 So today we are going to start with differentiation of class 12th, differentiation of class 12th. Okay, differentiation, you would have got a basic idea. I will in fact revise that also for you pretty quickly and not spend much time on it. But differentiation of 12th has the following subtopics. So I'll just give you an overview of whatever we are going to do under differentiation. We are going to talk about differentiation, differentiation of implicit functions. Implicit functions. Now what is an implicit function? Just to give you a very simple definition about implicit function, any function where you are not able to split the variables out easily, that is called an implicit function. Okay, for example, let's say if I have x square y plus xy cube is equal to 5. Okay, and I want to find dy by dx from here. So I cannot make y the subject of the formula or for that matter x the subject of the formula very easily over here because they are so intrinsically linked to each other. So this is an example of an implicit function. And how do we find dy by dx for such functions is what we are going to learn. In fact, this is the topic which I am going to start with today. We are also going to talk about differentiation, differentiation using logarithms. Okay, and differentiation of, I'll write it down as the next topic, differentiation of infinite series, infinite series type of functions. We are also going to talk about differentiation of parametric functions. Okay, so all these concepts which I am writing over here, they are very important for your board exam as well, not only from J point of view, but from your school or board exam point of view also. Then you are going to talk about differentiation of inverse trigonometric function, inverse trigonometric function. Let me write parametric functions. Okay, now this would require me to give you a bit of idea about ITF. So with a brief, with a briefing of ITFs, with a briefing on ITFs, inverse trigonometric function. So this may take a bit of time. Probably I will be able to do it in the second class for you. Okay, then comes, then comes differentiation of a function with respect to another function, differentiation of a function with respect to another function, another function, another function. So normally we talk about dy by dx or dy by any variable or any variable by any variable. So you are differentiating a particular thing with respect to some variable, but here you will be differentiating it with respect to a function itself. So how does one function change when another function changes, which involves the same set of variables. So that is what we are going to cover under differentiation of a function with respect to another function. We are also going to talk about higher order derivatives, higher order derivatives, okay. Also called as successive derivatives, successive derivatives. Now here in this particular part, I'm going to give you certain formulas, which you will see actually in your undergrad also. Okay, so you will be well equipped to handle difficult questions also. This is the most tested part for school exams. Okay, so there are a lot of questions which will come based on this concept, especially formulation of a differential equation which involves second order derivatives. So those will be the very frequently asked questions in schools. Okay, and finally we will talk about differentiation of inverse of a function, differentiation of inverse of a function and differentiation of a differentiation involving, I would say not of differentiation involving, involving functional equations, functional equations. So it's a great depth that we are going to cover the topic. So please do not have any inhibition that anything is going to be left out. No, I'm not going to leave out even iota of syllabus. Okay, reduce syllabus is not in my dictionary, the word is not in my dictionary. I do not want you to suffer for the mistake that CVSE people are making. Okay, till I am there as your math teacher, I will make sure you know more than what you are supposed to know. Okay, so these are the topics which we are going to cover up. So let us start today's journey with implicit function differentiation of implicit function, differentiation of implicit function. Implicit is basically a word in the dictionary, right? And you can search for the meaning of it. Explicit is something which is very clear implicit, which is not very clear. So implicit means hidden. Okay, so when we say there is a function, there is a function f of x, y equal to a constant where there is a difficulty in separating the variables out where separation of variables is difficult or challenging. I would not use the word impossible, but it is difficult. And actually we do not need to do so. And in these type of functions, let's say somebody asks me to find out what is dy by dx? Okay, so how do you find the derivative of y with respect to x or x with respect to y or x with respect to y? So all those concepts will be covered under this topic. Okay, now for this topic, you just need to know your chain rule properly. Now just a quick recap of the rules for everybody. Okay, so just a recap of the methods of differentiation. Methods of differentiation. Just quickly, hardly it will take five minutes for me. So we are aware of the sum difference rule. What is it? Sum difference rule. So if you want to find the derivative of sum or difference of two functions with respect to any variable, it is as good as taking sum or difference of the derivatives of these two functions. Okay, this is called the sum difference rule. Then we learn the product rule. So what is the product rule? So if you have the derivative of product of two functions or in fact, who are more than two functions will also talk. It is nothing but it's the derivative of one function multiplied to the other plus and vice versa with the other one also. Okay, and let's say if you have multiple functions involved, let's say if you have f, g, h function involved. By the way, I'm not writing that x, x in the bracket because that will just make it look a little bit ugly. So you differentiate it one at a time. So for example here, differentiate it with respect to x keeping f and g as it is. Then differentiate g keeping f and h it is as it is. And then differentiate and then differentiate and then differentiate f keeping g and h as it is. Okay, so this trend continues no matter how many functions are involved in the product. Okay, so this is something which is already known to you. Next is your quotient rule. What's the quotient rule? This is where many people slightly mistake the formula. So if you want to find the derivative of quotient of f by g, it is g times the derivative of f minus f times the derivative of g by g square. This is the quotient rule. And finally coming to the chain rule. Okay, what does chain rule say? Chain rule says if you want to differentiate a composite function. Let's say there is a composite function. Composite means one function is spread to the other. Okay, like sine of log x or e to the power tan x. So this is examples of composition of function. So this rule says that first differentiate f of g with respect to g and then multiply the derivative of g with respect to x. Okay, just to give you a simple example because this is the one where people do some mistakes here. For example, let's say if you want to differentiate sine of ln x with respect to x. So what do we do? We first differentiate this with respect to ln x. So let me just write it down as an expression. So this we differentiate with respect to ln x. And then we multiply derivative of ln x with respect to x. Okay, so for the sake of simplicity, normally people will write this as a t or something. You may call it as a t. So it's like you are doing d sine of t by dt and multiplying it with derivative of ln x with respect to x which we all know is 1 by x. Derivative of sine t is known to be cos of t and t is nothing but ln of x. So this is how you end up doing the chain rule. I'm just giving you a simple recall of the formulas. Okay, I will go into much depth of this. Probably it will blow up your mind so much of depth will go up. Now one important thing I would like you to highlight over here that this rule that you have written. Okay, it is very, very important rule when it comes to you applying the derivative to a function with respect to a variable in which it has not been expressed. Now let me give an example of what I am saying. Let's say I want to differentiate sine y with respect to x. Even this requires chain rule, right? Many of you must be thinking, sir, this is not a composite function. No, it is a composite function. It is basically made up of sine of another function. Okay, but that is written in terms of y. So how do you differentiate a function which is written in terms of another variable with respect to some other variable? So for this also we use chain rule only. We say, okay, let me differentiate sine y with respect to y first and then multiply dy by dx. The idea is to actually realize that there will be an indirect cancellation of dy and dy. And you will be left with d sine y by dx, which is the left-hand side expression. But we cannot do this activity directly because I do not know how sine y will change with respect to x. I know how it will change with respect to y. That formula was already given to us in 11. Okay, so we'll use that as an intermediate step to achieve this result. So this is nothing but it is again a chain rule concept only. Okay, so this is not coming on. So sorry, I just raised it by mistake. Okay, so this is not a rule which is coming from sky. It is coming from chain rule only. Sir Dheera sir kept saying you have learned chain rule, but you are confused what he's talking about. Oh, okay. So this is nothing but cos y into dy by dx. Okay, now this dy by dx you cannot do much about it. So that will stay as it is in the expression. Okay, so anybody has any doubt in this understanding? I'll take more examples. Don't worry about it. I'll take more examples. So let's say I want to do the derivative of let's say x square y cube with respect to x. How will I do that? How will I do this? See, chain rule is a very intuitive rule. I'll just give you a simple example. Somebody was asking how do you derive chain rule? Let's say I have a very simple analogy for this. Let's say Anurag and let's say Gaurav and let's say Hariharan. These are three sports people. Okay, they're running for, let's say they want to participate in some, you know, race competition. I know Anurag is speed of Anurag is two times speed of Gaurav. Okay, and I also know speed of Gaurav is three times speed of Hariharan. Okay, so if somebody comes and says or asks you this question, how fast is Anurag with respect to Hariharan? What will your answer be? What will your answer be? Six times? Yes or no? Correct? He will be six times. So this question mark will be six. Okay, so what you have actually done over here is nothing but you have done a chain rule only. You know how fast is Anurag with respect to Gaurav. So how is the differential of A with respect to differential of G, right? It is known to you. You also know how fast is Gaurav with respect to Hariharan. So this was two and this was three. And if you want to find out how fast is Anurag with respect to Hariharan, you just multiply dA by dG into dG by dH. Correct? That is nothing but two into three and you get six times faster. So in the same way, if you want to know how fast a function or what is the differential change in this function with respect to X, then you first know how fast will this function change with respect to G of X and how fast G of X is changing with respect to X. Correct? So think as if this is Anurag with respect to Hariharan. This is Anurag with respect to let's say Gaurav and Gaurav with respect to Hariharan. Correct? So there is no like, you know, very hard and fast, you can say rocket science or very rigorous proof hidden in chain rule. It is just a change of one variable with respect to another. That is what you are trying to link. Got it Anurag? Very simple proof. It doesn't require out of box thinking at all. Okay. Anyways, now my question is still unaddressed. So let us talk about this. I'll erase this all because I need space to solve it. Yes, how would you solve this question? Now, this is the question where you would require, of course, your rules of differentiation, which is your product rule and your quotient rule, sorry, and your chain rule. But here you need to understand we have a product of two functions. Okay. Now, let me differentiate them one by one. So let me keep y cube as it is. And I have to differentiate x square with respect to x. Right. And then I'll keep x square as it is. And then I have to differentiate y cube with respect to x. Get this clear. So basically I've started with the product rule. Right. Now, derivative of x square with respect to x, everybody knows 2x. No doubt about it. Correct. But derivative of y cube with respect to x. Now, don't say 3 y square because that is not the answer for derivative of y cube with respect to x. That is the answer of derivative of y cube with respect to y. Are you getting my point? So here you would require chain rule, my dear. Now all of you please see you have to differentiate y cube with respect to y because this is what you know. Right. It is think as if you want to know how fast was Anurag with respect to Hariharan. So now you know how fast is Anurag with respect to Gaurav and how fast is Gaurav with respect to Hariharan. Getting my point. So whatever you know, you try to express it in that those terms. Correct. So your answer will finally become y cube into 2x and this will become x square times 3 y square dy by dx. Okay. So the derivative of the derivative of the function x square y cube with respect to x will be this. Now this dy by dx will remain dy by dx. Still any other information is given to you about y. Okay. You can talk about, yes, you can compare it with that also Hariharan. Is this fine? Any questions here? So this rules must be clear in your mind before we start talking about the implicit differentiation. I'm sure you would want me to do another example. Okay. This time I'll give you a complicated one. Tell me the derivative of x cube plus 2y square x upon xy plus 3x square. Okay. Let me elongate this bracket. It was smaller one. Okay. With respect to y. With respect to y. Everybody please try it out. Now why is y a part of the final answer? Why not? Now gone are those days, gone are those days where your derivative used to only have functions of x or a single variable. Okay. So when you have a multivariate function, as of now we have a function in two variables. So my dy by dx or whatever derivative we are looking for with respect to whatever variable, it will have different varieties of terms in them. Okay. So we are just broadening the scope of whatever you have learned in your previous classes. Can you explain the second step of the first question? Yes. Why not? Yes. See first step is clear. First step. What was happening? Product rule. I'll write it down over here. This was product rule. Correct. This is clear. Okay. Now. When you're finding the derivative of x square with respect to x, that is not a problem at all because this is just a simple power rule of differentiation. Correct. But when we're finding the derivative of y cube with respect to x, now see here, your function is in terms of y and you're finding the derivative with respect to x. So you cannot write 3 y square for that. That would be wrong. Why it will be wrong? Because 3 y square is the derivative of y cube with respect to y, not with respect to x. So here, immediately for this chain rule will come into your picture. Okay. So you'll have to differentiate it first with respect to y and then multiply it with dy by dx, thereby creating a scenario as if your this term and this term is getting cancelled. So ultimately, eventually you're doing dy dx of that term only, but not very, very you can say explicitly. You're doing it in a phased manner in these two steps or by multiplying these two, you can say expressions. What the point? So derivative of y cube with respect to y, let me just erase that cancellation else you will think actually I have cancelled it. No, I have not cancelled it. This is just for explaining you. So derivative of y cube with respect to y, that is 3 y square and this dy by dx, this will be as it is. It'll come down as it is. Okay. Yes, my dear students, who is done with this? Done. Just say done if you're done. Yeah. Let your answer be in terms of dx by dy only. Okay. Done. Advik done. Adid done. Pratik done. Imanchu. Shitesh. Why is it so bad? Okay. We'll do this. So the look in the feel of it tells you that there is a quotient rule involved here. Right. So quotient rule is what we are going to start with now within quotient rule. I may require some rule. I may require chain rule. Okay. But the overall, the bigger picture, the bigger picture is we are going to start with the quotient rule because there is something divided by something. Correct. Yes or no. So let's start with our structure of the quotient rule. In structure of the quotient rule, we write the denominator times derivative. Now derivative with respect to what? We are doing it with respect to why? Derivative of the numerator. Okay. As if not just write it down. We will go into depth of it little later on when we are expanding it. Minus. I just some of you would have been trying to simplify it. Don't do it. As of now I am not interested in seeing your simplification. My whole and sole agenda here is you should understand the process as of now. Simplification you can do also right. It is up to you. It is in your hands. Okay. Yeah. Minus the denominator. Sorry. Minus the numerator which is x cube plus 2 y square x into derivative of the denominator. Divided by whole divided by the square of the denominator. Correct. This is what quotient rule has taught us. Okay. Yes or no. Is this rule is this step clear to everybody? Because if this step is not clear, there's no point going forward. Clear. No doubt. No question. No inhibitions in the step. Okay. Now let's start simplifying these two gentlemen. This guy and this guy because rest of all of them are static in nature. We don't have to do any operation on them. So here it will be x y plus 3 x square. Now derivative of this will you have to introduce this term in sight. So you'll have derivative of x cube with respect to y and then you will have here 2 times derivative of y square x with respect to y. Okay. Similarly here also this term let me copy as it is. Derivative of x y with respect to y and derivative of derivative of 3 x square with respect to y 3 you can take out if you want. Correct. Whole divided by whole divided by whole divided by the denominator square which is x y plus 3 x square the whole square. Okay. Now all of you please pay attention. All of you please pay attention. Look at this term. What is the answer of this going to be? Is it going to be 3 x square? Yes or no? No. Is it going to be 3 x square? Yes or no? Definitely no because I never asked the derivative of this with respect to x. If it was asked with respect to x, yes, it would have been 3 x square but I have asked the derivative with respect to y. So here chain rule will immediately come into picture. Chain rule will immediately come into picture. Correct. So this will be done in this way. You have to find the derivative of x cube with respect to x and then the derivative of x with respect to y which will actually lead to 3 x square dy sorry 3 x square dx by dy. Got the point. Very good. Many of you have got this right. Okay. Is this clear? So don't forget this dx by dy else it will become completely wrong. Zero marks will be awarded. Okay. Similarly this guy. Okay. Now here you will apply a product rule. So first you will differentiate let's say y square which will be 2y with respect to y. y square derivative with respect to y is 2y. Do you want me to write it in a more broken down manner or is it fine? Fine. Okay. Then y square derivative of x will be dx by dy. Is this clear or not? Anybody has any doubt? Immediately stop me because we are going to go into much details much depth and then it will be very difficult for me to come back. Okay. All right. So coming now here this term again this term again x derivative of y is going to be one because you are differentiating with respect to y and y derivative of x is going to be dx by dy. Any questions here? Any questions? Which part are you? Which part you want me to explain you again? This guy y square x1. Okay. I'll do it separately for you. Y square x1 right? Okay. Yes. Okay. Now focus here. I'll just make a demarcation over here. You are finding the derivative of y square x with respect to y. Correct. Now this is two terms. So I will immediately come into action with respect to the product rule. So I will say okay. Let me first differentiate y square keeping x as it is and then keeping y square it is as it is I will differentiate x. This is what product rule teaches us. I hope this is clear to you. Okay. Now derivative of y square with respect to y is 2y. This x is as it is. Similarly y square will be as it is dx by dy. We can't do anything about it because we don't know whether x how x is related to y. So we have to write it as dx by dy. This is what I wrote over here. Is that clear? Now is it clear? Dear students, let's have no doubt in our mind because this is the building block of calculus for you. And as you can see, I've already given you a brief of your classroom. It is calculus, calculus, calculus everywhere. So if you're thinking that okay, I'll go and understand it from my daddy or my elder brother or some tuition master of mine. Don't wait till that. I'm there to help you. Right. My whole and sole purpose in off-life is to help you. Getting the point. Yes. Any other doubt? Okay. Now, I've already written these terms. If you can actually collate it and write it, I will be happy. But let me do the honors. This into this term will be 3x square dx by dy. This term will be two times of 2yx or 2xy, whatever you want to call it. Okay, don't simplify. I have no interest in seeing the simplified version of it. Sorry. I just want you to understand the process and, you know, have a good command on it. This guy was this guy was x plus y dx by dy. And this guy as I've already missed out, this is actually three times 2x dx by dy. Okay. Whole thing, whole thing, whole thing here divided by let me use a let me use a let me use a tool from here, which will help me to underline it. Yeah. This divided by xy plus 3x square to the power of 2. Okay. Probably I gave you a slightly difficult example to begin with, but I think the process is more or less understood. Okay. Now, in implicit differentiation, we will have to know these concepts very well. So let me start with a simple question that you would be getting very, very simple example. Let me just take a question as an example for you. Where is the question gone? Differentiation. I'll give an example for me. Let's say I have xy cube minus yx cube equal to x. Okay. So this is an implicit function. And in this question, I want or in this function, I want to find dy by dx. I would request everybody first to attempt this out. Everybody first take a serious attempt at it and then we'll discuss it. I'm giving you around 90 seconds for it. Everybody first write out find dy by dx from this. I know some of you may know how may not have any idea how to do it. I'm just trying to, you know, give a question which is from your untreated domain. I want to see if you are able to attempt it. Hurry up. Try it out. I'm definitely there to help you out. Just want to see whether you are able to understand in these kind of scenario. How do we find dy by dx? Just a second, guys. I think let's use our understanding. No, no, no, no. Don't type in anything. Just say we are done. Just say we are done. Done. Okay. So let me take this up. So when you're solving this, you don't have to find y in terms of x at all. You can start differentiating from left to right. Whatever terms you start getting, you differentiate, you know, every term that comes in your way. So let us differentiate. Let us differentiate with respect to x on both the sides. Okay. What I'm doing, I'm going to differentiate with respect to x both the sides of my equation. Okay. So dy dx on both the sides. Now, when I'm doing this activity, just remember whatever exercises we had done a little while ago, recall all those concepts. So this will be derivative of this. Okay. Derivative of this. Guys, initially I'm doing it in a very, very staged manner in a very, very, you can say granular level, but I will not be doing it to such a great detail when I'm actually solving the subsequent question. It is because it is your initial days of learning. I'm writing every step possible just because you should not miss out on the basic understanding. Later on, you yourself and start skipping a lot of steps. Some of the things you will do mentally only. Let's say I read this is some mental thing I will do it. Okay. Okay. Now, all of you please pay attention here. When you're differentiating this, you are going to use product rule. Correct. So here, x will be as it is, what will be the derivative of y cube with respect to x? Now again, chain rule will come in your mind. Y cube derivative with respect to x. What will be the answer? What will be the answer? d y square d y by d x. Yes or no? Correct. Plus y cube. That is the second term. Basically, I'm following the product rule formula x derivative with respect to x is going to be one. Okay. You may not want to write one there because one into anything is the same thing. So this is the first part. Okay. Put it in brackets. Minus. Let's talk about this guy. Again, product rule. So first I will keep y as it is derivative of x cube with respect to x. That is pure and simple 3 x square. No doubt about it. By the way, I'll tell you an inside story. When you find the derivative of x cube with respect to x, you know what you do actually, you actually write 3 x square into d x by d x, which is actually becomes the one. So nobody's like nobody writes it actually. This is the inside story. Are you getting my point? Okay. So this is something which we normally do in a very, very hidden way. So your chain rule that you use is actually used everywhere. In some cases it is it can be seen. In some cases it is hidden. Okay. So now x cube I will keep as it is derivative of y with respect to x will be d y. d x. Okay. On the right hand side, there will be a one. Now what was my question asking me? The question was asking me to find d y by d x. So for finding d y by d x now I will have to collate. I will have to see how many d y by d x are coming and we have to take it at one place. So what I normally do. I write d y by d x. I put a bracket over here just to collect all the terms which are associated with d y by d x. And other terms will be kept over here. Okay. So d y by d x terms will be in this. It will be in this. So I will be collecting them. So it will be 3 x y square and from here I will get an x cube. Okay. No doubt about it. I'm just collecting my d y by d x terms and other terms will be y cube from here and minus 3 x square y from here. Okay. By the way, it's a very healthy practice to always write x before y because that is how they alphabetically also come. Is it fine? Oh, thanks for correcting that minus x cube. Thank you. Any questions? Any questions so far? No questions. So what I'm going to do is I'm going to send these, this guy to the other side. So let me just write down the next step. This and girls. Again, I'm saying I'm repeating. I'm doing it in a very, very detailed way. But I may not be able to do it in so much of detail for the subsequent problems. It is just the first problem and hence I'm taking a bit of time to do it. So this becomes your final answer. Now many people say, sir, d y by d x has a y term in it. Yes, it can have y term. Why not? You can have a y term in d y by d x. Why didn't we cancel x in the first step? Oh, you can. I mean, of course I can understand you want to cancel out x. You can. It's fine. Okay. But let's say if there is no such case like cancellation, then this is the rule that you can follow. Is it right? Any question with the approach? Please immediately highlight. Immediately highlight this. Okay. By the way, many people will think, sir, this long we'll have to do to solve this. Don't worry for J main and advanced exams. I will give you a direct formula for it. Aditya. But d y by d x, we're looking for how y changes with respect to x. Yes. So Aditya. Okay. You may not be able to write y in terms of x in implicit functions or it will not be very convenient for you to write early in this example. I'm not sure whether you are able to do so. Let's say even if you cancel it out, I don't think so. You'll be able to write y completely in terms of x. So this approach, whether you are able to write y in terms of x or not able to write y in terms of x, it will work for both the scenarios. Please do not have any wrong inhibitions that this will work only when there is a function where x and y cannot be separated. No. I can also do it when x and y can be separated. Are you getting my point? For example, let's say somebody says find d y by d x in this expression. Okay. Find d y by d x. I'll give you another simple example. Okay. Now I could do y as 3 by x and then differentiate it or I can apply my differentiation from the word go itself. So I can start differentiating with respect to x. So when I do that, I can write it like this d by dx of x y is equal to d by dx of 3. Okay. Of course, a constant derivative will be 0. Here I will use product rule x d y by dx plus this will become how much? y into 1. So d y by dx will become negative y by x and your y itself would be written as 3 by x. So this will become minus 3 by x square. So both are right actually. Right. So if somebody decides to write this like this and then does d y by dx, he will eventually get the same answer. He will eventually get the same answer. So don't get me wrong that these processes are only meant when x and y are different. When x and y are difficult to separate. Of course that is meant for it but it could be used for any scenario you want because the rules are universal rules. Is it clear anybody? Okay. So there was one question from Aditya which was I think I did not answer. Yeah Aditya. So y can come in your answer. See d y by dx when you're finding your finding is what is the rate of change of y with respect to x and that may depend on the instantaneous value of x and y where you are finding the rate of change. So y value can be used there. Are you getting my point? So because of our class 11th differentiation we are so used to not seeing a y in d y by dx that we think that it can never appear. No that is wrong. d y by dx can have y terms in it. Okay. How did you get? Oh my God. How did you get that term? Yes. My dear good friend. 3 into 1 by x if you want to differentiate it with respect to x. Okay. Can I write it as 3 into derivative of x to the power minus 1. x to the power minus 1 is following x to the power n derivative. So it is going to be n x to the power n minus 1. So that's going to be negative 3 by x squared. Don't ask such a question in this stage. This is class 11th power rule of differentiation. Revise it. Shares it is for you to note down that I need to revise my basic differentiation. Okay. Yes. We cannot proceed with difficult problems or higher level concepts if our 11th are slightly shaky. So please revise it before the class. You already know the chapter is going to be this covered for the next few weeks. Okay. So please revise it and come. Is it fine? Any questions here? Okay. Let's have another question. Let's have another question. Let's get good with our basics first. So first question is what we already did. So no need to do it. This is already done. Try this one out. Find d y by d x if x cube plus y cube is equal to 3 a x y. By the way, this is a very interesting curve. Okay. This curve is called Folium of Descartes. Okay. Folium of Descartes. I would actually show you the graph of this also. Okay. Let's go to by the way, people who are attending these sessions for the first time. Okay. There is a graphing software that we normally use and that is called the GeoGebra graphing software. Okay. I'll just reopen it. You can always have one on your phone also, which is Desmos. If you want to have one on your laptop or desktop, you can have GeoGebra. So GeoGebra.org. If you visit that site, you can download GeoGebra classic six there. So this is called Folium of Descartes because you'll see the graph of it. I'll just take a simple value off. Yeah. As you can see, there's a folium being created over here. Okay. A knot. It's a very interesting curve. Why? Because if you see, I mean, I'll talk about it when the right time comes in detail. If you see the X, the X axis is actually tangent and normal simultaneously. You see what? How is that possible? Yeah. So the X axis here is tangent to the curve and normal to the curve simultaneously. Okay. How it is tangent. It is because you can see it is touching this part of the curve. Correct. And how it is normal, it is because I just change the color. It is perpendicular to this part of the curve. Okay. It's a very interesting curve. Okay. It's not a real valued function. So how do I see the graph of it, my dear? If it is not a real valued function, how can I see the graph? Real value function graph only I can see. Okay. Now I started going to question how it is normal. Yep. This part is basically being cut by the X axis at right angles. Okay. So this is the right angle. Don't worry. I'll talk about this in the chapter tangents and normals with you, which is your application of derivatives concept. Let's come back. Let's come back to this question. Just to tell you the name, it is called the folium of Descartes. I know it. I know all of you are aware of the famous French mathematician Rune Descartes. Rune Descartes. The pronunciation is Rune Descartes. Not Rene Descartes. That many people say Rene Descartes. Yes. Done, my dear. Second one. Anybody done? Just say done. Done. Nishant done. Shrita is done. Advik done. Gurman done. Aditi done. Done. Done. Done. Done. Okay. Very good. Let's discuss. So what are we going to do is we are going to start differentiating it with respect to X on both the sides. Okay. Let's differentiate with respect to X on both the sides. Now, many people ask me, sir, can you differentiate with respect to Y on both the sides? Yeah, you can. Let me do that. Okay. Let's take a different approach. Okay. Unconvention approach. Let me differentiate with respect to Y on both the sides. Right. Why only X should get that special treatment? Why not Y? Let's differentiate with respect to Y. So now, if you differentiate X cube with respect to Y, you will be writing 3X square DX by DY. Correct. Any questions here? Anybody? No questions. If you differentiate Y cube with respect to Y, you get 3Y square. Any questions? No questions. When you differentiate XY with respect to Y, again, there is a product rule. 3A is a constant. So you can bring it outside. So X into derivative of Y will be 1. Y into derivative of X with respect to Y will be DX by DY. Any doubt in this step immediately stop me. Let me not move any further. Okay. Yes. That's why I chose a different approach in Iraq because I wanted to show that you can still find DY by DX even if you're differentiating with respect to Y. Okay. Anyways, let's collect all the terms having DX by DY in them. So DX by DY will have 3X square from here and it will have minus 3AY coming from here and you will have 3AX minus 3Y square as the other term. Let's do one thing. Why not we drop the threes unnecessarily? We are carrying this burden of 3, 3, 3, 3 when it can be cancelled easily. Correct. And from here we can always bring this DX by DY to the right side and this is what you are going to see as your answer for DY by DX. Is this fine? Any question? Any concerns? Anybody? Do let me know. Any kasta? Any concerns? Okay. Time to take a break. I know everybody would be earning for a break. Sir, mala bhuk lagli, mala break baya. So let's have a break. Let's have a break. It's always 6-5. Normally, for the new students who have joined us, we normally give a break at around 6-5 or 15 minutes sharply, not more, not less. Okay. So enjoy your break for the next 20 minutes. Okay. And on the other side of the break, I'm going to give you a very interesting formula to do all these things pretty quickly. Okay. Fine. Can we all take a break now? Any question anybody has? Anybody? Any question? Any question? No question? Shalo. Alright. So my camera is normally off. So for the very same concept, now I'm going to tell you a very simple straightforward formula, even though it comes from your higher understanding of calculus, which probably you will study in your first year of graduation of undergrad. So let's discuss about it. See, actually, what are we doing? I have been given with, I'll just give you a formula for implicit differentiation. Implicit differentiation. Okay. So here, our concern is if we have been given a bivariate function, let's say equal to some constant, and I want to find dy by dx, how do we actually do it in a pretty fast, you know, in a pretty easy step in a more efficient way. Now try to understand this. Whenever there is a function which has got multiple variables. Okay. So let's say if I want to find out the derivative of this function with respect to t. Okay. There is a rule in calculus which is called total differentiation. So what is this rule? This rule is called total differentiation, which anyways, you would not have heard of it, neither from your school textbooks, neither from your J textbooks. So this is the advantage that Sentom brings to you as compared to any other institute. So df by dt is nothing but it is doh f by doh x into dx by dt. For example, here there are two variables or two independent variables making that function. So we find what is the change in the function with respect to these variables and what is the change in these variables with respect to time like this. And had there been more variables involved, let's say there was a z also, then you would have doh f by doh z into dz by dt and on and on and on. Now let me explain this doh f by doh x term or for that matter, what is the meaning of a partial derivative? Some of you already know partial derivative. What does it mean? It means you are differentiating the function only keeping x as the variable. So this means differentiating f keeping only x as constant, keeping only x as variable and rest will be constant. Rest will be constant. Just to give you a simple idea, let's say somebody says that there is a, you know, function x y which looks like x y square plus x square y. Okay. And he says, okay, tell me doh f by doh x. Okay. Tell me doh f by doh x. Means he wants you to differentiate this keeping only x as the variable and y is a constant getting my point. So this guy will just become y square because y square is a constant. Derivative of x is going to be 1. Okay. And this guy is going to become 2x into y. Is it clear? This is called the partial derivative of f with respect to x. We'll talk about t. Don't worry. I'm going to replace that t. Not to worry at all. Okay. Just understand this rule. First of all, what is the meaning of partial derivative? First of all, and then we'll try to connect the dots. Which one? Got it. This one? Total differentiation? The example. Okay. See, when you say you are partially differentiating, now the symbol here that represents partial derivative is doh. And again, partial derivative is not a subject matter of your class 11th and 12th. It is primarily going to be taught in your first year of undergrad. But I found this formula to be very effective. That is why I'm sharing it with you. And you'll see in some time that how convenient or how easy it becomes to solve questions related to implicit differentiation when you use this formula. So now the process is if you want to do a partial derivative of this with respect to x, means start treating as if your y square is a constant, y is a constant. That means only x will be treated as a variable. For example, somebody says x into 3 and let's say x square into 5. How will you differentiate this? Say 3 only. And here you will say 2x into 5 only. Correct? No? Gaurav? That is how you are going to differentiate it? Yes or no? Gaurav? Are you there? Yeah. Gaurav is there. So, similarly here I'm going to treat only my x as a variable. Y square is just like a constant for me. So, y square constants never, you know, when we have a product of a constant with a variable, constant we normally take out and differentiate the variable. Y square is taken out, derivative of the variable is x, x derivative is 1. Similarly, y is taken out, derivative of x square is 2x. If the process understood, how do we do partial derivative? Now, same example I will show you, partial derivative with respect to y. Now, when you are partially differentiating with respect to y, x will be treated like a constant. In fact, x, x square, whatever terms of x that will be treated as a constant, only y will be treated like a variable. Similarly, where x will be like a constant, y square derivative will be 2y. Similarly, this x square will be a constant, this y derivative will be 1. Now, understood what is the meaning of partial derivative? Very good Gaurav. Any other doubt anybody has? Shares? Now, is that clear to you? Shares k? Partial derivative part is clear, no? Other part I will make sense in some time. Don't worry about it. I'm just putting a barricade over here because this is a separate discussion altogether. Now, all of you please pay attention here. All of you please pay attention. Somebody was asking what is this t doing over here? This t is just a variable which I have introduced because it could be any variable. I want to study how this function changes with respect to time for that matter. I could also find out how it changes with respect to x. I could also find out how it changes with respect to y. Now, what I'm going to do, listen to this. In this step, I'm going to replace my t with a y. So let t be replaced with a y. Oh, sorry. Let it be replaced with an x. You can do it y also and it doesn't make any difference. So when you do that, this becomes d df by dx is equal to dou f by dou x. Now remember, dx by dx will actually become a 1. Correct? Dou f by dou y, let it be as it is. Dou f by dou y, let it be as it is. This will become dy by dx. Now, since your f is a constant, can I say since your fx,y or whatever f, it is a constant, can I say df by dx will be actually a 0? So I can put a 0 over here. Correct? So this will be equal to dou f by dou x plus dou f by dou y into dy by dx. Guys, you don't need to know the process. Only the final result is what we need to keep in our mind and the final result is right now in front of us. The final result is make dy by dx the subject of the formula. So you will get minus dou f by dou x divided by dou f by dou y. This formula you need to remember, that's it. There is no other steps that you need to keep in mind. So please make a note of it and please remember it. It is going to make your life super easy. It is going to at least save two minutes, at least in solving the question. Okay, I will demonstrate the efficacy of this formula through the previous example that we have done. So how exactly was the first part found? See, you are talking about this formula, the total differentiation formula. Okay, see, normally in your field of physical world, you have a function which can depend upon several independent parameters. For example, I'll give you weather. Okay, weather is a function of so many things. It is a function of temperature of the surrounding. It is a temperature, it is a function of the humidity or the water content carried by the air. It is a function of the wind velocity. All those factors are there. Now you want to see how does this temperature, how does this weather as a function change with respect to something, let's say time. Okay, then what do you do? You see how does weather change with respect to the temperature and then temperature changes with respect to time. How weather changes with respect to let's say wind velocity and how wind velocity changes with respect to time and so on and so forth depending upon how many parameters it depends upon. Okay, when you were trying to study your y is equal to f of x, there are only few things involved. So your function has only a function of one variable that is x, where all these things become simplified to just df by dx. Are you getting my point? So there all these things do not feature out because you are mostly dealing with single variable function. But now this is the first time that we are trying to deal with the multivariate function where two variables are involved. So now these formulas will start surfacing out. These formulas will start coming out for you. Okay, anyway this is a subject matter of your undergrad discussion. I am not going to waste time on it. Okay, what will happen if there was a z? Very good question that you have asked. If there was a z, let's say I will just write it down somewhere over here. If you had a function in x, y, z and let's say I wanted to find out what is the change of this function with respect to some variable, let's say I want to find w. Okay, so what I will do is I will find out the partial derivative of this with respect to x into dx by dw then partial derivative of this with respect to y into dy by dw and partial derivative of this with respect to z into dz by dw. Got the point? Okay, if you want you can make w as any one of the variables also. You can make it as x also. Getting the point. So this is how we basically deal with total differentiation, which is again don't worry too much about it and don't even try to use it in school, please. Okay, there has been our old students who tried to show their extra knowledge in the school exam and they lost marks because of that. Okay, don't do this. Don't try to go beyond your NCRT curriculum. That is very important. Sincere request. Show your extra knowledge to your friends in the JEE paper. I will be more than happy to see that but don't do it in school. Okay, so does it matter what we replace w with? No, you can put any variable there Okay, anyways, so guys cutting the matter short here this is what we need to learn. Now, how efficient is this formula? Let's recall our problems. Let's go to the next slide. Let's call back our questions that we had done. Okay, let's say the first question. Okay, I hope all of you remember the amount of time we took to solve it. So first what I'm going to do is step number one, I'm going to write it and I'm going to write it as a constant. So zero has to come over here. So basically I'm trying to match this part with fx, y and I'm trying to match this part with a constant. So this was the beginning part of our derivation so I brought it to this. Okay, so this is our function of x, y. Now one step is all you need to solve this question. dy by dx is minus dof by dox by dof by doy. The problem is solved but don't do a mistake in evaluating them. So minus sign, let it keep, let's keep outside. Dof by dox. Now if you want to differentiate this with respect to, partially with respect to x, remember y will be constant here. So here it will be y cube, y cube. Here it will be minus 3x square y. And this will be minus 1, right? Divided by, partially with respect to y, partially with respect to y means x will be like a constant. So x into derivative of this is 3y square and then x cube will be like a constant. Derivative of y is going to be 1, correct? And x derivative is going to be 0. So this is going to be your answer. Your answer will be 1 plus 3x square y minus y cube by 3x y square minus x cube. Was this the answer that we had got in the previous question? Check, all of you please check. It was fast, okay? So these are the tricks which normally I'll keep. Arithra, I was so sweet. It looks like a illegal activity and it does look like, and please do not use it in school. Not to be used in school. Don't try this in school. Don't try in school, okay? School teachers will penalize you, telling you. Does it? No, it should not. It should not give you a different answer. Okay, anyways, is this clear to you? So you can use this in your competitive level exam to save a lot of time, okay? Let's do one thing. Everybody please try the second question also. Apply it on the second question also. So second question, everybody try it out. And see, just somebody was asking me, if you're cancelling things out, Hariharan, you may lose the actually, see, there is a lot of vagueness there actually because sometimes we normally solve it by doing simplification. Sometimes we don't. I would suggest don't interfere with it. I would say don't interfere with the expression. Anurag has a question. Will you be able to derive the total differentiation formula? For that, you need to know partial derivative concept actually. It is intuitive also, no? Anurag, don't you think it is intuitive? If you want to see how a function, which itself is made up of several variables, changes with respect to a variable, you need to see how every variable is changing with respect to that particular, let's say t, I was talking about t. So how every variable is changing with respect to t, it's like an extended version of the chain rule. But I would say keep these research work after you pass out, after you do your 12s, get into IITs, get into whatever colleges you feel like and then do this research work. You are young cells, you would definitely like to chase this thing. Don't worry, don't go into too much of details. Okay, very easy. I think you must be done. I'd not come out to be zero. No, then you have done a mistake, Arithra. See, first of all, bring it to this form. Correct? Okay, so now your formula is dy by dx is minus dou f by dou x by dou f by dou y, correct? So minus, don't forget this minus. Some people, they forget this minus somehow. I don't know. It may happen with you also. Now, what are the partial derivative of this with respect to x? Now, the moment you're partially differentiating with x, you should start treating y like a constant. Okay, so 3x square for this, y is a constant, so 0 will be coming over here. This will be minus 3a y. Because it's like some constant multiplied to x, like 5x, 5x derivative. 5 over, so constant will only be left off, okay? Divided by whole things partial derivative with respect to y. Now, for y, you will treat x as a constant, so this will be 0, this will be 3y square, and this will be minus 3ax. If you want, you can drop the factor of 3, so this will become y, sorry, a y minus x square divided by y square minus ax. Was this the answer that we had got for this question, or there was some different answer? I don't think so. Okay, it is fast, it is fast. So this basically does away with the need to, you know, segregate your dy by dx and all those things, and then finally frame it out as a subject of the formula. So those steps will become really, really fast, okay? Anyways, so let's continue with few more questions. Let's have few more questions. I think the third question is what we can take up next. Now that you know the formula, you can actually take the third question. Everybody please try the third question. Find dy by dx for this. You can use the formula if you want. Okay, how do you differentiate when the base is not e? Okay, good question. Let us say I want to differentiate, I want to differentiate log x to the base of 2, okay, with respect to x. All of you please pay attention. Here we use the change of base property. So we write this as log of x to the base e divided by log of 2 to the base e. Okay, this is called the change of base property. I'll write it here. Change of base property. Okay, change of base property. Now log 2 to the base e which is ln 2 is like a constant so it will come out, okay? And it is as good as differentiating ln x. So ln x differentiation we all know is 1 by x. So it's 1 by x ln 2. Make sense? Any question? Okay, this is just, I think this is the class 11 stuff which probably you would have not taken very seriously. Done, is it done? Just write done on the chat box. Done, Himanshu is done. Very good. Pranav is done. Shruthish is done. Adhvayak, Nishant, Gaurav, Aditi, Kashav, awesome. Okay, so first of all I would write it in our small, simple ln language, okay? This is my f, okay? The f which I use in the formula. So dy by dx, keep writing it so that you don't forget it is minus dou f by dou x by dou f by dou y. Okay, pronunciation is dou, d o w dou, okay? So negative, now this entire thing you have to partially differentiate with respect to x. Means y and any associated function of y will be constant only. So this is like x into constant. Getting my point. So how do you differentiate x into a constant? That constant only will remain, no? Right, convinced? Now here, y is a constant, ln x is a variable so it will become y into derivative of ln x which is 1 by x. Constant anyways will be not featuring in, that will be gone. Divided by dou f by dou y. Okay, now change the rule now. y will be a variable, x will be a constant. So this will be x into 1 by y. Correct me if I'm wrong. And again this will be ln of x. If you have any doubt in the way I'm doing these partial derivatives please feel free to stop me. Ask me question n number of times. I will be answering you back to n number of times. Don't go from this class with a doubt in your mind. Okay, so that is the service that you can do for me. Okay, don't go with a doubt in your mind. Other than how that formula and all come because they are not important at all. That's not waste time. We have very limited amount of time. Partial derivative is clear how this form, I mean the use of the formula is clear. If anything related to this is there please highlight. Okay, now it is up to you to further simplify it right in the proper way but I'm not going to do that because that's not my agenda in this class. Okay, let's have more questions. Now this is a single choice, single option correct question. If sin xy plus sin cos xy is equal to 0 which of the following is an answer for dy by dx. Now I'm putting the poll on for this. Once you're done put the poll, put forth your response on the poll so that I can know how many of you have voted for which options. Oh, somebody has already given the answer also, good. Which of the following is your dy by dx? One minute gone, three people have responded so far, others freeze back up, I'll give you around one more minute to do this. It's just a two minute question, not more than that. When you see the solution you would realize that it was actually a 15-20 second question. Good, good, eight of you, nine of you, 11 of you. Last 30 seconds, last 30 seconds, those who know me, they always give a question along with the time running because you should not solve a question as per your sweet timings because your speed will suffer. Okay, countdown begins, five, four, three, two, one. So 32 out of 39 or 40 of you have responded and out of that 73% or 72% Janta is saying option number B. Okay, let's see, B for Bombay. See, if I have to solve this question, it is pretty straightforward. I know you, a lot of formulas will be coming in your mind. That's where common sense takes the backseat. So first of all, this is nothing but tan xy is equal to minus one. Okay, by the way, I am not sure how many of you would be aware of the value of tan inverse minus one. Any guesses for tan inverse minus one value? Let me see if anybody has could guess this value. Tan inverse minus one, which angle? Okay, good try, good try. Some of you have said 135. Unfortunately, this answer is minus five by four. But anyways, not required. This part is not required right now. Not required to know right now. All I need to know is that this guy is a constant. That's it. Whether it is minus five by four or how do I care about it? I don't need it. Okay, anyways, once you've got this, either you can use your normal differentiation or you can start using your heavy formula also. It's not a heavy formula. So dy by dx is minus dou f by dou x by dou f by dou y. I'll use this also. So minus, this is like your f equal to constant. So compare this scenario, compare this scenario with f equal to, or fx, y, fx, y equal to a constant. Okay, so now the partial derivative of this with respect to x will be just a y. Okay, partial derivative of y with respect to y will be just an x. So your answer will be minus y by x. Option number B, B for Bombay is absolutely correct. Absolutely correct. Okay, you can directly use implicit also. That will also give you the right formula. Let us try it out. So this is one way to do it. Number one way, I'll give you an alliter, which is the direct formula that you have learned. So treat itself, this guy itself as your function. So let me write it down. So treat this guy itself as your f. By the way, remember, when you're choosing your f, there should not be any function of x on the other side. There can be a constant, whether zero or non-zero, constant chalega, but koi function of x ya y nahi bachna chahiye o the. There should not be any function of x and y left on the other side. You have to bring everything to one side. Getting the point. This is another mistake which people do. And then they will say, sir, your formula did not work. Okay, so dy by dx will be minus doh f by doh x. Now what is doh f by doh x? What is the derivative of this with respect to x? y will be treated like a constant. So see here. So this will be cos x y. Okay. And derivative of x y, keeping y as a constant will just give you an y my dear. Correct? Any question with respect to this, please stop me. Basically I'm using chain rule when I'm doing the derivative. So sign something is cos something and derivative of this part and you're treating your y as a constant there. So it'll be just y. Can you see me? I'm a audible dear students. Okay. So let's complete this. Plus derivative of this is going to be derivative of this is going to be negative sign x y into now partial derivative of x y with respect to x will again be a y. No doubt about it. Okay. So numerator part is clear. No doubt with the numerator part. Denominator part, the partial derivative of this with respect to y. So this will be again cos x y. Now partial derivative with respect to y will be I'm treating y as a variable but x is like a constant. So x y derivative partially with respect to y will just yield an x. Okay. Similarly this guy will give you minus sign x y into an x. As you can see here as you can see here you can actually pull out a y common from the numerator and of course x from the denominator and you'll be getting cos x y minus sign x y which will happily get cancelled with cos x y minus sign x y in the denominator yielding you with the very same answer minus y by x. I believe this approach was much faster as compared to the first one. But anyhow the result has to be negative y by x. Let's take few more questions. I think you would have done this question and already waiting. Again, apologies for the for this power cut. Let's take this one. If x square plus 2 h x y plus by square is equal to 0 then find dy by dx Let me put the poll on. Guys take some time to solve this question. Look at the options. And one of the options sitting is none of these so that may try to deceive you. All of you are missing out. Let's try to see that. So let's go back to our expression a x square plus 2 h x y plus by square is equal to 0. Here I can do one thing. I can do one thing. Here I can do one thing. I can break this up as 2 h x y I am breaking it up as like this. So a x square h x y is equal to negative by square plus h x y. Now all of you please pay attention. If I take an x common from here I will get a x plus h y and if I take minus y from here I will get by plus h x. In other words minus ax plus h y divided by divided by h x plus by is actually y upon x. And this guy is sitting right over here my dear students. This guy is sitting right over here. So I will replace this with y by x and your answer would be one of the options that you have here which is actually option A. This is what we call as a deceiving question. Many of you would basically thought that there is no A, B, H and you know in my answer so it cannot be any one of A, B and C it has to be none of these but no. This is a question with a twist. Okay so this is a question where you require to do more than what was expected from you and most of your questions in your exam may be of this nature. Yeah so Anurag just to answer your question how do I come to know when I look for solutions for these kind of questions you have to have some kind of an experience at your back to know that it could be a deceiving question. Okay so it could be a deceiving question. So option number A was correct. Let me check how many of actually voted for A. Just three of you said A. Okay good. Let's take another one. Yeah Oshik I am connected to hotspot only right now. Because the power is still coming and going. Let's take this. Let me put the poll on. If you use the formula you will find this question very easy. So try using the formula dy by dx is negative dou F by dou x divided by dou F by dou y. There is no restriction to the use of this formula but let me tell you when you are using this formula don't expect your answer to exactly come out what is sitting in your options. Sometimes your answer in the options may be very different from what you are getting. You need to do further more simplification in those type of questions. So the previous question was an exact scenario where the answer was not coming out as the question as the formula gave us. Okay 2 of you 3 of you have responded very good. Last 30 seconds 5 1 So I could see most of you have gone with option number A almost 50% of you have said option number A. Let's check this should not be a difficult problem to solve. We can directly use our formula for dy by dx which is nothing but negative dou F by dou x by dou F by dou y So negative of Now what is the partial derivative of this with respect to x. Now remember e to the power y will be like a constant so it will be 2x e to the power y plus the derivative of this Now see 2y is like a constant so 2y will come out x into e to the power x you have to apply a product rule you have to apply a product rule so it will be e to the power x into 1 and x into e to the power x okay 13 of course will be 0 so we will not you know care writing about it is it fine whole divided by dou F by dou y now again we have to differentiate the same thing but this time partially with respect to y means all x related terms will be treated like a constant so x square will be like a constant e to the power y derivative is e to the power y okay here 2x e to the power x will be treated like a constant y derivative is going to be 1 itself yes or no derivative everybody knows it's going to be 0 anyways right now if you look at your options your options has a different picture all together in the denominator basically you will see some different kind of a term sitting over here which is not matching with it does it mean we have solved it incorrectly no they probably would have divided the answer by e to the power x so you divide your numerator by e to the power x so when I do that when I do that I will get e to the power y minus x here I will completely lose out e to the power x it will become 2y x plus 1 okay whole thing is in bracket and down in the denominator I will get x square e to the power y minus x plus 2x where if you take an x common if you take an x common if you take an x common this is what you are going to see which clearly matches with option number a well done I mean most of you voted for option number a okay any question with respect to the approach and while you while you are applying this formula you can very well see how efficient is this formula how efficient is this formula for us any questions any concerns okay since we have lost time a bit we will take one more problem before we end the day okay sorry it was the problem from my side but you have to wait for few more minutes this problem which I am going to take is the one which is the favorite of the schools sir we divided by e to the power x how do we know what to divide with okay guy 3 you look at the options didn't I say your option is not directly matching with whatever is the result I got then I observed that they have done a division by e to the power x yes okay let's take this question if under root of 1 minus x square plus under root of 1 minus y square is equal to a x minus y then prove that let me write it in a different line then prove that dy by dx is under root of 1 minus y square okay 1 minus x square this is a very important problem from your school point of view important school level question also now in this question there is actually a trick if you start applying whatever formulas you have learned you may not be able to get to the desired result okay so this is not as easy as what people think it to be this question basically requires you to do certain substitution okay and before that before I start you know doing that there are certain things that I would like to discuss with you ah Pranav you can try it out all of you can give your best shot to solve this okay but before before I proceed with this question there are certain things that I would like to discuss so some prerequisites okay what are the prerequisites for this question let me discuss about it how many of you are aware about the derivative of your basic inverse trick functions like sine inverse cos inverse and tan inverse okay now some of you may be already be aware of it because you have been using it some of you may not be don't worry about it we will you know definitely discuss it in more detail when I am doing the inverse trick differentiation but as of now I would just give you a brief idea that the derivative of these functions are given by this okay we will take it up in more detail more detail under differentiation of inverse trick functions okay I will talk more in detail about this okay now all of a sudden you must be wondering why is sir talking about derivative of inverse trick functions over here when this is not even seen in the question okay now I think the presence of sine etc will actually tell you that there is some substitution which I am going to do over here so all of you please pay attention so I am going to rewrite the question once again under root of 1-x square under root of 1-y square is equal to ax-y so let's say this was my question so here what I am going to do is I am going to substitute x as let's say sine theta okay and y is let's say sine phi now why am I doing it we will discuss it in few minutes time so when I do that this guy becomes cos theta this guy becomes cos phi and on the right hand side I get a sine theta minus sine phi now we have already learned our transformation formulas of trigonometry everybody I think knows the formula of 2 cos a plus cos b which is 2 cos a plus b by 2 into cos a minus b by 2 similarly everybody knows the formula of sine a minus sine b which is 2 cos a plus b by 2 into sine a minus b by 2 so let's cancel out few things like 2 and 2 will get cancelled this and this gets cancelled you bring this sign below this term so it will become cot of theta minus phi by 2 is equal to a okay ultimately what do you get is theta minus phi by 2 is equal to cot inverse a that means theta minus phi is equal to 2 cot inverse a now here is the beginning of our story so this theta can I write it as sine inverse x so this theta I can write it as sine inverse x this phi I can write it as sine inverse y okay and now think as if this guy is a constant because a is a constant so 2 cot inverse is a constant now here you can apply your formula directly so you can say dy by dx so treat this as your function okay so you can say dy by dx is negative dou f by dou x by dou f by dou y so I can write it as negative dou f by dou x means you are partially differentiating it only keeping x as a variable this is going to be a constant so that will be gone divided by partial derivative with respect to y so this will be 0 this will be minus 1 by under root 1 minus y square so when you simplify this when you simplify this negative negative will get cancelled off and you will be ending up getting under root of 1 minus y square by under root of 1 minus x square so as you can see this is what we wanted to prove this is what we wanted to prove now let me tell you if you are trying any other method let me tell you it is going to take you very long to solve it no don't use this formula use your normal differentiation to get dy by dx I asked you not to use this formula for school I never said you cannot do it without the formula don't get me wrong you could differentiate directly also you could use direct differentiation also to get your dy by dx got the point hariyaran so pranav got it by using your method very good pranav so this is one of the very important problems you can mark it with a star also now aditya that is a very good question he asked aditya said sir how did you come to know about these substitutions so basically there is a topic that we will be doing in our differentiation only which is called differentiation of inverse trig functions under that I will tell you there are certain trigonometric substitutions which will make your life very very easy so actually when I do that concept I take it up before this problem so many people are aware of it but again you will slowly learn with more and more exposure so we will stop here and I think because of this power on and off issue because the building was on fire there was a lot of electrical issues happening let me tell you those who were with us last year this is one of the cases it has never gone so many times ok so thanks a lot and thanks for bearing with me through this power cut I will send you an assignment on the group with respect to this topic and share the class notes and recording whatever has been done in today's class till then take care stay safe and again sorry for what happened during the power cut today ok good night to all of you bye bye take care take care bye bye bye everyone good night good night thanks a lot