A Level Maths Made Easy - Implicit Differentiation

hey guys welcome back to another a level maths Made Easy video this time we're looking at implicit differentiation and again this is another topic that isn't always taught particularly well instead they normally just teach you the rules and then that's it but I'm going to show you exactly where the rules Come From as well because that will lead to a bigger understanding or deeper understanding sorry um and that'll actually make you get this better and it make you be able to answer questions a lot more easily because the questions for this once you understand what you're doing are extremely easy so the differentiation that we do is is really minor in fact it's probably easier than the other differentiation that we've heard about before this also super useful in real life so if any of you going on to do some kind of math based degree or medicine something like that this will be used as well so before we get into it uh just to let you guys know most of the people that watch these videos are not subscribed so if you could consider subscribing if you found any of this useful or any of my other videos useful that i' be perfect but let me go into what is the difference between implicit and explicit well when we have something that is explicit that means that it is very obvious and is clearly defined so for example an explicit function will look something like this you'll have y = let's say x^2 + 2x + 4 I've clearly defined what Y is and what x is clearly an implicit function is a function that is not clearly defined so let's say I had something like well what I'm actually going to do is I'm going to rearrange this equation I'm move everything over to the left hand side so y - x^2 - 2x - 4al 0 now it's not clear what y or X depends on or if it's y or X that we're actually that's the actual function right so we have here y's and X's all mixed together the reason why this is a little bit trickier is when I differentiate this I mean if I differentiate using the usual rules you'll notice that we don't actually get the same thing as what we do up there so if I differentiate all of this with respect to X I will get zero here and zero here and I'm left with just Min - 2x - 2 whereas here we actually get a slightly different function first of all we get Dy by DX we don't get Dy by DX over here the reason is because when you differentiate constants you get zero that might be something that you're aware of so when I differentiate four you get zero the thing is when I differentiate y I don't know that it doesn't depend on X so for example because I've given you the original equation Y depends on X in which case when we differentiate it it doesn't equal zero that's where the Dy by DX comes from because we're saying that we'd have to differentiate y with respect to X and that gives us you know 2x + 2 in this case but in this case here because it's not clearly defined what Y is if it is a function of X or if it's a constant so we don't know if it's a constant or function and because we don't know that when we differentiate we can make mistakes now the good news is there is a way that we can actually differentiate this and do it all nice and neatly the first thing I want to do is I want to differentiate the top just to prove to you that this gives us actually the exact same thing we get 2x + 2 so we're looking for Dy by DX = 2x + 2 now if I differentiate both sides of this let's have a look at what happens the first thing I'm going to do is I'm going to differentiate everything that isn't y so we get - 2x - 2 = 0 and of course if we move this over to the other side we get 0 = 2x + 2 which is not correct so how do we get um the Dy by DX well here's what we do I'm going to call y a function and I'm going to call it let's say f f of y equal y when I differentiate this we get DF dy I get the number one now I don't want DF by dy I want to work out what DF by DX is so and again I'm going to explain why do I want DF by DX I want DF by DX because I'm trying to differentiate this function with respect to X right logically so DF by DX well on the numerator we would have to have the DF by Dy but then we have to somehow cancel out remember this this is what I was trying to teach you guys last lesson and with the chain rule as well is what do I need well I want DF by DX so I need to cancel out the Dy so I'm going to put Dy on the top and I want a DX on the on the denominator so I'm going to put DX there so that gives me that DF by DX is equal to 1 * Dy by DX so in other words when I differentiate y I get Dy by DX that's where the Dy by DX comes from then if you just move everything over to the other side you get Dy by DX = 2x plus 2 which is the exact same thing as we get when we have an explicit function which is the point so when you have a bunch of y's and X's because it might not just be one y and 1 x it might be a bunch of different ones and they're not expressed as y equals something or x equals something you have to use implicit differentiation because you don't know if Y and X depend on each other we can't assume that they don't which again is really useful for a lot of real life stuff so what's this rule that I was talking about the rule that I was talking about it follows like this if you ever differentiate a function in fact I'm going to write this slightly differently with respect to X but that function is a function of Y then all you do is you have DF by Dy time d d y by DX now that might look a bit weird so we're going to go through a few practice questions of course but essentially whenever you differentiate something that's y you differentiate normally and you times it by Dy by DX that's it that is the only thing that you have to do so let's go through a couple of examples just to make sure this makes sense so we're going to do a ton of examples here okay now the first thing you might notice is that there is no like equation as such right it's not y equals or anything like that because again this is implicit so remember what we do we differentiate it as if it's normal so we're going to make it four if I differentiate 4 y with respect to Y I get four then we just times it by Dy by DX that's what that expression becomes again remember we're using this original thing here if we're differentiating a function of Y we differentiate with respect to Y and then we times it by Dy by DX so same thing here when you differentiate sign you get positive cos right uh well you get two in this case so you get two cos y cos 2 y time Dy by DX that's it that's all we're doing and again if we wanted to use the chain rule to do that again remember we have the function of y equals sin 2 y so we differentiate with respect to Y and then again DF by DX will equal DF by Dy time Dy by DX so we just times that by Dy by DX and this is true anytime you differentiate anything that has Y in it the next type of question this is honestly the most common type of question is find Dy by DX in terms of X and Y in each case so here again we're going to differentiate this with respect to X so we get 2x well the next thing be differentiate with respect to Y so it becomes 2 Y and then times it by Dy by DX x equal 0 now they want Dy by DX so we're going to rearrange for Dy by DX so I'm going to minus 2x on both sides we get 2 y Dy by DX = -2X I'm going to divide both sides by 2 y we get 2 y dy over DX equals well we have 2/ 2 so that's just negative now X over y so you can always rearrange to actually get what Dy by DX is and you may need to factorize so I'm going to show you one of those a bit later as well so again we're finding this is just part B of the question here find Dy by DX so we differentiate with respect to X so here we'd have 4 y cubed Dy DX = 2x - 6 so divide both sides by 4 y cubed - 6 over 4 y Cub we could divide everything by two as well just in case they wanted us to simplify we get get x - 3 over 2 y Cub for Dy by DX which is fairly nice so as promised I'm going to give you one that's slightly harder now I want you to think for a second what makes this more tricky why am I saying this is difficult and I'm going to explain it to you in three two and one this uses the product rule why because X and Y timesing each other remember they're both functions of X so we have to actually use the product rule twice for this question so we're going to use it for the left hand side so we're going to say that u = x v = sin y so that means DX is 1 DV DX is well it goes to cos right but then it's cos y Dy by DX because again anytime we differentiate y with respect to X we need to add that Dy by DX to make sure we're not missing anything and again we do it with the other side as well so we have U = x^2 V = cos y DX is 2x dvdx is netive sin Y and again Dy by DX so we've done the product rule twice so it's going to be the cross term multiplying so it's going to just be sin Y and then it's going to be plus x s cos y sorry Dy by DX I'm going to have a plus again we're going to do the cross terms so we have 2X cos Y and we have negative x^2 sin y Dy by DX equals well differentiating that is nice and easy just equ = 0 so now we have to actually don't forget the question was work out what Dy by DX is so how do we do that well what we need to do is we need to put everything with Dy by DX on one side so it's already on the left hand side so I'm not going to bother changing it I'm going to move the sin Y and the 2x cos y over so we're going to have X cos y cuz I want to keep that here Dy DX - x^2 sin y Dy DX equals and it's just going to be a bunch of negative so - 2x cos y minus sin Y and hopefully this kind of brings you back to gcsc when you had rearranging harder formulas where you're moving everything that you want to one side and now we're going to factorize out Dy by DX so factorize Dy DX and you get X cos y - x^2 sin y equals this disgusting stuff and then just divide it over and we're finally done now yes this is absolutely bloody hideous but remember that's okay right maths can be hideous sometimes but this is the correct answer we're doing everything great and that's it now just so you guys are well I need to take a quick aside in mark schemes if you ever see an answer that has all the same bits as you but different signs so instead it's 2x cos y plus sin y overx cos y plus x^2 sin y or of course the denominator could be swapped around so x^2 sin y - x cos y these two things are indeed equal to each other all they've done if you'd like is they've times the top and bottom by minus one which is an equivalent fraction so it's totally okay now this is actually something that's quite interesting and some of you that have done or will do University interviews for something related to maths or physics you may get given something like this to differentiate so this is actually quite a pretty it's pretty good question it's only ever come up once but it shows that they can give you differentials that you've never seen before and that is completely allowed that's completely fine and I'm going to show you how you do that now luckily in this question they say by first taking taking logarithms as a bit of a hint now I will say in the actual exam it just said this differentiate this and show that it equals blah blah blah blah blah so I'm going to show you how to do this by first taking logarithm so it makes sense to do that first I would advise just taking Ln and now what can we do before we get started well we can use the power rule to bring that forward and now we can do the um implicit differentiation LNA a is just a constant so we're going to treat it as if it was the number three in front of X in fact I'm actually going to rewrite this to make it super super clear as LNA X so we can treat that exactly like we would treat a number three a number four a five whatever because it's just a number it's just a constant because it says here a is a positive constant so now implicit differentiation when you differentiate this with respect to Y you get one over y but remember we need to times it by Dy DX and that just equals well LNA and that's it so all we're going to do now is we're going to DX we're going to times both sides by Y and we get Dy by DX equal y Ln a now if we actually have a look at the question it just says find an expression for DX in terms of a and X so a and X only well Y is the same as a^ X so that means Dy by DX is the same as a^ X Ln a and that's where that result comes from if you've ever seen that before um it is something that's hopefully taught fairly regularly but here's another way that implicit differentiation helps because differentiating this normally is a pain in the butt I can tell you that now the reason why I mentioned that you may have seen that result before is these are two other questions based on the same thing where it says says differentiate 3^ X and the funny thing is in the mark scheme it literally just said it just used this result so it says 3^ X Ln 3 so you could literally just write and get the right answer because this is a standard result that we've derived which is very nice right we derived a result how cool how exciting but if you follow the exact same steps you'll get the exact same thing but for Part B I'm actually going to show you how to do something like that so again we're going to say that y equal 6^ 2x we're going to take the natural log just like we did before 2X and I'm going to bring the 2X x to the front but then I'm also going to just uh shove it at the back which is you know exciting so we have Ln y equals ln6 and I'm only doing this just to make super clear that Ln 6 * 2 is just a constant so when we differentiate again differentiate with respect to Y we get one over y then we need to Times by Dy by DX because it is a function of x equals just no it doesn't equals just uh let's call it 2 ln6 yeah that works times Y and again we're just going to sub in what Y is which they told us at the beginning well we said at the beginning that Y is equal to 6 ^ 2X and you are allowed to do this you're allowed to say that y equals this because all we're saying is is that 6^ 2x is a function of X which it is obviously we could have called it any other letter in the in the alphabet if we wanted to but because we're doing implicit differentiation I thought it'd make more sense just to leave it as dydx just to you know not confuse anyone um that will become two I'm going to put this in Brackets because something that I see sometimes which uh I'm not going to lie it makes me physically ill is people would write this as 12 ^ x that is completely wrong do not do that this is 2 * 6^ X just leave it as that again you can sub in some numbers to prove it if you want if x was 2 you'd have 6 s is 36 * 2 which is 72 12 the^ 2 is 144 not the same same at all so be really careful with that and yeah let's move on so once we've done that I'm hoping this makes a bit more sense again I do think that in in some cases just from when I've tuted other students and things like that this isn't taught super well instead it's just given as a rule and the problem with rules is they don't always make a lot of sense and when they don't make sense it's very hard to to learn them whereas if you understand where it comes from you probably realized that you know the rule without actually memorizing the rule because you understand it you understand where it comes from so hopefully that's helped you a little bit if you have found it useful again please consider subscribing CU it does help me out and my link to my website and to my email address is in the description in case you have any questions or want tutoring or anything like that but I will leave you to your exam revision and I'll see you later