 So last class, we had a very, very useful and very, very elaborative discussion on finding derivatives by using first principles, right? So we had seen a lot of examples of functions where we found out the derivative of the function with respect to x using first principles, okay? But as we all know that first principle is a lengthy way. It is a, you can say, a complicated way of evaluating the derivative of a function provided it is differentiable. So there are certain methods that we are going to revisit. In fact, we have already done that in our bridge course. But for the benefit of those who joined us a little late, as we always tell us that the bridge course is just a showcase of what is going to come. It is not the, you know, completion of the topic. So bridge course just keeps you ready just to, you know, face whatever is going to come up in your future course of the topic. So we are going to repeat those concepts once again. So I'll be starting with methods of differentiation, methods of differentiation, which will make our life pretty easy while we are finding the derivative of a function with respect to x. Okay, so let's look into the methods of differentiation. The first method is what we call as the sum or difference rule, sum slash difference rule, difference rule. So what does this method say? It says that the derivative of sum or difference of two or more functions is as good as sum or difference of their respective derivatives, sum or difference of their respective derivatives. So I have basically chosen both the addition and subtraction symbol over here because in either case it is going to work fine. Now, what are the proof for this rule? How does this rule come about? Very simple. Let us say I take one of the case. Let's say I take one of the case of f of x plus g of x. Okay, and I call that function by a special name, let's say k of x. Okay. Now what do you want to find out here? We want to find out the derivative of f of x plus g of x. Okay, by using first principles by our ab initio method. In other words, I'm finding out the derivative of k of x with respect to x by our ab initio method. So as for the definition, the derivative of this function can be found out by using this expression limit h tending to zero. k of x plus h minus k of x by h. Correct. And what is k of x plus h? In fact, what is k of x itself? k of x itself is f of x plus g of x. Isn't it? So can I not write this as, can I not write this as f of x plus h plus g of x plus h? So this term, I'm just talking about this term. Can I not write that as f of x plus h plus g of x plus h? We definitely can, right? Minus k of x, k of x is f of x plus g of x. Let's put a bracket around this. Okay. Now let me split open this term and pair certain terms in this fashion. So I'm splitting this expression and I'm pairing certain terms in this fashion. So I take this fellow and this fellow together. Okay. And I decide to take this fellow and this fellow together. Okay. And whole thing is divided by it. Now what I'm going to do is I'm going to split this as two separate limits because our algebra of limits, they allow us or they permit us to write down the limit of some of two functions as separate limits provided both these limits exist. So f of x and g of x here, both are differentiable functions. So these limits will definitely exist. So I'll write it down and like this. Okay. So what happens, what happens to this term? What happens to this term? You will say, sir, this is the derivative of f by using first principles, isn't it? And what happens to this term? This is the derivative of g by first principles, isn't it? So can I say this results into f dash x or derivative of x with respect to x plus derivative of g with respect to x? Okay. Nothing but f dash x plus g dash x. Is it fine? Same will be true even if you want to prove for the difference of two functions. So even if there was a minus sign in between, the approach would not change much. Is this fine? So this is called sum or difference rule. Is it fine? Any questions? Any concerns? Let me see if I have an example based on the same. I might be having some question. Oh, okay. We can. Did I give you the list of the derivative of these functions officially? I'll come back to this example. Did I give you a list of the derivative of the functions, standard functions officially? Oh, in the bridge course, I have not given you in the class. Oh, I'm so sorry. So first we'll take that list and then we'll come back to this. Okay. So extremely sorry. I'll be going back to the, you know, one slide before this and I'll be giving you a list of the derivatives of certain functions. Okay. Just, just allow me to go back to the previous slide. In fact, I'll take a subsequent slide. No issues. Okay. Because without that, you will not be able to apply these rules of derivatives. Okay. So we need to, you know, understand the derivatives of certain standard functions, derivatives of some standard functions. Because without this, without knowing the derivatives of these standard functions, how would you actually apply any kind of a rule? Okay. So let us first understand how does a car run? What is a clutch? What is a break? Okay. And then I will tell you the traffic rules. So first learn to drive a car, then we will talk about the traffic rules. Okay. So let's learn the derivatives of certain standard functions. Now all these results, which I'm going to write down, they are all derived from the first principles or they can all be derived from the first principles. Okay. So a few of the results I might ask you to derive right now. Okay. And then we'll continue with the rules. So sorry, I was not aware that I had not completed the derivatives of certain standard functions. Okay. So never too late. Let's get started. So derivative of a constant. This will always be zero. Okay. Why very simple. If you talk about use of first principles to prove it. If let's say this is your f of x. Okay. So what is f of x plus h then? I think it will still be a C because there's no x in the function itself. So derivative of C with respect to any variable, whether it is x or whether it is why or whether it is any variable under the earth. Under the sun, this will become C minus C upon h. Correct. So f of x plus h will also be C f of x will also be C. So there will be no change in the output, even if there is a small change in the input. So this will become zero by h. Please let me tell you, this is not an indeterminate form by the way, indeterminate form happens when there is a tending to zero by tending to zero. This is a case of exact zero by tending to zero exact zero by tending to zero is just considered to be zero in mathematics. Okay. So f of a constant with respect to any function is going to be zero. Is it fine. Any questions. Derivative of x is one derivative of x square is to x derivative of x cube is three x square now all these results can easily be proved by first principles. What I want you to observe in this pattern is that if you differentiate any x to the power n and being a real number. This result would be n x to the power n minus one. Okay, this is what we call as the power rule of differentiation. This is called the power rule of differentiation power rule of differentiation. Now let us try to prove this. Let us try to prove this is it. So as I told you, we will be proving certain results by first principles. Okay, so as per our first principles, the derivative of x to the power n will be x plus h to the power n. Please note that this is your f of x. Okay, so I have to write f of x plus h f of x plus h will be nothing but just remove your x and place x plus h in place of that. So this is going to become x plus h whole to the power n minus x to the power n by edge. Okay, now how do I solve such kind of a problem very simple. I'll be writing, I'll be writing this expression in a slightly familiar way to you. Okay. And now I'll be calling x plus h as a y. Okay, so as h tends to zero, you can see that y will tend to x. And now let's also call x as some a. Okay, that means this term can be written as y tending to a. Now why am I doing all these things because I wanted to write this expression something like this, why tending to a y to the power n minus a to the power n by y minus a. So what I'm doing I'm calling this term x plus h as a y. Next I'm calling it as a a. So as h tends to zero why tends to a because why will tend to x, isn't it. Now this is very much familiar to you in our standard limits remember algebraic limits with it. Right, the result in this case used to be an a to the power n minus one. But here remember a is x. Now we're going to become n x to the power n minus one. Okay, and this is how we prove the, the power rule of differentiation. Is it fine any questions, any questions any concerns, let me know. Is this fine. So people whose ut's were going on, how was the paper, how was the math paper. Somebody send me the paper on the group. So I just have anybody please I would request you to share your maturity paper with us. You can put it on the group. Okay, who's taking responsibility for it, because everybody's responsibility is nobody's responsibility. One person has to take. Who's taking responsibility of sending me the paper. Thank you. Others, other schools, also if your ut has happened, kindly share your paper with us because we can analyze the level of questions that has been asked and in fact we can forward it to other nps schools in case they need some practice. Okay. So any any doubt related to the power rule of differentiation, any doubt related to the proof of the power rule. Let me know. Okay, this is one of the most widely used in a rule in differentiation and you'll be requiring it very, very commonly. Now one important point to be noted down over here is that don't confuse power rule with differentiation of exponential functions. So let me take the, what number was it I'm so sorry I missed out the number. This was second. Okay, so we'll start with the third one now. So don't confuse power rule with differentiation of exponential functions. So when you have an exponential function of the nature a to the power x. Okay, many people think oh I can write this as x a to the power x minus one. No, this doesn't work like that. This doesn't work when there is a variable raised to a constant power. Right. In case of exponential function that rule is not going to work. Simply because it will not be supported by your first principles. Okay first principle must support this fact. Okay, so this is not going to work. So what is going to work. Okay, let's figure it out so I will not write the result here but I will try to prove it. So first principles, the derivative of the derivative of a to the power x with respect to x will be what f of x plus h f of x plus such will be a to the power x plus h minus a to the power x patch. Okay, take a to the power x common. Okay, you will have a to the power h minus one by h. Okay. Now remember your limit is applied to edge. A to the power x can be treated like a constant because it has no h in it. You can bring it always outside the purview of your limit. Right. Now what is this limit anybody can tell me on the chat a to the power x, sorry, a to the power h minus one by h limit h tending to zero what does this give you anybody can write it out. No, it's, it's long a it's long a a log a to the base e. So this is, is it fine. Any questions, any concerns. So this is the result for derivative of a to the power x please note it down a to the power x, long a long a or natural log of a. So you can find any questions and of course, needless to say that the base of the exponential function should be a quantity which is greater than zero. Any questions any concerns. Okay, under exponential function, there is a special case when you're a is e. So what happens when a is e. So when your a is e the derivative becomes e to the power x l and e l and e is or long e is one itself. It is just e to the power x. Is this fine. So derivative of e to the power x is e to the power x itself. Okay, that means on e to the power x graph at any value of x if you sketch a tangent, the slope of the tangent will match with the value of the function at that point. That means let's add e to the power five. Let's say at x equal to five. That means that e to the power five comma e to the power five you are sketching a tangent. So let me just draw it for you. So this is e to the power x graph. Okay, so at any point. Let's say this point is five comma e to the power five. If I sketch a tangent. The slope of this tangent will also be e to the power five. That is what the that is what is the meaning of the fact that derivative of e to the power x is same as e to the power x. Is it fine. Any questions, any concerns. If you have any questions if you want to copy down anything please do so. And do let me know if I can help you to understand anything once again. I'll be more than happy to help you. Yeah, so basically I was trying to explain. What is the meaning of this. So many people say sir, even on differentiation e to the power x doesn't change. What does it implies it. I'm not able to understand this pictorially. So it means that on e to the power x graph let me show you through GeoGibra. Let me just go to GeoGibra and show you something from there. Yeah. Can you see the GeoGibra here. Okay, so I'll make y is equal to e to the power x graph. Okay, e to the power x. Yeah, this is the graph of y is equal to e to the power x. Now, at any point on this graph, let's say I choose a point. Let me choose a point. Let's say I choose a point here. So as you can see, let me just slightly bring it towards one or let me take a point at two. Yeah, so you can see there's a point two comma seven point three nine fine. Now at this point I'm sketching a tangent. Let's take a tangent. Okay, do you see a tangent. Okay, now I'll find the slope of this tangent. Do you see the slope value and the y coordinate will match. Do you see that? Seven point three nine. See my cursor moving in. Yeah, seven point three nine. That is the value of the function at two. That is e squared seven point three nine is e squared. It matches with the slope of the tangent drawn at that point. Okay, and see the fun. I'll be moving this point around and you'll see that. You'll see that this y coordinate and the slope will match. Please, please put your eyes on. I'm wearing it. Whatever is this value? See now it is five point one six and slope is also five point one six. See four point four eight slope is also four point four eight three point nine slope is also three point nine two point five one slope is also two point five. Okay, so wherever you go, the value of the function at that point and the slope of the tangent will match. It's a very interesting phenomena by the way. Clear everybody. What are the meaning of the derivative of e to the power x is e to the power x. Okay, no problem. Let's discuss further. Now, next we'll move on to the logarithmic function. So this was our exponential function. So let us move on to the logarithmic function. Which number was this third number? Okay, so let's move on to the fourth one. You may have all done this in school. You may have all done this in school and in school you might have also proven these results through first principles, haven't you? Where you made to prove these results in school? Anybody in anybody's school it was done. Yes, no, a little bit. Okay. Yes, some of the definitely. So next is the derivative of log x to the base e. The derivative of log x to the base e is given as 1 by x. Okay. Now, from where does this is outcome? Let's prove it. So let's find out the derivative of log x to the base e by first principles. So by first principles, you will say limit x tending to zero f of x plus h. This will be your f of x plus h. Okay. This will be your f of x by h, h tending to zero. Okay. Now what I'm going to do here is I'm going to use my log properties log a minus log b is log a by b. Am I right. And the numerator can be further written as one log of one plus h by x upon h. Yes or no. As any doubt so far. Now, go back to when we had done our logarithmic limits. I told you that if you have log of one plus some function divided by the same function, and that function is tending to zero, then this answer is going to be one. Well, here if you see it is log of one plus some function of h, but unfortunately the same function is not down in the denominator. It is just h over it. I should have h by x. It doesn't matter much. I will just produce x by multiplying and dividing with it. So now this part is going to meet our standard logarithmic limit requirement. This part is going to meet our standard logarithmic limit requirement. And the answer here in this case is going to be one. So answer is one upon x. That is what is our given result. Is it fine. Any questions, any questions, any concerns. How many people ask me this question. So what about log x to some other ways, will it be still one max. No, the answer to that will be different. We will take that up. Once we have done some properties. In fact, I will give it to you right now. No, no, no, no, it doesn't work like that. I am holding thank you. Thank you for bringing data. I strongly advise you not to extrapolate any formulae. I strongly advise you not to extrapolate your any formulae he extrapolate means, since this is this, hence this is this. Don't do that mistake. It doesn't work like that in calculus. It is not a unitary method. process is completely different. Since log x derivative is one by x, it doesn't mean log x square derivative will be one by x square. No, it doesn't work like that. It doesn't work like that. So please stop extrapolating any formulae. Whatever formulae you are basically studying right now, they are sacrosanct means any deviation from there will lead to a different result will lead to a different result. Even if I make, you know, minor changes, it may result into a different type of expression. So don't try to extrapolate blindly. Is there any other function where the slope of the graph is equal to the value of the function at that point? At that point means at every point on the graph, you're trying to say right? Okay. Okay, so if that is your question, shortly, you can easily answer your question by writing a differential equation like this. You're saying derivative matches with the value of the function. Correct? Yes or no? That means differential of the function is f of x into d of x. Put it down over here like this. Okay. Now see, let's say I call this as a y kind of a thing. So dy by y is dx. Integrate both sides. Okay. So you would have done a bit of integration, I'm sure. So this is ln y equal to x plus c. Correct? Right? So in this case, your y is e to the power x into c. Okay. So any function of this nature, any function of this nature, in fact, I could have written it as e to the power c only, but e to the power c is another constant called that call it a c dash if you want. So any function which is e to the power x multiplied to a constant will also follow the same suit. Yes, it will be like e to the power x only. So let's say two e to the power x or five e to the power x or one by 10 e to the power x. They will also follow the same property. Okay. All right. Let's go back here. So my question was if the base is changed from e to some other base, some other legitimate base, will my result remain the same or will my result differ? Okay. Let's try to understand this. So let us say if I have log of x to the base b, and I want to differentiate it with respect to x. What happens to my answer? Let's go to the first principles to prove this result or find this result. So by first principles, I will write limit h tending to zero log of x plus h to the base b minus log x to the base b upon h. Correct. So f of x plus h minus f of x by h h tending to zero, that will be applicable. Okay. So you will have, sorry, limit h tending to zero. You will have log of x plus h whole divided by x by our log rules. Okay. So this is going to become limit h tending to zero log of one plus h by x upon h. Now, the problem with this expression is number one problem. This base is not e. So nothing to worry. I can convert it to e by using our change of base property. So I'll put e here, and I will put log base log b to the base e also. Okay. So that helps us to change the property. Another thing that is missing here, there is an h by x over here, but there's an only h over here. So what I'm going to do the same way as what I did for the previous case, I'm going to multiply and divide with an x. Now, this whole thing is going to become a one. Do you see this chapati? This is going to become a one, isn't it? So overall, this becomes one upon log b to the base e times x. Okay. So please note this is a noun. If your base is something other than e, your answer becomes one by x log b or log b to the base e. Is this fine? Any questions? Any questions? Any concerns? Is it fine? Any questions? Okay, great. So with this, we are now heading towards trigonometric differentiation or differentiation of standard trigonometric functions. So let's get started with derivatives of standard trigonometric functions. Which number was this, by the way? I think it was number five or number four. So we'll start with number five. Yeah. So derivative of sine of x. Okay, that is cos of x. Now again, I have proven this result while I was doing the derivatives by using first principles. Okay. Quickly, can we prove it? Okay, let me give you more lists and then I'll ask you to prove one of them. Let's prove this one. Let's prove this one. Derivative of cos x with respect to x is negative sine x. Everybody try this out. I'm giving you around one and a half minutes for it. Please prove this result. Done. Should I discuss it out? Okay. So by first principles, it will be cos of x plus h minus cos x by h. We can use our transformation formula cos a minus cos b is two cos sorry, two sine a plus b by two. So a plus b by two, that means x plus h plus x by two into sine b minus a by two, b minus a by two. Okay. Whole divided by h. So the first term is going to simply reduce to sine. Now two x plus h by two is just x plus h by two. Correct. And this is going to be sine minus h by two. By the way, minus can always be pulled outside. Now this two that you have, I'm going to bring it down over here. So what is signed h by two divided by h by two when h tends to zero. This term is actually a one. And this term, when you put h as zero just becomes a sine x. So overall your answer will be minus sine x minus coming from here, sine x coming from here and one coming from it. So it's negative sine x. That's the derivative of cos x. Not easy. Okay. I will not ask you to prove all the results. Maybe if I feel that there is one interesting function coming up. Okay. And I feel people will get stuck while finding the derivative of it by first principles. Only those I will ask you to try out. Okay. So please note down derivative of sine and derivative of cos. By the way, I'm categorizing this entire derivative under one serial number. Okay. So that you have all the trigonometric functions at one place. Okay. So moving on, derivative of tan x with respect to x is secant square x. This list is already, this problem we already did in our previous session. Derivative of cosec x with respect to x is minus cosec x cot x. Derivative of cot x is negative cosecant square x. Okay. And last but not the least, derivative of derivative of cosec x is cosec x tan x. Okay. Let us do a small exercise. Can we all find the derivative of cosec x? So please try this out. Try it now. So find the derivative of cosec x by first principles and prove that it is minus cosec x cot x. And do let me know with the done on the chat box once you're done. And then I can discuss it out. Oh, sorry, Situ, I missed your message. Change your base property you want. See, change your base property works like, oh, you got it. Okay. Nevertheless, I'll tell you once again. Change your base property says log m to the base n can be written as log m to some common base. You can choose b to be a common base divided by log n to some common base. Okay, this is called the change of base property. Change of base property. Everybody has tried it out. Did you get minus cosec x cot x? Yes, no, still doing. What is the status? Okay, I will also try along with you all doing. Okay, fine. In fact, if you want, you can I can wait for a few more seconds. Question is done. Very good, question. Very good. Very good. Most of you are saying done done done done done done very good. Okay, so let's write cosec in terms of reciprocal of sine. So it's one by sine x plus h upon one by sine x whole divided by h. So this becomes sine x minus sine x plus h by sine x plus h into sine x into h. Okay, now we can relieve ourselves from a lot of hard work over here by doing this. First of all, I like to take a minus sign out. Okay, when I do that, the numerator turns into this expression. Okay, now let's split this limit like this. So what I did, if you carefully see the denominator terms I have written as a separate limit one by the denominator terms. And after taking sine, whatever term I got on the numerator, sorry, after taking negative sine outside, whatever term I got common on the whatever term I got on the numerator, that I club with an h. So what is the advantage here? This basically is a result which is already known to us. It's the derivative of sine with respect to x and that is nothing but cos x. And this is the case of simple substitution. So this will be sine x into sine x. Yes or no? Simple. So now let's write down this as a cot x and this as a cosec x. So it will become minus cosec x cot x done. Problem is solved. Is it fine? Any questions? Okay, great. Alright, so I'm not going to actually stop here. I'm going to give you certain more differentiations or certain more derivatives of some trig functions, especially the inverse trig functions. Okay, because in the DPPs that you might be getting, you will be seeing those kind of questions. Okay, so I will call them as serial number six. So this will include derivatives of certain inverse trigonometric functions. Sine inverse x being the first one derivative of that is 1 by under root 1 minus x squared. Now actually this topic is a subject matter of class 12th. In class 12th, we talk when we talk about differentiation chapter. Are you sir? In 12th also we have differentiation. Yes, very much you have in 12th as well. So this is just an introduction part of derivatives. In 12th you have the main you can say chunk coming up where you learn about derivatives of inverse trig functions, what we call as ITF inverse of derivative of ITFs. We'll be talking about derivatives of parametrically defined curves. We'll be talking about derivatives of functions by taking logarithms. We'll talk about derivatives of a function with respect to another function. We'll talk about implicitly defined curves derivatives. We'll talk about higher order derivatives. We'll talk about derivatives of infinitely series functions expressed as an infinite series. So there are plethora of topics which are left to be covered under this. And one of the concept is the derivative of inverse trig functions, which I'm going to give you the result as of now. Now, how does this result actually come? Since you have already only learned first principles, I will be using first principles to derive this result. So all of you please pay attention. All of you please pay attention. So by first principles, the derivative of sin inverse x will be given by limit h tending to 0, sin inverse x plus h, that is your f of x plus h, minus f of x, which is your sin inverse x whole divided by h. Now, all of you please pay attention. This is how I'm going to do it. I'm going to call sin inverse x plus h as let's say theta. That means sin of theta, sin of theta is x plus h. By the way, everybody knows a bit of inverse trigonometric functions. Inverse trig function basically is a function which does opposite of what trigonometric functions do. So what does trigonometric function do? They take some angle and give you some ratio. For example, sin pi by six is half inverse trigonometric function does exactly the reverse of it. Or you can say exactly the inverse of it by taking in the input as some ratio and giving you some angle. So like sin inverse half, that is pi by six. Got the point? It doesn't mean reciprocal of sin. Please don't get me wrong here. Many of the student things said, doesn't it mean one by sin x plus h? No. One by sin x plus h is cosec x plus h. It is not written as sin inverse. Sin inverse is a reserved symbol. This symbol has been reserved. Reserved means like how you reserve your tables in a restaurant. So even if somebody arrives before you, he will not be given that table because you have reserved it earlier on. So sin inverse is a reserved symbol which does opposite of what sin function does. So sin pi by six is a half. So sin inverse half is pi by six. Got the point? Now here, I will also call sin inverse x to be some angle phi. That means sin phi is x. Okay. Now see what I'm going to do. This entire expression, I'm going to write it once again like this. Okay. So down in the denominator, I will just write x plus h minus x. And sin inverse x plus h is a theta. Sin inverse x is a phi. X plus h is sin theta. And x is sin phi. Okay. Now, all of you please pay attention. When h tends to zero, when h tends to zero, can I say theta will tend to phi? Do you all agree with me on that or not? Write a yes on the chat box if you agree with me on this. Write a yes on the chat box if you agree with me on this. Only Karthik Sanoj agrees with me. Nobody else. See, as h tends to zero, this will become sin inverse x. This will also become sin. This is anyway the sin inverse x. So can I say theta will tend to phi? Yes or no? Yes or no? Right? Yes. Without you writing yes on the chat box, I will not proceed. Yes. Okay. Thank you, Aselesha. Aselesha. Hey, somebody has using a wrong name or something. Or is it a new student? Oh, okay. You're a new student. Okay. Sorry. I didn't, okay. Is it fine? Okay. Now, from here on, how do I proceed? Very simple. This type of limit problems are very familiar to us. So what I'm going to do is I'm going to put theta as phi plus k. Sorry, I'm using a different notation now here. Earlier I used to use phi plus h, but h has already been used up. Okay. So before, now I have come to phi plus k. So can I say, when theta tends to phi, k will tend to zero? Yes or no? Yes or no? So the entire expression would now head towards something like this. All of you, please pay attention. Okay. So instead of theta, I will put phi plus k. Okay. So it becomes phi plus k minus phi. And this is nothing but sine of phi plus k minus sine phi. So phi phi goes off. What do we get? We get limit k tending to zero k upon sine phi plus k minus sine phi. Now who will tell me what is this expression equal to? In fact, I will write it down again just for the benefit of everybody to recognize it. What is this expression equal to? When you say sine phi plus k minus sine phi by k, k tending to zero, what does this expression actually give you? What does this expression actually give you? It gives you the derivative of sine calculated at phi. So this will give you 1 by cos phi. Are you getting my point? Is this okay with everybody? So this will give you 1 by cos phi. And cos phi, if I am not mistaken, you can all check here. Cos phi, let's derive it from here. So cos phi from here would be under root of 1 minus x square. If sine phi is x, cos phi is under root of 1 minus x square. Yes or no? Basic trigonometry? Yes or no? Just like we say cos theta is under root of 1 minus sine square theta. So can I not replace this phi, sorry, cos phi with under root of 1 minus x square? And there you go. We got the derivative of sine inverse x with respect to x. Okay. So I would request all of you to include this in your formula list and remember it because you will be needing this at so many places, especially in 12th, when your board is at stake. Is it fine anywhere you want me to take the screen just to show you how I carried out the whole process? Do let me know. Okay. Left, right, up, down, wherever you want me, I will put the screen so that you can copy it. So Aselisha, we are actually in the second class of derivatives. Our first class on derivatives happened last week. So if you want, I can share with you all the class notes and the videos of the previous classes. So do let me know. You can ping me personally. I will share you all the class notes and videos. Okay. All right. So I hope you are, you know, you are from Miss Cool. You are able to understand what is, you know, happening or you are from HSR only. Okay. Good. Okay. So everybody has noted this down. Great. So let me carry on with the list of inversed functions. So I'm sure you would be interested in knowing what is the derivative of cos inverse, tan inverse, cos inverse, cot inverse, seek inverse. Okay. So all those derivatives I'm going to list down over here, but I will not be deriving those results. In fact, in our class 12th differentiation, we will take these concepts once again and that time I'll be deriving it, but in a smarter way, not by using first principles because it is slightly lengthy and cumbersome. So let me just continue with this list. So derivative of cos inverse x, derivative of cos inverse x is negative 1 by under root 1 minus x squared. So wow, that makes it actually simpler to learn because whatever was a derivative of sin inverse, the negative of that result becomes a derivative of cos inverse. So it makes us, you know, remember this formula easily. Similarly, derivative of tan inverse x, please note this down. It is 1 by 1 plus x squared, 1 by 1 plus x squared. Try proving this as a homework question. Try to prove this by first principles, by first principles, like the way I did for sin inverse x. If you're stuck, do let me know. I'll help you out. Next, derivative of cos c inverse x, derivative, let me just take c inverse x that will be interesting. C inverse x derivative is 1 by x under root x square minus 1. Okay, so just keep noting this down. We will be proving this in our most smarter way a little later on. Okay, derivative of cos c inverse x, derivative of cos c inverse x is negative of this result. And by the way, I forgot to mention that derivative of cot inverse x, so I'll place it over here, is negative of 1 by 1 plus x squared. Okay, so as you can see, there are three pairs, sin inverse cos inverse, tan inverse cot inverse, and sec inverse cos sec inverse. So these pairs, the derivatives are actually negatives of each other. Okay, so let me write down sin inverse once again here. So derivative of sin inverse, which I already proved in the previous page. So if you can see these two are pairs, they are negatives of each other, the derivatives, these two are in pairs, and these two are in pairs. Okay, so that makes us learn only three set of formulas here, not six on the time. Is this fine? Now in class 11 school exam, you will not require this. To be very frank, you will not need this immediately in class 11. But again, we are not only preparing ourselves for the school exams, we are also equally serious about our comparative exams. So yes, in comparative exams, this is going to be useful, but not immediately in class 11. Is this fine? Any questions? Okay. Now that we have done the sum rule, and we have learned the derivatives of these functions. So let us take one quick question, and then we will move on to the further rules, which I abruptly stopped in the beginning. Sorry, again, apologies, I was not aware that we had not done this list. Note it down, and when we are ready, when you are ready, do let us know. Do let me know. Tomorrow's class will be on introduction to 3D geometry. Okay. So tomorrow's class would be on introduction to 3D geometry. Okay, 430 to 730, as most of you are feeling convenient with that time. So by the way, this is just an extra class which I'm taking only for three weeks, because your seniors are busy with their first semester exam. So I'm utilizing their class time to take your classes so that you are benefited. Okay. So only for three weeks, you will be called for one extra maths class. Okay. So you'll have two maths classes in a week for just three weeks. Is it fine? Not like KVPY. By the way, KVPY, all of you would have got the notification that it is now being conducted on, conducted on, conducted on, right down, right down, right down. Let me see how aware you are. How aware you are. Where is the KVPY exam shifted to January 9th? Yes. See, Nikhil is serious. So he knows. Yes, it has been shifted to January 9th. So those of you who have filled the application form, please be ready. I think we are already sitting on 1st of December. So just one month, eight days more. All right. Let's take a question. First, let us do this question. It's a simple question. If Y is equal to log X to the base 10, please note log X to the base 10 is normally called LG. Okay. So this is also sometimes written as LG. Okay. This is log 10 to the base C. This is log X to the base X. This is log 10 to the base 10. Okay. Now, these are basically, you can say a sum of functions and you want to find out the derivative of it. So what is the derivative of this? Quickly tell me, write down your response on the chat box. It's a very simple question. Give me a response on the chat box. By the way, the answer is only going to have one expression in it. That's it. You can take that as a hint. Nobody. Absolutely, Nikhil. Let's write answer. Anybody else? Okay, Vashna. Okay. It's actually a touch and go question. Okay. It can be solved within five seconds. See, all of you please square engine. This term is log X to some base. Correct. And that base is not E. So let's go back to that formula where we had discussed what was the derivative of log X to the base B. So we had learned that this is going to be one by X ln B or ln B. So this first term derivative is going to be one by X ln 10. Right. And please remember all these terms that you see here, this, this, this, they all fall under the category of constants. Why? Log 10 to the base E is a constant. Okay. And most of you would be familiar with that result. 2.3, 2.301. Have you used this 2.301 somewhere in ideal gas equation? 2. Yes, in chemistry. So log 10 to the base E is actually that 2.301. Okay, anyways. So log 10 to the base E, log X to the base X is what? One only log 10 to the base N is also one. So they will all be like constants. What is the derivative of a constant? Zero. Need not write it. Correct. So this is the only answer. So presumably why, why that extra one by 10? Not required. No. Is it fine? Any questions here? Any questions? Okay, all good. So we'll now move on to the other properties. So we already did some in difference soon, right in the beginning of today's session. So I will move on to, let me write down the rules of differentiation continued or methods of differentiation, rules, methods, whatever you call it, rules slash methods of differentiation continued. So second rule that we are going to talk about is your product rule. So product rule says if you're finding the derivative of product of two functions with respect to X, okay, then this will give you the result as f of X into derivative of G, okay, plus G of X into derivative of F. We had already done this in our bridge course. Okay. And I don't exactly recall whether I derived this result for you guys. If no, then don't not to worry. I'm going to derive it once again for you. So first of all, how does this rule actually come about? What is the genesis? What is the reason for this rule to come out? So let's prove this. Let's prove this result. So the derivative of F of X into G of X by first principles will be nothing but this whole thing you replace, this whole thing you replace X with X plus H. So it'll become F of X plus H times G of X plus H minus, okay, now what I'm going to do is I'm going to write minus F of X G of X at a slightly distance apart from this first term. Okay. So the reason for making such a big gap between them is because I'm going to put in certain term in between. Okay. So first of all, this is our definition of the derivative of any function by first principles. So my function is F of X into G of X. So in that, I first replace my X with X plus H, I get something minus the old function divided by the change in X value, which is H and you're extending to zero. This is from our very basic first principles definition. Now what I'm going to do is I'm going to introduce, I'm going to introduce these terms in between. Let me write them down in white. Sorry. I'm writing the same thing in white now minus F of X plus H into G of X plus F of X plus H into G of X. Okay. So what I did, I subtracted and added the same expression, thereby I have not disturbed the entire expression. So entire expression still remains to be the same first principle definition. I just added and subtracted something. Okay. Why did I do that? There's a reason behind it. So if you see, let's group these two terms and let's group these two terms, let's see what evolves out from that grouping. So from these two, if I take F of X plus H common, I'll get G of X plus H minus G of X. And from these two terms, if I take G of X common, I'll get F of X plus H minus F of X. Okay. All divided by H. Now I'm going to split this limit as separate limits. One limit being this limit. Okay. Plus other limit being this limit. Clear any confusion, any concerns, any conceptual voids, still this position, please highlight it right now. I'll find everybody's happy. Okay. Great. Now I'm going to further split this like this. So this first limit, I'm going to split up like this, this into this. Is it fine? I can do that very much. I can do that. The algebra of limit allows me to do it. Now this G of X, I can just simply translate outside. Okay. And I can write the remaining term like this. Okay. Now let's try to understand. What is this limit going to give us? You will say, sir, simple F of X only because just substitute H as zero. That is going to be a limit. What will this limit give us? You'll say, sir, this appears to be the derivative of G with respect to X. Yes, exactly. It is the derivative of G with respect to X. That is G dash X or D by DX of G of X. This is anyways G of X. So let it be as it is. What about this term? This term is again the derivative of F of X by first principles. And there you go. This is the result which I wanted to prove hence proved. Is this fine? Even though this derivation is not going to be of help used for you in your school exams, but still you should be aware how these results of product rule, some difference rule and a few more rules are going to come in future, how they actually come up. So please note down whatever you want to and let me know if you need my explanation again at any point of the solution or this derivation. Everybody's happy? Okay. So now a natural question will arise in everybody's mind is that instead of two functions if I had a product of three functions or four functions or five functions, how will this rule actually work? Is this rule extendable to let's say more than two functions? Yes. This rule is extendable to more than two functions. So in case you're having more than two functions, so please note this down. So this rule basically works under the principle that you need to differentiate one at a time. That is the baseline of this rule. Okay. So the rule basically works on one at a time base rule. One at a time means you keep one as it is differentiate the other one and then repeat it. So for example, if you had three functions, so please note this down. If you had three functions, let's say f of x, g of x, h of x. Okay. How would this derivative take place? So basically the derivative of this will be keep two of them as it is. Let's say I keep f and g as it is, differentiate the last one. Then keep f of x and h of x as it is, differentiate the middle one. Then keep the last two as it is, differentiate the first one. So the bottom line is one at a time. You can't differentiate two functions at the same time. So two of them would be as it is, no change in them. And the third one will be differentiated. So this is what we say differentiation one at a time. So you can extend this to even four functions, five functions depending upon whatever is your requirement. Is this fine? Okay. So basically one at a time. Remember this phrase you will always remember the approach one at a time. One at a time means derivative of one function at a time. Other two will be or others will be undisturbed. Is this fine? Any question? Any concerns? Now let's take some problems based on the same. Ready? Should we take up a problem now? Simple problem to begin with. Why is e to the power x tan x plus x ln x? Find d by d x. Please don't simplify your answer. Leave it in the law form only. Don't waste time simplifying it. But yes, please do some basic simplification for your school exams. If at all you get those kinds of questions. Okay. Say two. Very good. I didn't say that. Let me check. It's actually correct. Okay, let's discuss it. So this is actually a mix of product tool as well as some difference rule. Why some difference rule? Because they're two functions which are added up this and this. Why product rule? Because in between there are products or functions. Okay. So when you're differentiating it, first understand that let's say if you call this as f and you call this as a g, you need to write it as derivative of f plus derivative of g. Now this is just for your internal understanding. I'm writing it. I may not write it in future. Once you become mature with the concept, I will not be writing these intermittent steps. Okay. Now, what is this f? f is made up of product of two functions. So in order to find the derivative of it, so I'm just doing the first part, I need to write first function e to the power x into derivative of the second plus second function as it is into derivative of the first, which is e to the power x. Let me just separate the brackets. So this is done. Derivative of the first again. Second day again, g is itself made up of two functions x and ln x. So it is x into derivative of ln x, which is 1 by x plus ln x into derivative of x, which is 1. And there you go. You can simplify it to a certain extent. This is going to be 1. This is going to be ln x. This is going to be your answer. Is this fine? Any questions? Any concerns? We'll come back. We'll take more questions on these rules, but let me finish off few more rules or methods which are pending so that we are ready to tackle any type of questions by the use of those rules. So let's move on to the third rule, which is called the quotient rule. Meanwhile, if you have any questions with respect to this, please do let me know. Exactly, Prasim. I think Aarav also got it. Who else? Vashna. Very good. Okay. All set. So let's take the third rule, which is called the quotient rule, quotient rule. So quotient rule helps us to differentiate when one function is divided by the other. Okay. So as you can see, you would have now figured out that after having given you the derivatives of standard functions, now I'm telling you that if those standard functions are associating with each other like a sum or a difference or a product or a quotient, later on, we'll talk about composition also. How do we deal with the derivatives of those type of cases? Okay. So that is what is the whole and sole purpose of introducing these methods. So this rule says that the derivative of f of x by g of x will be g of x into derivative of f of x minus f of x into derivative of g whole divided by whole divided by g of x the whole square. Okay. Please note this down. Initially, many people do a mistake in this formula. The most commonly seen mistake is stopping the positions of these two terms on the numerator. Obviously, if you do that mistake, your answer will become negative of what is actually the answer. So please note this down and commit this to your, of course, you will end up committing it to your memory once you practice more and more questions. So don't switch the position of the two terms over here. So it is g f dash then minus f g dash by g square. Now what is the proof for this? Let's quickly do the proof. So as per our first principles, it will be limit h tending to zero. Now this function, change your x with x plus h everywhere. Okay. Minus the original function whole divided by h, h tending to zero. Okay. This is our first principles. Now let's take the LCM. Let's do some minor simplifications here. So f of x plus h into g of x minus f of x into g x plus h. Now as you can see, I have written them slightly spaced apart. Okay, because I'm going to fill certain expressions in between. And of course, the denominator will become like this. Is it fine? Any questions, any concerns so far? Any questions, any concerns so far? Do let me know. No questions. Okay. Now I'm going to introduce certain terms in between minus f of x into g of x plus f of x into g of x. In short, I have not introduced anything at all because it is as good as a zero. Okay. But I've written it like this. So my total expression or my entire expression is undisturbed because of this. Okay. Now what I'm going to do is I'm going to club these two and I'm going to club these two. Okay. Let's see what happens when we do that. So from these two, if I take a g of x common, I'll end up getting f of x plus h minus f of x. And from these two, I'm going to take minus f of x common, whole divided by h. And of course, there was a g of x plus h also into g of x in the denominator. Fine. Now all of you, please pay attention. I'm now going to start splitting off this entire limit. So the first split which I'm doing is with these two terms, one by g of x plus h into g of x. Okay. Second split which I'm going to do is this term f of x plus h minus f of x. Okay. Of course, there's a g of x that will also come out. Whole divided by h. Okay. So what I did, this term I'm keeping as one single expression. This I already separated out commonly and this I'm now writing it like this. Okay. Similarly, this term I'm making it as a one single limit expression like this. Okay. Any doubt, any concerns, any discomfort, anywhere, please, please, please feel free to stop me. Let me write it properly. So I will use it and write it. Yeah. All fine. All set. Any confusions? Do let me know. So now this limit is as good as one by g of x into g of x, which is g of x square. And inside over here, this is g of x and what is this term? This term you will say, sir, it is the derivative of f by first principles. So that is f dash x. Similarly, f of x and what is this term? This is the derivative of g by first principles. So that is nothing but g dash x. So overall, this entire expression boils down to g f dash minus f g dash upon g square, hence proved clear. Any questions? Did you all do the proofs in the school for all these methods, quotient rule, product rule, was the proof given in school? No. Okay. Okay. So time for some questions. Let's do some questions based on the same. Can I go to the next slide? Let's go to the next slide. Okay. Let me begin with let me begin with a few questions, which are simpler ones. Okay. Let me begin with this one. Very easy question. So that everybody is on the same page with respect to the understanding of the formula. Fine derivative of x by x square plus one with respect to x, that means fine dy by dx. Now just write it done. No need to give me the answer on the chat box. Just say I'm done. Oh, Vaishnav is done. Pishcham is done. Okay. Okay. Siddharth, Karthik is done. Karthik Sanovich is done. Anybody else? Done. Okay. Great. Should we discuss it now? They just said he also done. Okay. So this is our first function. Let's say I call it as f of x. Okay. And this is our g of x function. Okay. So as per our formula, it will be g into derivative of f minus f into derivative of g. Now, g itself has two functions involved. So both are derivatives you need to add. So x square derivative is two x, zero derivative, sorry, one derivative is zero. Okay. See, going forward, initially I'm writing all these things in detail, but going forward, I will not write this zero and all. Okay. In fact, you will, you yourself would not like writing those details. Okay. So initially, I mean, since this is our only second class in derivatives, I'm writing all these steps to detail, but later on you yourself will get very used to it. So let's simplify this a bit. So on simplification, this is going to give you one minus x square upon x square plus one, the whole square. Is this fine? Everybody got the same result? Any questions, any concerns? That's also done. Okay. Now let's look into a few more questions. So let's start with this question. Why is basically a continued product of one plus x, one plus x square, one plus x to the power four, till one plus x to the power n, that means they are basically having n terms multiplied. Okay. You need to find out the derivative of this function at x equal to zero. See, what is the meaning of derivative of the function at x equal to zero? It means differentiate the function and then put x as zero, not the other way around. Please get this, you know, very, very straight here. When somebody says derivative of a function at x equal to zero, differentiate the function first and then substitute x equal to zero in your result. Okay. If you substitute and then differentiate, you will always get a zero answer also and marks also. Of course, because you are differentiating a wrong result. Okay. So please don't do that mistake. How many times this expression is also written by the exam, let's say I call this as a function. So I can also call this as f dash zero also. And many students interpreted in the wrong way. They think that I have to put zero and then differentiate. No, no, don't do that mistake. Please. This means differentiate first and then put x equal to zero. Okay. I'll just give you some synonymous terms for this. Many times we also write it like this, dy by dx, you put us verticals, you can say small line and then write x equal to zero. This also means the same thing. So instead of writing this at in between, we just write it like this. So they all mean the same things. Okay. Some books will also write it like this, y dash zero. That also means the same thing. Okay. So find the derivative and put x as zero. Do this for all of them. Tell me what do you get as your answer? Okay. Nikhil has already given the answer. Well done, Nikhil. Okay. So three people have answered and all of them are getting different, different results. Okay. Anybody else who would like to give it a try? Okay. I can give one more minute and then we can discuss it out. So please try to wrap up your things, wrap up your solution in the next one minute. All right. So let's discuss this out now. There are two ways to do this problem. I'll discuss both the ways with you. The first is like most of you would have, you know, done it thinking this is a product of, you know, n functions multiplied to it. Okay. So when we perform the product derivative, we do it one at a time. That was the bottom line. Right. Remember, I told you one at a time. Remember that. So if you differentiate first one, you get a one. Keep others as it is. Don't disturb the others. Okay. I'll not be writing a lot of terms here. Don't worry. Plus differentiate the second one. Keep others as it is. Now, mind you, when you differentiate the second one, you get to X and others are as it is. Similarly, here, the third function you differentiate, keep others as it is and continue doing this till you reach the derivative of this last of the function. That is going to be 2n X to the power 2n minus 1. Okay. Now, once you have got this, once you have got this, yes, you say that you can change your answer if you want. Once you can, once you've got this, now put X as zero, you realize that the first term will give you a one, but rest all the terms will give you 0, 0, 0, 0, 0 like that. Isn't it? So the answer to this question is a one. Let's take, let's take this one. Here, four options have been provided to you and I'll put the poll on as well. If Y is equal to C kicks minus tan X upon C kicks plus tan X, then dy by dx is which of the following expression? Easy question. I'll give you around two minutes for this, two to one half minutes, not more than that. Later on, when we learn the chain rule, this is going to become an easier and a problem to solve. Two minutes gone. Only two people have responded. Okay. I can give maximum one more minute here. Okay. Five, four, three, two, one. Please vote everybody. Don't worry. I don't get to see who is voting for what option. Okay. I just want you to just participate in the problem solving. Okay. So I'll be closing the poll now. Just 11 of you have voted, by the way, out of which maximum people say option number B. Okay. I think B, C and D, B, C are very close to each other, equal, almost equal votes. Let's check whether B is correct or not. So as per our quotient rule, let's say the upper function is F, lower function is G. So it is G into derivative of, derivative of F, which is going to be C kicks tan X minus secant square X. Okay. Minus, minus F into derivative of G, which is going to be C kicks tan X plus secant square X, whole divided by C kicks plus tan X, the whole square. Okay. Now, all of you please pay attention. Many of you would have wasted a lot of time simplifying it. Correct? Now something very interesting. If I take a C kicks common from this term, let's say I take a C kicks common from this term. So let me just take it out. You would realize that this, this and one of the powers here will get canceled off. Correct? So you will, you will end up getting a simpler expression to deal with rather than a complicated one initially. So this term will be left. And from here, just multiply C back. So it'll be secant square X minus C kicks tan X. Okay. And you'll just have a C kicks plus tan X left. Okay. Now, again, if you open the brackets, if you open the brackets, you would realize that you will be having two times, two times C kicks tan X and secant square X. Okay. Now take a C kicks common again out. Take a C kicks common again out. Okay. And most of you would already know that please recall your class 10th dignometry. Many of you would recall that one by a C kicks plus tan X is actually C kicks minus tan X, right? Because of our simple derived Pythagorean identity that this is equal to one. So from here, I can simply write this as C kicks plus tan X into C kicks minus tan X equal to one. So one by C kicks plus tan X is nothing but C kicks minus tan X. So I can do one thing. I can write, I can write, let me write it like this. I can write this term as C kicks minus tan X. So overall, if you see, you get to take a negative sign out. Okay. You get C kicks tan minus tan X whole square and there's a C kicks sitting over here. So that is nothing but this. So this becomes your final answer. Let me write it more clearly here. Yeah. This becomes your final answer. So minus two C kicks, C kicks minus tan X whole square, whichever option says it, which is clearly option number B. Is it fine? Later on, when I teach you the chain rule, you would realize that this problem can be solved even in more easier way. Is it okay? Any questions? Any concerns? Okay. Should we go to the next question? If you have any doubts, any concerns, please highlight. Okay. Let's go to the next question. Another MCQ type question here. Y is one plus X square plus X to the power four upon one plus X plus X square. D Y by DX is AX plus B. So when you differentiate this function, you get something like AX plus B kind of a thing. What are the values of A and B? So these are your options given to you. And again, I will put the poll for you all to give your response. Poll is on. Almost going to be two minutes. Two people have responded so far. Very good. Four of you have responded. I can give you one more minute maximum. Okay. Let's begin the countdown. Five, four, three, two, one. So just nine of you responded. Now, the reason why many of you have taken a longer time to solve this question is you have directly started differentiating this by using quotient rule. Remember, what did I tell you? Sometimes simplification before differentiation helps. This is one such example where if you simplify before differentiating, that is going to be helpful. And most of you would be thinking how simplification. Let's do that. By the way, six of you, that is the maximum 67% of you have said option number C. Okay. Let's check. See here, the term that you have on the numerator, one plus x square, let me write it as one plus two x square plus x to the power four minus x square. Okay. So the term in the numerator, I have written it like this. Okay. Which clearly is one plus x square whole square. This guy, at least this fellow is what one plus x square, the whole square, isn't it? Minus x square. Now you can use a square minus b square on the numerator. So a square minus b square is a plus b into a minus b. Yes or no? Now, what do you realize when you write this is that this term and this term are just going to get canceled off. And you will be left with a simple expression, which is one plus x square minus x. So what is the derivative of this? Easy to find out zero to x minus one. And you're comparing it with a x plus b. So a has to be two and b has to be minus one. Okay. This will come with a little bit more practice, more exposure to different types of questions. Okay. So don't worry. Initially, it might be a learning experience for you. But as you grow old in problem solving, you realize that certain things, you'll start making up a kind of a pattern and start applying it. Is it fine? Any questions, any concerns? Which step you're talking about? Second step? What is the issue with it? This term, it should be x square. Why? a square and this is your b. This term is your b. This term is your a. So what did it be? a minus b. See properly. Say to this term is a, this term is b. This whole, this term, which I'm circling out. So it's a square minus b square. So it's a plus b a minus b. Okay. Fine now. All your confusion is gone. Okay. So we'll move on to one more question before we take a break. All right. So I'm putting the poll on for this as well. It's an MCQ question with single option correct. If y is equal to this expression, find dy by dx at e to the power m to the power n to the power p. The two of you have responded well within a minute. Four of you have responded so far. Last minute, if you feel you need more time, you can always put it in the chat box, maybe like, you know, 30 seconds more or one minute more. I can always give you don't worry about time. Why I'm keeping the same restriction is because this is this, this is the time that you will be getting in your competitive exams to solve this question. So let us try to honor the side of respect that time limit. We don't have infinite amount of time to solve questions. Infinite time will make any problem look easy. Last 30 seconds. Okay. Five, one, and now I'm going to end the poll. Anybody wants to vote last minute? I mean last second. Okay. All right. So there has been a very, very confusing response. And if you see on your screen, C has got most number of votes. Okay. D has got least number of votes. Okay. I'll come back again to this poll, but let me solve this question first. See again, probably most of you did not take my advice of simplification before differentiation seriously. Okay. That's why you took longer to solve this problem. And in fact, most of you got it wrong as well. Look at the first term. One by one plus x to the power. Now, if you write it like this, n by x to the power m and x to the power p by x to the power m, you would realize that this whole term simplifies to, let's multiply with x to the power m throughout this whole term simplifies to x to the power m, x to the power n, and x to the power p. Okay. I found this on the web for sorry about that. My Siri got activated. Yeah. Right. If you see the second term, second term, this is your first term. Let me call this as the first term. Okay. This is your first term. This is your second term. Second term in a similar way would look like this x to the power n by x to the power m, x to the power n, x to the power p. This is your second term. Let's say this is the third term. Third term will become now something like this x to the power p, x to the power m again, x to the power n, x to the power p. Okay. Now, what are you doing? You're adding these three terms to make a white. So when you add these three terms, you yourself would realize that the denominators are same. So when you're adding it, just add the numerators and to our surprise numerator and denominator will become same. You're actually getting a one. So this expression that has been given to you is just a complicated way of writing a one. Isn't it? So what are the derivative of one with respect to x zero? So answer to this question is none of these and let me go back to the poll. Just one person got this right. Out of 11 of you who voted only one of you was right. Right. You probably got carried away with all these, you know, confusing things that the question center gave you like e to the power m to the power n to the power p etc etc etc. Okay. So always now this is a very, you know, important learning for us that we should never jump into application of these rules blindly. If at all there is a possibility to simplify the expression, go for it. Do it. It'll only make your life easy. All right. So with this problem, we are now going to take a small break. Normally, after two hours of class, we take a small break of 15 minutes. So we'll be taking that break right now. 608 we will meet exactly at 623 pm as per my watch. I hope you can see my watch over here. 623 will be meeting once again. Okay. Till then, enjoy your break, eat something, have a glass of water, eat something and come back. The rule is the rule which is going to help us when the two functions are related to each other like a composite function. Up till now if you see, if you just have a glance at whatever rules we did, the very first rule was talking about some and different rules when two functions are added or subtracted. Second rule was product rule when two functions were multiplied. Okay. So I'll write that down. So we talked about some or different rule for such cases. Right. So what helped us to differentiate such operations? Some indifference rule. Correct. When two functions were multiplied, what helped us to differentiate them? The product rule. When one function was divided by the other, what helped us to differentiate them? Caution rule. Right. But these are not the only operations which are possible between two functions. The third operation that you're going to see right now is a composition or function. When one function is fed to another function, this is called a composition or composite function, just like you form a compound in chemistry. Okay. So they react in such a way that they give a different, you know, compound or the different, you can say, product altogether, which has a different property than the reactants themselves. So f of g of x is like a compound, which is made up of two elements, like f and g. Okay. So this type of function is called a composite function. I'll give you some example of composite functions. But what are we going to learn here? The chain rule that is going to help us to do the derivative of such kind of functions. And trust me, more than 80% of the questions that you will be getting would be involving composite functions only. Okay. So some example of composite functions, which I can give you. Composite functions, examples. Okay. Let's say sine of log x. This is a composite function. As you can see, it has been made up of two functions. One is sine x and other is ln x. So what are you doing? You're feeding ln x within the sine x function. That means this goes and sits in place of x. Okay. If you do that, that is what will generate f of g of x. Right. So this is how it is pronounced f of g of x, which is in this case, sine of ln x. Now don't get this wrong. This doesn't mean you are multiplying sine x into ln x. No, it is not a multiplication of sine x into ln x. It is altogether a different variety of function. Sine of ln x. Are you getting my point? Okay. So in the same example, can somebody tell me, for the same example, can somebody tell me what will be, what will be g of f of x? Write down on the chat box, g of f of x. Very good. Lawn of sine x. That will become g of f of x. Okay. Tell me for the very same example, f of f of x. What will be f of f of x? Sine of sine x. Right. Absolutely. Tell me one more. G of g of x. ln of ln x. Absolutely. Well done. So you can have multiple compositions. You can have, let's say, three functions involved. Right. Let me give you another example. Let's say I give you three functions. One is x cube. Other is, let's say tan x. And let's say another one is maybe let's say 2 to the power of x. Okay. Now let me ask you this question. Create f of g of h of x. What will it be? So see here, the way to read this is first h of x is fed to g. And whatever comes out, that is fed to f. Fed to f means in place of x, you're putting that particular expression. So write down this on the chat box. Let me see who all are getting this right. Very good. Very good. So first, if you feed this to g and then whatever you obtain, that you are feeding to x over here. So it'll give you, let me just write down the result over here and match your answer. If you have got it, it'll give you tan of 2 to the power x whole cube. Or you can say tan cube of 2 to the power x. Got it. Excellent. Very good Nikhil, Ashleisha, Siddharth. Excellent. Okay. Try this one out. Tell me g of f of h of x. G of f of h of x. Anybody? Okay. So first this h of x, you feed it to f. So it'll become 2 to the power x whole cube. That is 2 to the power 3 x. Okay. 2 to the power x cube is 2 to the power 3 x. And that you're feeding to g. So it's tan of 2 to the power 3 x. Please note, there's a lot of difference between these two functions. Absolutely right. Absolutely right. Ashleisha, Siddharth. Very good. Is this fine? Now understood what is composite function? So in order to differentiate such kind of function, we have a rule called the chain rule. Now how does this chain rule actually work? Let's try to understand the rule first and then we will apply it. In fact, those who attended our bridge course program in the beginning of the year, you're already familiar with this rule, but just a quick recap for you all as well. So new people will learn it for the first time and old people, for them it will be a revision kind of a thing. So when you want to differentiate f of g of x with respect to x, chain rule says first differentiate f of g of x with respect to g of x and then multiply derivative of g of x with respect to x. As a symbol, we write this as f dash g of x. And this is what we already know, g dash x. Okay. So f of g of x whole derivative, that means dash of the whole expression is going to be this. This is what we call as the chain rule. Let me give you a demonstration for this. Note it down first of all, everybody. See the difference in the symbol and the symbol. So please everybody note down the difference here and here. Okay. This is a symbol. This one is a symbol where you are differentiating. If you read the symbol, you should interpret it like you are differentiating f of g of x with respect to x. But if somebody writes like this, f dash g of x means he is differentiating f of g of x with respect to g of x. Get this point. Okay. Let me give you a difference between these two. See, if let's say I give you a function f of x square. Okay. I say I am finding the derivative of this with respect to x. I will write it like this. So this is d by dx of f of x square. Okay. But if I write something like this, f dash x square. It is read as derivative of f of x square with respect to x square. Please notice the difference. Please note the difference. Getting the point. Is it fine? Any questions? All right. So let's see. Let's see how it works here. If you at all you want to find the derivative of, I would take that example of sine of log x because I started with this. Okay. If I want to differentiate this. So first I will differentiate sine of log x with respect to log x. Okay. Into derivative of log x with respect to x. Right. Now, why do we do this? It is because if you see this, let's say I call this log x as u. Okay. This also I will call it as a u. Okay. So it is like differentiating sine u with respect to u. And this result is very much known to us. It is going to be cos u, isn't it? And not only that, we know the result of derivative of log x also with respect to x, which is 1 by x. So overall, this becomes cos u into 1 by x, which means cos of log x into 1 by x. Is it fine? So in the beginning of this chapter, I was telling you don't extrapolate any formula. And this is the reason why I was saying don't extrapolate because if you think sine x derivative is cos x. So sine log x derivative should be cos of log x. You will become wrong there because the answer is not cos of log x. It has got some extra term also multiplied. So it is cos of log x into 1 by x. Are you getting my point? So I will just make this formula appear very easy for you. This is the process that normally I follow and I recommend people to follow. You start with the outer function. The outer function here is sine. So next, the outer function is sine and the inner function is ln x here. So start with the outer function. Whatever is inside that outer function, just call it as something. Okay, just label it as something word. So I would do this question like this. So I'll say, oh, sine something I need to differentiate. So cos something first I will write it down. Then that sine thing I will remove from my mind. Then whatever I see next, log x, I will write down the derivative here. Are you getting this idea how I do it? Let me give you one more example. Let me give you one more example. Let's say I want to differentiate sine of tan of 2 to the power x. Okay, so I purposely took a composition of three functions, sine, cos and 2 to the power x. Now if I have to do this question, if I have to do this question, a simple mechanism I will follow. Just give me a second. Yeah. So here what I'm going to do, first is sine something, whatever is inside, just read it at something. No need to read it completely. Just say sine something. So sine something is cos something, correct, into, now forget the sine. Then you have tan something. Tan something is secant square something. Okay. Now forget the tan. 2 to the power x derivative is 2 to the power x ln 2. Okay. And the moment you have basically covered all the function means your result is done, this is going to be your answer. So this is the working rule. Okay. So even though, I mean, I have given you the chain rule, we do not write the entire expression because it looks very, you know, time cumbersome. It looks like in a lengthy expression. So it'll be taking time to write those expressions down. So this is the working rule that we follow. Okay. And the name chain rule has come because as you can see it is forming a chain kind of a thing. Okay. So from outer to inner, you are going slowly, slowly by forming a chain kind of a structure. Make sense? Any questions? Okay. Would you like to try one more? What about the proof? See the proof for this is actually very intuitive. What is the meaning of, I think Nikhil is asking us for the proof. So proof of this, what is the meaning of this? The meaning of this is what is the change in f of g of x with respect to x? It is as good as saying change of f of g of x with respect to g of x. Okay. Into change in g of x with respect to x. So literally speaking, these terms will actually cancel off. So you're not doing anything to prove this thing. You're just rewriting it in a different way. For example, if I write five as five by two into two, what is the proof for this you're asking? There's no proof for this. It is just like rewriting the expression in a slightly different way. Now, let me give you an example which you will understand. Let's say there are three people, Ram, Shyam and Ghan Shyam. Okay. Now, you want to find, now let's say these are the 300 meters, athletes. They are 100 meters. Let's say Ram is twice as faster as Shyam. So two x faster than Shyam. Shyam is, let's say, three x faster than Ghan Shyam. Okay. Now, I'm asking you how much faster is Ram with respect to Ghan Shyam? What will your answer be? What will your answer be? You'll say, sir, six times faster. Isn't it? So how did you get this? It is because you were trying to compare Ram's speed. Let's say I talk about Ram's speed with respect to Ghan Shyam. So what do you wrote this as? You wrote it as Ram's speed with respect to Shyam into Shyam's speed with respect to Ghan Shyam. Right? This is the, you know, you can say in a nutshell you are doing over here, something very similar you are doing over here. So you're trying to see what is the change in this function first with respect to G into what is the change of G with respect to x. So ultimately, you'll end up getting change in f of G of x with respect to x. So chain rule, there is no proof per se, but it is just a, you can say a rearranging of the expression. Why do we need to rearrange it? That is the main question to understand. The need to rearrange this is because finding this is easy, finding this is easy, but finding this was not easy because our formula never taught us how to differentiate sign of LNX, did we? In any of the formula list, did I ever teach you or did I ever give you a formula for sign of LNX derivative? No, I never gave you. I gave you the derivative of sine x with respect to x or I gave you the derivative of sine of a variable with respect to the same variable. So when the functions become complicated like this, we need to break this or we need to rearrange the expression so that this and this are easy to find out. That is the whole and sole purpose. Are you getting my point? The same thing you used to do in your limits concept also, the same thing you used to do in your limits concept also. See, I'll just give you an example here. Let's say I want to find out the derivative of, the derivative of sine x square with respect to x. So as per the first principles, what would you write for it? Tell me. This is what you're going to write for it. Okay, please correct me. If anything is missing from it. Yes or no? This is what you're going to write. Now, what you're going to do here is that, what you're going to do here is that you are going to, you are going to simplify this expression. Okay, something like this. Okay, so all of you please notice this, what I'm going to do here is I'm going to call, I'm going to call, let me write this down. I'm going to call, let's say x plus h. Okay, this whole term, x plus h, the whole square. Okay, or let me write it as production formula here, transformation formula here. So this is going to be to sine a minus sine b, that is cos a plus b by 2 into, into sine a minus b by 2. Correct? Whole divided by h. Now all of you please pay attention. This h by 2, I will bring it down over here or let it be here. No issues. Okay, no issues. So you'll see here that what you're doing is something of this nature. This term, if you simplify it, this term if you simplify it, okay, or even if you don't simplify it, multiply and divide by the same expression down in the denominator. Okay, multiply and divide with the same expression down in the denominator. I think I'm finding lack of space over here to write it. Let me do one thing. Let me just copy this to the next slide. Just give me one second. I'll just copy it to the next slide. I'll just rewrite the whole stuff in the next page. So d by dx of sine of x square. So I'm writing it like this d by, sorry, limit of h tending to 0 x plus h the whole square minus sine of x square upon h. So now this term is to cos a plus b by 2 into sine a minus b by 2. Okay, and down we have an h. So what I'm doing is I'm now multiplying and dividing with x plus h whole square minus x square by 2 down in the denominator and up in the numerator. Now, see here, everybody please pay attention. I will rearrange these terms like this. This and this, I will club together. Okay, this, this term, 2, I'll cancel out with this 2 and this h, I will place it over here. Okay, let's see what has actually happened. This is going to be a standard limit, sine something by something where that something is tending to 0. Okay, that is going to be 1. This is cos of x plus h whole square plus x square by 2. Okay, h tending to 0. And this is going to be another limit expression. Okay, you can see for yourself that this will result into cos of x square. And this is nothing but the derivative of x square by first principles, which is actually 2x. Do you see what is happening? So the first outer function gets differentiated with respect to x square first, which is nothing but cos x square. And then the inside function gets differentiated over here, which is your x square derivative, which is your 2x. Okay, so while you are writing the first principles for it, you will automatically understand what is the reason that they get multiplied like this, because while you are evaluating them as a limit, the expressions will formulate in such a way that it will match with this expression. Getting the point. So this is the working rule, which I want everybody to be very, very clear with. The working rule is you start from the outer most function and start proceeding towards the inner most function. Okay, and while you proceed to the inner most function, keep canceling these things from your mind at least, or maybe take a pencil and keep scratching it off. Let me take more examples. I think when we solve more and more questions, idea will be clearer. Let's say I want to find the derivative of under root of let's say x cube plus 2x. Okay, let's take this simple example. Now the problem is, I know the formula of under root of x or x to the power half. I know how to differentiate it. Power rule, right? But I don't know how to differentiate something like x cube plus 2x, which has been raised to a power of half. So for such kind of a question, chain rule basically helps you. It says, okay, if you have something like this, first read it like something to the power of half. Okay, so this root can be read as something to the power of half. Okay, so something to the power half derivative is half something to the power half minus one, which is actually minus half. I would not like to write it as half minus one. Let's directly write minus half. So that's something is x cube plus 2x. Okay, now this half power is taken care of. Scratch it off from your mind. Of course, don't scratch it on your this thing. Solution part, just from your mind, forget that there was a half. Remove that half. What do you see next? You see x cube plus 2x. So multiply, put a multiplication sign. Derivative of x cube plus 2x is 3x square plus 2. That's it. Over. This problem is solved. Okay, and just putting this half back just to not disturb the question. The question was actually this. Is the idea clear how it works? Any confusion with respect to the solution here? Let's try more questions. Let's try more questions. Take another example and I would request everybody to work this out. What is the derivative of e to the power of tan of x to the power 10? Tell me. If you wish, you can write down your answer on the chat box also. It is not going to be too lengthy. Don't worry. So start from outer function. Outer function is e to the power something. Okay, so e to the power something is e to the power something only. Correct? Then just scratch this from your mind. Think as if there was just tan something. Tan something is what secant square something. One more mistake that people make is they start multiplying on the power of e. No, you have to always multiply at the same, you can say level. Okay, don't start multiplying on the power. Again, remove tan from your mind. What do you see next? x to the power 10. What is the derivative of x to the power 10? 10 x to the power 9. That's it. Excellent. Excellent. I think some of you are why it is just secant square x. It is secant square x to the power 10. I think you forgot writing. Okay, that's everybody has got it. Correct? Clear? Everybody's clear with this? Okay, let's take one more example. I'll keep taking examples because we need to be very, very clear about. Okay, let's take a simpler one. Costs of ln x. Okay, whole raise to the power of let's say 100. Okay, cost of ln x whole raise to the power of 100. If you feel it is too lengthy to type it out, just write it down on the chat box. Even that is fine for me. If you feel you have solved it, just write it down. Then we'll discuss and then you can match your answer. Okay, good, good, good. Very good, Siddharth. Very good, Ashlisha. Okay, now first something to the power 100. This is the way to read it. Something to the power of 100 is the outermost function. Okay, so something to the power 100 is 100 something. This is your something, by the way, to the power 99. Okay, into. Now this 100 should be scored off. Now cost something. Cost something is minus sign something. Got it? Now cost also you score off. Next is ln x. ln x derivative is 1 by x. Once you score everything off means your problem is completed. Excellent, excellent, good. So you guys have a certain command on chain rule also. Okay, let's take a question on the next page. Okay, I would request you to do number two question from here. Second one, this one. Under root of this is your ln actually. By the way, if the base of a log is not mentioned, it is to be taken with the base of E. This is something which I've already discussed with you. If the base of the log is not mentioned, it is to be taken as E. Done, very good, excellent. Excellent, very good. Okay, so first of all, first of all, let's call this as y for the timing. Yeah, so dy by dx. First of all, something to the power half, something to the power half is half, something to the power minus half. Okay, so now put a multiplication and just score off this under root. Think as if there was no under root. Okay, now next, ln something. ln something is one by something. I'm just using different color codes so that you are able to understand. So now ln is also scored off. Remove this. Okay, yellow. Now next thing that you have is sign something. Sign something is cost something. Okay, sign is also gone. Next, x square by three minus one. Now treat this together. Okay, don't think like, okay, next is x square. No, x square by three minus one will come together at the same time. Okay, so that will become, I'll use a blue color here, two x by three minus zero. Minus zero, you need not write. So this is also scored off. So everything is now taken care of. Is the idea clear how it works? Any questions? Is it fine? Alright, can we try one more? Maybe can we try the fifth one? Everybody? Fifth one, let's take the fifth one. Let's call it as white. So I'm just waiting for one minute for you to solve and then we'll discuss it and let me know by saying a done that you're done. Sign root of x square plus one. Done. Very good. So that's also done. Okay, so let's let's do this. Should we do it? Anybody wants a little bit more time? Very good. Okay, so this is how you read this. Ellen something first. Ellen something is one by something. And the thing that you have called something is sign of x square plus one. Put a multiplication, score off Ellen. Sign something is cause something. Correct. Score of sign under root of x square plus one is half something to the power minus half. So under root also you score off. Next you see x square plus one. That is nothing but two x plus zero. Of course, you need not write zero. Okay, but this is just for your understanding that I have written it. Okay, and some simplifications you can always do. For example, you can cancel this two with the two, you know, and rewrite this expression. So it actually becomes caught off x square plus one divided by under root of x square plus one into x. So this becomes your final answer. Of course, I've just simplified it. Is this fine? Any questions? Any concerns? Okay, now I'll tell you something very interesting. Actually, every function is actually a composite function. For example, if I say sine x itself. Okay, so sine x function is actually a composition of sine x and x. So you can apply a chain rule even on such simpler functions. See how it works. So sine something, derivative is cause something. Okay, now remove this sign x. What is the derivative of x with respect to x? One. That we anyways know that sine x derivative is cause x. Right. But ideally, we don't write such trivial things because it just waste our time. Okay. So in reality, so in reality, every function in this world is actually a composite function or actually can be written as a composite function. So your chain rule should work on all the functions. You know, that exists. Okay, right. Later on, we will scale up this concept to finding the derivatives of a function which is written in some other variable with respect to some different variable. For example, let's say if I want to differentiate sine y with respect to x, chain rule can be used here also. See how sine something derivative is cause something. Okay, into score this off y derivative with respect to is dy by dx. In class 12, you would require to know this type of chain rule also. But as of now, I'm not putting that much of emphasis on this, because you're anyways going to learn this in class 12 to large extent. In fact, the moment your 11th is over, your school teachers are going to start differentiation like this because they will be requiring it for your classroom topics. Is it fine? All right. Let's take some more problems. So these are some conceptual problems which are basically related to chain rule. If f dash x is under root of 2x square minus one and y is f of x square, find dy by dx at x equal to one. Find dy by dx at x equal to one. So these are your options to one minus two minus one. I'm putting the poll on. Almost about to be two minutes. Three of you have responded so far. Very few of you have responded. Okay, let me start the countdown. Very easy question actually should be done within 30 seconds. Okay, five, four, three, one, go. Okay, Siddharth, I'll help you out. Let me end the poll as of now. Most of you have gone with option B, which is one in this case. Okay, let's discuss whether B was correct or some other option was correct. See, I'll do it very, very, you know, you can say in a self-explanatory way. When you're finding derivative of y with respect to x, okay, or dy by dx, what do you call it? It is the derivative of this term, right? Because your y is f of x square, right? Which means you are finding the derivative of this whole thing with respect to x. Symbolically, this is what we write it. And as per our chain rule, this will be written as f dash of x square into derivative of x square, which is 2x. Is everybody fine with this step? Yes or no? Do let me know. Both are fine. Both are welcome to me. Even if you don't understand it, let me know it. Even if you have understood it, then also let me know it. Just like a yes if you are convinced. Great, most of you have written yes. So now see. So dy by dx is f dash x square. Now, f dash x is given to you. What is f dash x square? You will say, sir, simple replace x with x square. So it'll become under root of 2x square minus 1. So in order to get this, what do I have to do? Just a simple activity. Just change this x wherever you have with x square. So here if you put x square, here if you put an x square, here also you have to put an x square because whatever you feed there, the same thing will appear over here, right? It's like a machine. Into 2x. This 2x, I'm just copying it as it is. Now, at x equal to 1 means wherever there is a x, you have to put a 1. So it becomes under root of 2 that this is x to the power 4. x to the power 4 is like 1 to the power 4 minus 1 into 2 into 1. Correct? So what is this answer? This answer is going to be 2 minus 1 under root into 2, which is actually a 2. So most of you got this problem wrong. Six of you, I'm again sharing that is it. Out of nine who voted, six got it wrong. Only three got it right. Are you clear? This is how the competitive level question will be framed. Okay, this chapter is very easy, but competitive exams are known to frame difficult questions on easy topics because they know people will take it easy. Okay, is it fine? Any questions you have to let me know. Clear? Okay, let's take few more questions. Yes, somebody has a question. Yes, Seethu, that only I told you in the little while ago. There's a difference between, again, I'll repeat it. Please note the difference. Note this difference. When somebody says f of x square dash, he means derivative of f of x square with respect to x square. But when somebody says f x square dash, then he means this. So earlier when I told you did not care about it, right? When you saw that I am making a mistake in a problem because I don't know this concept, then you realize the importance. That's the whole sole purpose of giving a problem because many people don't appreciate this difference in the beginning. Only when they start making continuous errors while solving questions, then they appreciate that this is the difference. Clear everybody? Okay, let's take another one. Let's take this one. f of x is log of log x to the base x. By the way, there is no base mentioned here. So as I told you in the beginning, when the base is not mentioned, you have to take it as e. So I'm relaunching the poll for this question. So once you have solved it, please choose the right option from the poll. Oh my god, three people, three different options. Okay, so one of the options has now taken a lead. Okay, Siddharth, if you are unable to see the poll, you can always put your response on the chat box also. Okay, last 30 seconds. Let the countdown begin now. 5, 4, 3, 2. Most of you have gone with option number A. Okay, let's check whether A is correct or not. So first of all, when I write this expression, I will use my change of base property and I will write it like this. Okay, in short, ln of ln x by ln x. Okay, so this is your expression. So this is a mix of, you can say, a quotient rule involved along with chain rule. So f dash x will be nothing but let's apply quotient rule first. ln x into, now derivative of ln of ln x, you will have to follow chain rule there by because the composite function ln of ln x. So ln something is one by something into ln x derivative, which is going to be one by x. Is this part clear? Okay, minus ln of ln x into derivative of ln x, which is one by x. So basically what I've done, g into f dash minus f into g dash divided by square of the denominator function. Yes or no? Now they're asking you at E. That means wherever you have an x, you have to put a E there. Okay, now see the simplification. ln E will become one. One by ln E will again become a one into one by E minus, now see here, ln one is zero. So everything will be coming as zero over it divided by one square. That is one itself. So over all this entire expression simplifies to just one by E, which is your option number A. So those who went for option number A, well done. That's the right answer. Is it clear? Any questions? How did it become a zero? Okay, so ln, ln E is ln of one. ln of one is zero, my dear. So this is how this becomes zero. Log of one to any base. That is always zero. Clear, Sethu? Is it clear to you? Yeah. Prishan, which part do you want me to explain once again? The whole thing or just a particular step? Okay, you got it now. Okay, great. Is this fine? Any questions you have? Please let me know. Okay, don't carry any doubt with yourself. Let's take another one. Yes, Sethu, tell me what doubt you have. Should I go back to the previous question or is the doubt in general? Why can't we do without changing base? Why can't we do without changing base? Do you know the derivative of log of some function with respect with the base of x? Is that result known to us? Then how do you plan to do it? My first principles. Don't even try. It will take you a long, long time to solve it. See, the entire process that we are following, it is because we are not aware of derivative of every function existing in this world. What we know is the derivative of those standard function. So we are basically depending on the results of those standard functions to get the answer for complicated functions by breaking it down. So change of base is one such step that we take to break it down to a simpler function. Okay, so the entire story is revolving around that we don't know the derivative of every function directly. So we are trying to calculate it from the derivatives of standard functions. Okay, so I'll be relaunching the poll here. I'll be relaunching the poll here. I'll be uploading the assignment and all those things by 9, 9, 30. Okay, so there will be small delay. So by 9, 30, you will get the assignments on the on the WhatsApp group. Okay, so why is f of 2x minus 1 upon x square plus 1 and f dash x is sin x, then dy by dx is which of the following? Yes, only two people have responded after two and a half minutes almost, I can give 30 seconds more. Same story. I mean, there's nothing different about this problem. Same story here to repeat. Okay, now I've got almost five responses. Differentiation is a gateway to understand calculus, right? So please, I'll request you again. It's my sincere request that don't keep any doubts, any confusions with respect to differentiation because that will prove very costly in your last final year, class 12th. Okay. Okay, let's begin the countdown 5, 1, and I'm ending the poll, most of you have gone with option number B, which is 57% poll. Okay, let's check. See here, dy by dx you want to find out, right? That means you want to find out, you want to find out f of 2x minus 1 x square plus 1 whole dash. See, this is the way I'm writing it whole dash. So as per our chain rule, it will be f dash. So this dash will just come in the middle like this into derivative of this inside function. Now derivative of this inside function will require quotient rule. So x square plus 1 into derivative of this, which is 2 minus this derivative of this, which is 2x by x square plus 1 whole square, correct? So on simplification, this will give you f dash 2x minus 1 by x square plus 1. So this will be, if I'm not mistaken, 2x square minus 4x square. Okay, that is going to be negative 2x square. Okay, and you'll have plus 2x, okay, and plus 2 divided by x square plus 1 the whole square. Take a 2 common, and one more thing I can do here is, please note f dash x is sine x. So if I change my x with this input, can I say the output will become sine of this term? See treat this like a function in itself. Okay, think as if this is some g of x. And I'm asking you to find g of 2x minus 1 by x square plus 1. So this x you just remove and put this whole fellow into that. Okay, so just become sine of 2x minus 1 by x square. Take a 2 common, if you want. Okay, so take a 2 common out from here. You'll end up getting 1 plus x minus x square upon x square plus 1 the whole square, which is clearly matching with option number b. So b is the right option. Most of you got it. I mean, just seven of you voted out of which four of you got b. Well done. Is it fine? Any questions? Any trouble you are experiencing in deciphering this expression? Do let me know. Okay, so we'll take a last question. Okay, let's take this one. This could be a multiple option, correct question. So please put your response on the chat box. So more than one option can be right here. Done. Okay, let's discuss this out. See, dy by dx. First you have to differentiate the first fellow. So this first fellow will give you e to the power root x into derivative of root x, which is 1 by 2 root x. Then plus derivative of the second fellow will be e to the power negative root x into derivative of this term will be minus 1 by 2 root x. In short, it will become e to the power root x minus e to the power minus root x by 2 root x, which is clearly option number a. But wait a second. One more option can be correct over here. Let me show you which one. See, this term e to the power this, if you square it, what will happen? It will become e to the power. This is going to be a square of this, which is I mean, I can write it as this also. Okay, plus two. Yes or no? Now, when you subtract the four, when you subtract the four, it becomes e to the power 2 root x plus e to the power minus 2 root x minus 2, right, which is actually e to the power root x minus e to the power minus 2 takes the whole square. In other words, under root of this term can be written as, now remember when you're under rooting both sides, this will become mod, isn't it? But we already know this term is greater than this term because here the power is a negative power. Here the power is a positive power. So it will be as good as this. Okay. In other words, this term can also be written as under root of y square minus 4 and this two root x, you can just copy paste as it is, which is also option number C. C is also correct. Is it fine? Any questions? Any concerns? Okay. So with this, we end this today's session here. We will meet again tomorrow with a new topic, which is which is Introduction to 3D Geometry. Okay. That's a one-class topic only. See you tomorrow, 4.30 to 7.30. I will put the time. Fine. Okay. See you. Bye-bye. Take care. Stay safe. Sethu, you can call me for that. Thank you.