right all to the power of n okay um so the derivative of that so DDX um will be we'll take the function out the front just like we do for the simple differentiation so it'll be n times F ofx to the^ of n minus one so so far it's like exactly like our x to the N kind of formula but in bigger version but the next thing is something that we need to do that's not in the X version and that is we also need to Times by the derivative of what's inside this function so we're going to also Times by f- of X which gets us to you know n f- of x f of x to the power of n minus one okay I think there's an example in next slide yeah we'll use that example in a sec let's do this one first um so my question for you guys is F ofx oh we'll just go y make life easier Y is equal to 2x^2 + 1 to the^ of 4 I'm going to give you a second to try this one on your own because it is quite like quite similar to our x to the power of N1 and I I have a feeling you guys will be able to do this on your own so I'll give you a minute e also I probably should have told you in the start of the lecture the reason why I'm barely looking at the camera it's cuz my computer's like I've got my type my writing pad down here and my screen like over here so I like never look at the camera sorry I've got an external camera on so I apologize if I'm like barely making eye contact with you guys um but yeah let me get started with the question so let's define our f ofx f ofx actually let me do it in different color that we're not confusing it with my like explanation up the top we'll go back to Yellow so f ofx is equal to 2x^2 + 1 so my f d of X using our basic differential rule gets me to 4X okay so Y dash will be equal to n which is the power so 4 * f- X so 4X * F x^ nus1 so the original F ofx so 2x^2 + 1 to the power of 3 so my final answer is 16 x 2x^2 + 1 to the^ of 3 okay I hope that made sense we'll go through another example real soon as well okay so the chain rule is used for composite functions so I kind of went through this already um but it's also known as a function of a function rule okay which I think there might be another explanation never mind no explanation I'm going to give you a second to do this question on your own because it's simpler than the one that I went through before and I also believe that you guys can do on your own while you guys are going through that I actually want to explain to you that the X to the power of n Rule still follows the chain rule example so let's do X2 okay so Y is equal X2 I'm going to set my f ofx to just be you know what's underneath the power so it'll just be X so F Das of X is just equal to one so doing my um like plain example my Y dash will be equal to bring the F the power not the flower um to the front so two time um f- of X so 1 time f of x so just X to the^ of 2 - 1 so 1 which leaves me with just 2x so even the simpler rule is just the chain rule again but um a little bit nicely done because X X's derivative is just one okay um and that will nicely lead us into the explanation for this question differentiate Y is equal x - 4^ of 10 let's um Define our F ofx so F ofx isal to x - 4 so FD of X is equal to just one okay uh so let's do our final Y dash which will be equal to n so the power which is 10 time f of x which will just be 1 time f ofx x - 4 to the power of n - 1 so 10 - 1 which gets me 10 x - 4 to the^ of 9 so see it's a disgusting looking rule but honestly it's not that bad right it doesn't it's I think it's very much overhyped in that in the negative in the negative way it's very much I can't say underhyped I don't know what the negative version of overhyped is but yes it is that word okay now I have another question for you which is also the chain rule so Y is = to 2 x^2 + 5x - 3 I'm going to give you a minute to do this one on your own because it's a little more tricky but it's actually the exact same thing so take a second e okay um you'll probably need a bit more time but just so that we can get everything done in time I'm going to get you to pause the video if you need some more time and then unpause to take a look at my working okay um so let's go through this question so um let's oops let's define our F ofx which is whatever's inside the brackets right so X2 + 5x - 3 um so we have F Das of X will be equal to 2x + 5 all right now let's actually do our Y dash so going back to our original rule when I said that we have a constant times the function the constant itself the coefficient isn't being times so the two it just chills out the front and we're going to times that by our n so four times f d of X so 2x + 5 times our f of x so x^2 + 5x - 3 all to the power of n - 1 so 4 - 1 which will be equal to 8 2x + 5 x^2 + 5 x - 3 to the^ of three and that's my final answer um I suppose if you wanted you could expand out the brackets for this um but honestly if you're leaving things in simplest form this is as simple as it gets so that is what the HSC would want you to do okay all right let's see if there's another question if I'm going to get you guys to do that one in a sec I want to write another question for you guys though so my question for you is differentiate the square root of x Cub + 2x plus one okay I'll give you a sec to do it um yeah okay I'll give you 30 more seconds and then I'll get started for okay okay shall we get started I think yes so let's I'm going to actually rewrite this um so most of you guys would know a little bit of your index laws from year 10 um so this can be Rewritten as X Cub + 2x + 1 all to the power of a half now from here I think you guys would be able to solve so f of x is equal to X Cub + 2x + 1 so f d of X is = 3x + 2 so my Y dash will be equal to um my n which is a half time my f- of X 3x + 2 let me back 3x + 2 time my f ofx XB + 2x + 1 all to the power of n - 1 so a half - 1 which gets me to a half 3x + 2 x Cub + 2 x + 1 to the power of half now if I'm going to write this in the my best mathematical manners because they didn't give me to the power of a half and stuff like that they gave me in my square root form so I'm going to do the same thing um so a half power is basically one on root blah blah blah so the way that this can be Rewritten is you know a half times we'll go we'll go like this so that I can visualize it a better for you guys so it's 3x + 2 onto * 1 onun XB + 2 x + 1 1 okay so my final answer will be 3x + 2 on 2 < TK X Cub + 2x + 1 and that's my final answer okay um now I have another question for you guys and that is to differentiate Y is = x - 4 ^ 5 * x - 5 okay um heads up this might take a couple couple of Rules from what we've learned already I'll give you a sec to try this one on your own okay just for the interest of time I'm going to get you guys to pause this video If you're not yet done and you want to give it a go and I'm going to get started on the working okay so let's define our two functions so I actually need to use the product rule and the chain rule for this um the product rule is the big rule that I'll need to be using though so I'm going to Define my two functions my U and my v um I always pick U to be my first function and V to be my second function um just because it makes a bit more sense and um it will help you in doing it that way for the quotient rule which will be the next rule that we learned okay so Oops why does that keep happening U is = to x - 4 to the^ of 5 so u-h using our chain rule will be 5 * x - 4 to the^ 4 and then f- of X inside the function is just one so we don't need terms anything else our V will be equal to x + 5 our Dash will be equal to 1 so keeping in mind that y Das is equal to um u- V plus v- u um let's give this one a go so it's going to start off with this cross so it'll be 5 x + 5 x - 4 ^ 4 plus this cross so 1 * x + 4 to the^ 5 okay um now I'm going to take out a common factor of x - 4 oops x - 4 to the^ of 4 which leaves me with x - 4 the^ of 4 which leaves me with 5 x + 5 plus x - 4 okay so that x - 4^ 4 * 5x + 25 + x - 4 which leaves me with x - 4 the^ of 4 um 6 X Plus 21 okay and that's my final answer I'm actually just going to chck this out the front because I like um my power function to be at the end um that's just my personal preference you can leave it as the line that I just did though 6 x + 21 * x - 4^ 4 and that's my final answer so it's definitely a little bit more disgusting than the questions that I've given you so far but um it's still achievable right I'll give you another example here we go so if f ofx is equal to X2 - 5 um * x + 1^ 7 then find f- of X I'm going to give you a minute to try this on your own okay e okay um for those of you that would like to try this on your own I'm going to get you to pause um so I'm going to get started so let's define our two functions as you can see it's two functions being multiplied together so I'm going to Define my first function to be U so U is = x^2 - 5 u- whoops U Dash will be equal to so X2 becomes 2x -5 becomes - 0 my V is = to x + 1 to the^ of 7 in this case my f ofx is equal to x + 1 my f- of X will be equal to 1 so v- will be equal to 7 x x + 1^ 6 so my f- of X using my equation of y- is equal to u- v um plus v-u gets me to 2X x + 1 to the^ 7 CU I'm timesing by this cross first and then I'm going to make this cross which will be + 7 x^2 - 5 x + 1 to the^ 6 I'm going to take out a common factor of x +1 to^ 6 like I did last time and this is just simply because I want to simplify as much as I can so x + 1 to ^ 6 * 2X x + 1 + 7 x^2 - 5 okay simplify further 2x^2 + 2x + 7 x 2 - 35 x + 1^ 6 2 nope 9 x^2 + 2x - 35 okay I'm trying to think maybe this quadratic could be simplified um so let's take a look and see if it can be simplified I don't actually know I haven't done this question before so um what we'll do is you can choose which method suits you best I think I might just use the quadratic formula so x equal to minus b +un b^ 2 - 4 a c on 2 a okay that gets me to - 2 + - 4 - 4 * 9 * -35 on 2 * 9 let's take a look at what the discriminant is okay it's not going to be a nice number because my discriminant is like - 2 +us 3555 on 18 so I'm actually just going to leave it like this because this is the simplest form that I can get it to okay awesome okay now the next question is for you guys to try on your own um I'm only going to give you maybe 30 seconds to try this one um cuz it's been on the slider pole for a while now so yeah 30 seconds I'll come back at an hour 17 e okay let's get started um so for this I only need to use the chain rule so what I'm going to do is I'm going to Define my f ofx or I'm going to put a like in quotation marks because it's just my mini f of x in this case right so it'll be 5x - 3 so my FD of X is equal to just five so my grand scheme f- of X will be equal to the three that's coefficient at the front that we can't do anything to so 3 * um n so 7 time F of sorry f- of X time f d of oh my gosh time f of x so 5 x - 3 to the power of 7 - 1 6 that's going to get me to whatever 3 * 35 is 105 5 x - 3^ 6 so anyone that got D good job all right now let's go into the quotient rule um so basically this is when we have a function being divided by another function okay um so pretty much like the opposite of the pretty much the opposite of the product to all um yeah so the notation that is a bit nicer to learn is this one in red um but the one that's in the formula sheet is this thing which just looks absolutely atrocious but it's just a disgusting way of writing exact same thing um so the method that I'll be showing you since they're both the same thing will be the same but the actual equation that I'll be showing you will just be this one to make your life a bit easier okay um let's go into an example so differentiate y y is = x - 2 on x + 7 I'll work through this one with you so I'm going to first just write the equation so y Das is equal um to u- vus v- U on v^2 okay that's all it is so let's define our U to be x - 2 u- is equal to 1 V is = to x + 7 v- is equal to 1 so F or will go y Das is equal to u- V so x + 7us v-u x - 2 all on v^ 2 so x + 7 all 2 okay so the one that the main thing that I want you guys to keep notice of is that we've got a minus over here so the actual um ones that you times first actually matters because if we put v-u in front of u-v then you're going to be getting the complete incorrect answer so it needs to be on this diagonal first and then this diagonal and only that way um awesome and now we're just going to simplify further which gets me to just 9 on x + 7 squ and that's my final answer okay um I have another question for you so this one I'm going to get you guys to try on your own I'll write the equation up again so Y dash is equal to u- vus v- U on v^ 2 I honestly don't even I mean yes please remember this but um I always find the V's in the U so confusing to like remember which one goes first so I always obviously I'll write down what U is I'll write down what V is um U is always the numerator V is always the denominator um and then just remember the crisscross that goes like this only in those two directions so the first Cris the second cross okay I'll give you a minute to try this on your own because I would love for you guys to be able to confidently independently do the question rule before your class even gets to it okay I'll give you like 90 or so more seconds e e okay I'm going to get started now so anyone that isn't yet done please just pause the video and then keep going once you ready to see my explanation all right so let's define our variables so as I was saying before the numerator will always be U so U will be equal to X cubed so u-h will be equal to bring the three out the front subtract by 1 so 3x^2 V will be equal to x + 1 V Dash will just be equal to one so I'm going to write my D DX which is just the same way of writing Y dash or what was my other one f- of X um so we're going to do this cross first so 3x^2 x + 1 - 1 * X Cub or all on x + 1 s okay so that's going to get me to 3x Cub + 3x 2 - x Cub on x + 1 2 which gets me to 2 x Cub + 3x2 on x + 1 2 I'm going to take out a common factor of X2 from the front so it'll be X2 2 x + 3 all on x + 1 s and that's my final answer okay um hopefully that wasn't too bad and I'm going to move on to um pretty much the last section of um like what we're going to be learning today before we go into like study tips and stuff so we're going to go into the notion of what is a tangent to the line and what is a what's the normal which a lot of you guys might not even know of the tangent to the line is basically um we have a line like like so and we're given a specific spot and we want to know what the tangent is it's basically the instantaneous gradient at that spot so it'll look like this it'll just be a straight line that just scrapes it touches it only once um the actual curve okay and the normal is pretty much the gradient except it's perpendicular so 90° to the tangent so it looks like this so you would say this is the gradient of the normal and you would say that this is the gradient to the tangent okay so we know the formula of the gradient is rise over run um so when we're given the angle we can apply basic trig to find our gradient right so since tan Theta is equal to opposite over adjacent it's equal to rise over one which is the gradient um which I'll give you an example of now so find the value of tan Theta from the following okay I'm going to give you a sec to try this one on your own um it is something that we went over in year 10 and possibly again already in year 11 um so hopefully this is just revision for you but if not I'll go through how to go do this again if you need a little bit more time looking at my previous slide I you can actually find the download of the slides I think to your right you're right that way isn't it oops for okay hopefully you've had enough time to give this one a go um but let's take a look at this so the gradient of the function is just the oops just a coefficient of two so the rise of a run is equal to 2 on 1 okay so tan Theta will be equal to 2 Theta will be tan inverse of 2 make sure that you're in degrees and not radians like I am and we'll go shift tan 2 and my angle of elevation will be equal to 63° if well 63° if you're run into degrees I'm going to run to minutes so it be 63° and 26 minutes okay to the nearest minute okay oh well technically I didn't even read the question properly they just want the value of tan Theta my actual answer is just this okay cools now let's go into a little bit more of the definitions of what we've gone through in maybe year 11 sorry maybe year 10 for some of you guys um I know some year 10 schools just don't teach everything that they should I don't know um but basically um the gradient of the tangent is equal to the gradient at the x coordinate at which um the tangent intercepts with the Curve okay that's like the definition of the tangent gradient um so the gradient of the normal is actually equal to the negative reciprocal of the gradient to the tangent okay so um from year 10 you'll know that if two lines are perpendicular then M1 * M2 is equal to1 AKA Mt * n MN is equal to1 therefore M oops that's a big M I want a small m m n is equal to one on Mt okay that's a rule that I want you all to write down so I'm going to give you 5 Seconds to write that down 5 4 3 2 1 okay hopefully you've written that down okay cool um yeah so in order to find the equation of the tangent or normal um of a curve at a specific point of X um we have to use something called the point gradient formula um which I hope you guys already know um is y - y1 is equal to M * x - X1 where m is the gradient and X1 XY is the coordinates of like a specific point okay um I think I have oh I've got the steps to work through it first okay so um let's do the actual methods so the first thing that you need to do is to actually differentiate the curve okay from there you're going to find the um tangent at the like a specific x coordinate so you're going to sub in so you find Y dash you're going to sub X is equal to I'm going to call it X1 um into Y dash okay your next step is to sub X isal to X1 into y to find to find y1 and then you're going to use the point gradient formula to obtain the um actual like equation to the tangent SL normal so y - y1 isal to m x - X1 okay so here is a question I would like for you guys to try this on your own I'm going to give you quite a few minutes to try it so find the equation of the normal to the function y = XB + 3x^2 - 12 x + 1 at X is equal to 2 if the gradient of the normal to the curve Y is equal of the curve f ofx is equal to 3x^2 - 4x is a/4 but find the point of contact of the normal okay so that's kind of like the inverse method of doing a I'm going to give you a few minutes for this e e e e okay hopefully everyone's had a start to question one I'm going to go through that now and then I'll give you a few minutes to try be by yourself as well all right so I need to find the equation of the normal of the function y is equal to F ofx at X cubed + 3x^2 - 12 x + 1 okay um I'm actually going to take a new slide for this CU there's a lot to be done okay sorry just needed to write it down so I don't forget all the numbers so find MN of f of x is equal 2 2x Cub + 3x^2 - 12 x + 1 at X is equal 2 so the first step is to find the gradient function so f- of x which will be 6 x^2 + 6 x - 12 now I'm going to find the gradient of the tangent at X is equal to 2 because that's just the first step to do it so I'm going to write f-2 so I'm going to sub in x equal to 2 that's just another way of writing it which is 6 * 2^ 2 + 6 * 2 - 12 which is 24 + 12 - 12 which is 24 I'm just going to double check with my calculator just in case I made a mistake no I'm good okay so that is the gradient of the tangent all right um give me a second how do we recall what the gradient of the normal is from the gradient of the tangent did you say that it is the negative reciprocal so MN is equal to1 on Mt which is -1 on 24 okay from here I need to sub in F of two so my y coordinate of um like my y1 basically from our step three the F of two actually I'm going to do it on the side over here so I don't confuse anyone or myself which will be equal to 2 * 2 cubed + 3 * 2 2 - 12 * 2 + 1 which my brain doesn't want to comprehend so 2 * 8 + 3 * 4 - 24 + 1 hope I got this right I got five okay so now we're just going to sub it into the formula y - y1 is equal to m x - X1 so y - y1 is equal to m x - X1 I'm going to take a step to just write down all my variables at this point so my X1 is equal to 2 my y1 is equal to 5 and my m is equal to -1 on 24 all right so y - 5 is = -1 on 24 x - 5 No 2 I'm going to show you how to do it um using the like Y is equal to MX plus b kind of method and I'm going to show you how to do it in the general formula as well um let's do MX plus b first actually no let's do General first um basically for the general equation you cannot have a fraction coefficient in front of X and your coefficient in front of X cannot be negative so what I'm going to do is I'm going to times everything by -4 so that my X is kind of like free on its own I'm going to show my lines of working time -4 * -4 which will oops which will get me to -4 y plus whatever 4 * 5 is 120 is equal to x - 2 now what I need to do is I need to just bring everything over to one side so it'll be a x + b y + C is equal to zero that's the that's like the what's the word the structure that it needs to be so my X as I was saying before needs to always be positive so everything needs to be brought over to the right hand hand side your right hand side's over here sorry so it'll be X plus 24 cuz I'm bringing the 24 to The Other Side Y um minus 2 - 1 20 so -1 22 is equal to Z and that is my final answer using the general equation now I'm going to do it using Y to MX plus b which we've got y - 5 isal -1 24 x - 2 um so I'm just going to expand those brackets so -1 on 24 oops 24 X Plus 1 on 12 I'm going to bring the five over to the other side so it's y isal to X on 24 plus whatever 5 + 1 12 is 61 on 12 okay and that's my answer using the Y is equal to MX plus b um like formula okay now I'm going to give you a sec to try doing B on your own so I'm only going to give you like 30 seconds or so so if you need more time please pause the video but if the gradient of the normal to the curve Y is equal to f of x is = 3x^2 - 4x is4 then find the point of contact of the normal okay I am going to get started oops oh I need to make more pages give me a quick sec perfect okay so the question is my f ofx is equal [Music] to 3x^2 - 4x um and my gradient of the normal is equal to A4 what is the x coordinate was it no it said what is X1 y1 equals what okay that's the question and I just need to double check that the actual equation I got right Perfect all right so the first thing that I need to do is I need to differentiate my function so f d of X will be equal to 6 x - 4 okay um now I know the gradient of the normal which means that I can find the gradient of the tangent right using the same like relationship that I had in my last question in that MN is equal to1 on Mt so likewise Mt is equal to 1 on MN which be 1 on A4 which is 4 okay so now I need to find the x value for which my tangent is equal to 4 so I'm going to write f- of x isal to 6x - 4 I'm going to change my pen color so I can like show my working to you guys a bit better which is equal to 4 6X will be equal 8 so X will be equal to 8 on 6 AKA 4 on 3 okay so that's my X1 X1 is equal to 4 on 3 now unfortunately that's not my full answer I need to also find my y1 AKA I need the coordinates of the point of intersection so what can I do I can substitute my X isal to 4 on 3 into my f of x function so what I'll write is f of 4 on 3 is equal to 3 * 4 on 3 2 not cubed - 4 * 4 on 3 okay which will get me 3 * 4 on 3 2 16 on 3 minus 16 on three which is zero okay so my y1 is equal to zero so therefore um the point of intersection is 4 on three 0 and that is my final answer okay so that's it for the actual like content for today's lecture um I'm going to spend the next few minutes just going through like um general study tips and then um yeah so I'll spend the next 13 or so minutes on that all right so my study tips for you is number one to ask questions um best time to ask questions is when they pop onto your head pop into your head so um when you're actually learning the content um but also when you're doing your study um for all of your subjects this is important um so I personally on my desk I have like a bunch of like scrap paper that I've like just cut into like a quarter of a piece of paper and I just have this big pile that's like next to all my papers pencils and whilst I'm writing down like all of my like I'm working through a question or I'm doing something and I get stuck on a question and I don't know how to answer it or I am reading the textbook and um for chemistry like I don't understand um you know how to properly balance this specific equation I'd write down the question or I'd write down the concept that I don't quite understand and I'd bring it to either my friend or better yet my teacher or tutor and I'd be like hey I got this question I don't know how to answer it or I've got this question that I don't quite understand can you please explain it to me and they will explain it to you um so yeah there's no there's no losing in asking questions um there's only gaining um second one is to find questions from a range of sources um so your textbook is lovely but your textbook isn't the only thing that like if you're just using your textbook or if you're just using past papers um or if you're just using one single resource you won't be properly prepared for your prelims and especially for the HSC um you do need to find a range of resources um your textbook obviously is one past HC papers and trial papers absolutely another um so past HC you just use the Nite I'm just search up you know 2021 hsd mat Advance you'll find it trial papers um there's a few websites um there's like HSC being one of them and a few others that I can't think of off the top of my head um a topic test is also really good so I personally because I did tutoring myself at um T Smart in year 12 I had access to Ed unlimited so I really took advantage of that fact and I did use all of the a notes like text guides and topic tests um so from a student perspective I can tell you that it's definitely worth um the subscription even if you're not like a to Smart student um another one that I discovered maybe a week before my trials not my trials sorry my prelims was the Cambridge checkpoint um like booklets so they're like an A5 booklet and they pretty much have every single past HSC question um on a topic and it's like specific like topic specific so it'll have chapters into it like chapter one a being whatever some specific like module is and it'll just have all of those so um I my local library was lucky enough to have a lot of these so for year 11 I just did it for physics and chemistry because that's what my school had sorry my library had and then ehw they also had one for Matt Advanced which was pretty good so um unfortunately by the time you do practice questions like that you'll probably already know a lot of those questions so it's not worth buying but if your library has it it's definitely worth taking a look at um make your own formula sheet is a massive good one so another thing that I didn't really mention at the start of the Q and frequently asked questions was um what was it oh how do what does studying for maths Advanced look like for me um something else that I do typically at the end of a month or at the end of like a a chapter in my textbook is I would have a single piece of paper so whoops double-sided lined a piece of paper like so and what I would do is I'd write down um any like specific key points that were like the biggest takeaways of the module and then the back side of the paper would be every formula that is relevant to that module and then I would highlight the ones that aren't on the form well I have like a color code system so one color would be that they are on the formula sheet one color would be that they aren't on the formula sheet but like um it'll probably be given to you or like it's kind of easy and then my other color which would normally like red or pink that would be it's not on the formula sheet and you are like expected to know it so for example for calculus from what we've learned today the only two rules that you don't have on the formula sheet is differentiation by first principles and the gradient of the normal so the MN is equal to one on M sorry negative one on Mt so my personal advice to you would be um try and make that sheet yourself maybe just based off of what you learned with me today um and yeah highlight the all the formula that are in the formula sheet one color the ones that aren't in the formula sheet in another color and um you know the ones that you reckon will be a bit more difficult to learn um in red or pink or something okay um and my fourth tip I've got more tips after this but my fourth tip is to be consistent and that is the biggest tip I can give you um because math is a skill it's a skills-based subject so what's amazing about maths especially when you're getting towards the HSC is that The Closer you get to exam time the less you actually need to study for it it's something that you can neglects a strong word but it like kind of let go of a little bit um because it's a skill right so if you don't play soccer for a week um the next week you'll still be pretty much just as good right um if you've been practicing for you know months on end beforehand um math I know it's such a weird analogy but it's just like playing soccer like the skills that you gain after a weeks time won't really be any different um if you've been practicing for months or years in advance so um when it comes to exam time um you will definitely be very glad with yourself if you've kept up with your homework if you've kept up with you know at the end of every module doing your little um like sheet or doing like what I was doing at the end of each week going through what you learned this week and what you found hard the week before and what you found hard all year so far so just being consistent so practice practice practice is definitely the biggest thing now another thing that I want you to take note of is that there is a little bit of like almost a bell curve to the time taken on a subject and the actual Mark that you receive um so yes I want you guys to work hard um but also I don't want you guys to burn out um that's one thing that um You probably hear a lot about burnout um it's definitely not something that you want to experience um but you can see that obviously putting in a little bit of time or a little bit of effort will barely make a difference once you start putting in the time and effort you'll definitely see literally exponential growth but if you overwork yourself and spend heaps and heaps and heaps of time on maths or chemistry or English or physics or whatever um they you'll come to a point where you plateau and you start going down so you need to basically find the perfect like the perfect effort and time needed to get this section right here and that will be different for each and every one of you some people have a you know really high stress threshold and Thrive of stress others um like me I get the best results out of being consistent so I'm not telling anyone to study last minute you will not do well if you study last minute um because you just like once again especially for math it a skill it is something that you can't just wrot learn the night before so um make sure you invest the time just don't invest too much time okay um and as I was saying before practice practice practice most important thing is practice practice practice find a bun resources keep going at it and practice will make I can't say perfect but near perfect okay and my last tips for you during the exam would be um use your reading time to your advantage there's so many people I feel like it's almost 50% of us that um will just spend the 10 minutes doing multiple choice as nice as multiple choice is um my like advice to you is to skim through the whole paper um take a look and be like oh that question over here in page 12 that's kind of hard and oh we're going to need some extra time on question 16 then you go to the last question you're like oh that's gross so um take a look look at each question really fast and be like oh I might need some help here here and here just so that you kind of have a mental map of where you might need to allocate more time so um that number one you don't run out of time during the exam but number two um you are kind of prepared for what kinds of questions you're going to be asked and then if you have more time do multiple choice or do one of the harder questions that you like found but I I feel like the harder questions might be a bit hard because you need to like write which reading time you can't use your pen so do whatever you can my next tip is for Mark maximization um never leave a question blank like your answer blank um especially multiple choice you have a quarter answer quarter chance of the right answer um but even if you like write down a little bit of scaffolding um write down equation or two that might be relevant you've got the chance of getting one Mark out of how many there are so write something rather than nothing um and going more into max Mark Maxim Mark maximization is um foll the framing technique that I was talking about earlier in today um you can read the slides you can download them and like write them down um third one is to have faith in yourself you definitely know more than you think that you do um I don't know it's just something that the brain does that like always underes estimates your ability so um you've done the hard work you've done your practice at least I hope you have and um once you get to the exam you just got to You' just got to do it you've you've done everything that you need to lead up to it okay and the last thing is to breathe um so many people they like hold their breath during like especially during the reading time people like stress out and they're like don't be like that um so you know take take take a second to like for me personally the very end of my reading time if I did have spare time that last 30 seconds I I closed the book I actually Clos my exam paper close my eyes and just take a few deep breaths that I don't know that always cleared up my head um and you know stopped the nerves so it's something that I have been advised to do and I also advise for you to do um so that's it for today today I really hope you guys enjoyed this lecture um please let me know down below what you thought um if you have any suggestions we are new to this whole pre-recorded lecture thing so any suggestions um will be taken and we love your feedback so thank you so much for today I will hopefully see you next holidays in the June July um lecture series um but until then I wish you all the very best for the next um term coming up and make sure you enjoy the rest of your holidays okay thank you bye