 Hey, hello guys, are you there? Are you able to hear me? And let me know if you can see the screen. Just typing yes or no, if you can see the screen and if you can hear me. Can you guys hear me? Okay, cool. So let us start. And I hope those who had missed the first differentiation session last time around, they have gone through it. And if you have any particular query, you can tell me. Otherwise we'll move ahead with the session, what we have planned today. So last class, we discussed basics of first principle of differentiation, if you remember, and why are we doing it? Because we want to give you a head start in differential calculus. And next year, when you're going to solve, you know, or let's say when you are going to deal with kinematics and dynamics in mechanics, you know, in the early days of your 11th grade, this basic knowledge of differentiation is going to help you a lot. So please be very, very thorough. I also give you some basic standard differentiation or differentiation of some standard functions. And then, yeah, can you now not see a blank screen and a whiteboard, sorry, a blackboard? So this session is for differentiation. Can you see this? Yeah, guys, please confirm if you can see this. Okay, in the last class, we discussed basics of differentiation. I talked about what? First, we talked about rate of change. If you rate of change, change of y with respect to with respect to x, if y is a function of x. Yeah, and then, you know, we learned that. We learned that and we also saw concepts of average rate of change, average rate of change as well as instantaneous rate of change. Instantaneous rate of change. And we saw the graphical interpretation of rate of change. We said that if we have, let's say, yeah, if we have, yeah. So let's say we had this x and y, yeah. So this was my x-axis. This was my y-axis and we had a, let's say a graph in it, a function, right? And then we discussed about the rate of change. If you see, what was the rate of change given as, average rate of change between, let's say, point x1. And let's say this was my x2 and corresponding y values were y1 and y2, y1, y2. And then we joined this, okay? So this was my card, which was defining the slope of which defined the average rate of change, isn't it? Yeah, so this was the card which defined the average rate of change, let's say this. Yeah, so this was my, this one. And how did I define average rate of change? So average rate of change was nothing but we just did this. So this portion was delta y and this was delta x. And average rate of change was given by nothing but delta y upon delta x. Now, what did we say? We said that as delta x tends to zero. So if we start reducing the gap, this delta x, if I start reducing, or let's say, if I start moving delta x towards zero and then I, in technical terms, I wrote this that limit delta x tends to zero and I said delta y upon delta x, delta, sorry, delta y upon delta x, limit delta x tends to zero, delta y upon delta x, this particular term was given a new notation and it was called dy by dx, right? Where delta x tends to zero and in this particular thing was called dy by dx and this is known as first derivative. First derivative of y with respect to x, to x and we said that don't treat it as d into y and d divided by d into x. Here we say that d upon dx is a differential operator for y with respect to x, okay? And in running this thing, we write dy by dx, like that, dy by dx. This is the first derivative and then what was its significance, physical significance was this, that as delta x tends to zero, as delta x tends to zero, this chord, this chord, let us say this chord was AB, this chord tends to become tangent at point A. So if you reduce delta x, so if you shorten this gap, if you reduce which delta x, this delta x, if you keep on reducing this, you will get, what will you get? This B, this point B starts moving towards point A and that the chord AB eventually turns out to be a tangent at point A. So hence we say this is nothing but slope of tangent, slope of the tangent, slope of the tangent at x one, right? So if you are doing in, you know, so at or at any given x, so don't write this at any given x. Okay, this was my definition of, or this is what we did or, we understood the concept of rate of average rate of change and as delta x, the gap in x window, the gap is reduced to zero, close or it moves to zero, it gets converted into instantaneous rate of change and hence the chord, average rate of change was given by chord slope and this is given by the tangent slope. So tangent was the limiting case of the chord. This is what we learned as the basic concept. Then we went to understand how do we do, how do we find this differentiation thing? So okay, let me use this space only. So if you see, how did we find the differentiation part? So we learned, next we learned something called second thing which, what was the number? This was third, I believe. So third thing which we learned was, third thing which we learned was differentiation by first principle, differentiation, differentiation, differentiation by first principle, okay? And what was that? We again, did the same thing. So what did we do? We had x, we had y, sorry, the other way round. So this was y and this was let's say x and we had a curve like that and we call it as let's say y is equal to x square. This curve is y equals to x square. And then what did we do? We took a random point x. This is, let's say this point is x, I'm sorry. And then we also took another point on the same x axis. Let's say this is x plus delta x where clearly this gap was delta x. This gap was delta x. And then corresponding to x was y. So this point was y and this point was y plus delta y. Yeah, and then what was y? Clearly y is nothing but x square. Why? Because it is x comma y is this point, isn't it? Let me write, this is x comma y. And this point is x plus delta x comma y plus delta x and these two points lie on the curve y is equal to x square. So clearly y plus delta y will be equal to x plus delta x whole square clearly clear. Now, how did we do the differentiation by first principle? We did this, what did we do? We said that, okay. So average rate of change will be nothing but again. So average rate of change between x and x plus delta x is the line joining these two points slope. So if you find a slope of this line, you'll get the average rate of change, isn't it? So what is average rate of change here? So you draw it like that and you draw it like that and find the slope of this line. This slope of what slope? So tan theta or slope, slope is here. What is, this is delta y and this is delta x. So average rate of change. So average rate of change was average rate of change. Rate of change of y with respect to y with respect to x was given by between what, which two points? So at the moment I talked about average, you must know which two points I'm talking about. So I'm talking about which two points between x is equal to x and x is equal to x plus delta x between these two points, I want to find out the average and it was given by delta y upon delta x. But now I'm interested in finding out instantaneous, instantaneous, instantaneous rate of change, instantaneous rate of change of y, rate of change of y with respect to x at x, at x, that means at this point. What is the instantaneous rate of change? So you know what to do. You take this point B and start moving towards A. Okay, that means you reduce delta x and hence technically we say, you reduce delta x and technically we say that, this is nothing but an instantaneous rate of change of y with respect to x at x is defined or let's say, denoted by dy by dx, right? So this D is for delta. So this is an operator. Again, I'm highlighting, don't get confused that it is some D into y or something, no. So D stands for delta, delta y upon delta x, but just to differentiate, this is our limiting case, isn't it? What is this limiting case we say? Limit delta x tends to zero delta y upon delta x. For all this, we will use only this terminology. We are not going to write this again and again. So hence for this, whenever we say dy by dx, it will mean limit delta x tends to zero delta y upon delta x. This is symbolized by dy by dx. And hence let us calculate this limit then. So this is nothing but limit delta x tends to zero. Now here my function is y is equal to x square. So I can write delta y can be written as y plus delta y minus y, isn't it? The numerator delta y is nothing but y plus delta change in y. So y plus delta y minus y divided by delta x is the limit. Okay, now, so I have to take it to this here. So please remember I'm going to write this again, this side also, so don't worry. So hence I'm saying dy by dx, dy by dx is equal to, dy by dx is equal to limit delta x tends to zero y plus delta y minus y divided by delta x. This is the instantaneous rate of change. Now this is equal to y plus delta y from the curve I can know, I know delta x, x plus delta x whole square. If you see, if you see what was y plus delta y if you see here, y plus delta y was x plus delta x whole square and y is x square. So remember this, okay, we are going to write it here. So hence what can I write? I can write minus x square divided by delta x. And then we used, okay, here this is missing limit delta x till we apply the limit we keep on writing this limit delta x tends to zero. So next step will be limit delta x tends to zero. It is nothing but if you expand the top one x square, x square plus two x delta x plus delta x square delta x whole square minus x square and this divided by delta x. So clearly I can strike off what this goes, this goes, so hence the resultant finally it is delta x. So again, you have to write this step. What step limit delta x tends to zero delta x common and it is two x plus delta x, I took delta x common because this is delta x square, this is delta x I can take delta x common divide by delta x, delta x, right? Now you can strike out this delta x because this is not zero because it's tending towards zero, it's not zero, it's not equal to zero now. So hence you will get what? Limit delta x tends to zero twice x plus delta x. What does this mean physically? What is the physical significance of this? Physical significance of this is that the rate of change of any variable y which is varying with x at any given value of x is nothing but twice the value of x plus the change in x. Can you see this? So what do I mean? The rate of change, how fast or how slow something is moving or y is moving with respect to x is given by, given by twice the value of x plus the small change in x which you are, in which you are trying to find out the average value but the instantaneous value will be where this delta x becomes zero is the limiting case so hence dy by dx, applying the limits. So applying the limits, apply the limits now. Applying the limits, applying the limits you will get what? So as delta x tends to zero that means the limiting value of two x plus delta x when delta x goes absolutely towards zero will be simply. So we write dy by dx is equal to twice x. So this is what we got as the derivative with first derivative of y is equal to x square is two x. So what does it mean? That means if y is equal to x square then dy by dx or instantaneous rate of change of anything will be twice of x. Let me take an example to illustrate this because last time when delta x is almost zero how can we cancel it from numerator and denominator? Ruby is asking this here delta x is tending towards zero it's not zero. So delta x is let's say numerically speaking it will be what? It will be 0.000001 into some twice into some value and then something n divided by 0.000001 like that. So this by this is one, this is not zero so you can always cancel delta x is not zero this doesn't mean it is equal to zero it is going towards zero that means it is very small value infinitesimally small value but it has a non-zero value. Right, so hence you can remove delta x from there. Okay and this value is so basically what this means? This means this is the limiting value of two x plus delta x. So what is the limiting value of two x plus delta x is two x only because two x plus delta x is never gonna breach the value of two x that's the last limit. So hence it is a limiting value of two x plus delta x is two x. So now if you see I have removed the limit signs and I am saying two x. Okay now so this is what does it mean? Let us take an example from your ninth grade syllabus if you remember then it will make more sense to you. So now let us go to kinematics of ninth grade if you remember my dear friends that their displacement so what is the case? There is a straight line motion. A body is moving in this straight line. Initial velocity was zero and acceleration is let's say A constant, acceleration is constant. Then let's say this was my zero point. So hence let's say it at a given time at time t the distance traveled is s then you know what was the equation? Second equation was s is equal to u t plus half a t square. Okay now applying these limiting conditions u is zero. So let it let u be zero. So s was given as half a t square. Why? Because u is equal to zero right? Now this was so if you now let me also parallely write whatever we discussed here. So we it is similar to let's say saying y is equal to half ax square right where y so you can you can draw the correspondence s is equal to y and t is x here right? So this is what we did here in the previous case and this is what we did in ninth grade. Now let us find out so if you remember you know so rate of change of displacement with respect to time will give you instantaneous velocity. So instantaneous velocity in this case will be given by ds by dt differentiating s with respect to t will give you the instantaneous rate of change of displacement with respect to time that is velocity. Now if you differentiate and we discussed in the previous session also that if there is a constant you can take out the constant and then differentiate the function. So basically differentiate dt square by dt. Now we just learned that if y is equal to x square then dy by dx dy by dx is twice of x isn't it? So hence if let's say the function is s is equal to t square. So ds by dt will be 2t ds by dt will be 2t. So hence this is what it is right now. So hence you can write half into a into 2t which will be at if you remember. Now if you remember what was velocity from the first equation v was equal to u plus at isn't it? And if u is zero v is equal to at and u is zero. Can you see you get the same result here? So when you differentiate this displacement with respect to time you'll get instantaneous velocity which you learned in your previous grade. So I hope this example you know you could relate to that. You know if s is a function of time so if you see here what was it? s was ft function of time. Okay and when you differentiate s with respect to time with respect to time then you will get instantaneous velocity if a was constant. If a was not constant we'll see how a has to be differentiated with respect to time again. So this was the case example of how differentiation yields to the rate of change. Okay fair enough then we learned I gave you some common standard differentiation rules and the rules were these and we had made a table also so if you guys could not do that that day you can do it now. So let's say I am going to write y here and I am going to write okay. Again another thing is this is an information please note down that dy by dx. dy by dx can also be represented as y prime. It's called y dash or y prime. First derivative whenever you'll see this this will mean this and let's say y was equal to fx. Let me write in a separate this thing. Wait a minute. I'll write it somewhere else so that you guys don't make mistakes, right? So let me just wipe it off totally. I'll write it first. Please have this understanding very very thoroughly so that you don't make or your concepts should be crystal clear now. Wait a minute. Yeah. Okay. No. So let's say if these are some terminologies please remember these. Yeah. Please remember what? You should remember that if y is equal to fx let's say. Now don't get confused if variables are changed. So you can also have v is equal to f of t. That means v is a function of t. You can also have s as a function of t. Yeah so these are just variables. So here you will write dy by dx is the first derivative. Here you will write dv by dt as first derivative. Here you will write ds by dt as first derivative. This can also be written as d of fx by dt. This can also be written as d of ft by dt. This can also be written as d of ft again ft by dt. These are all same things. And then the first derivative can be written as f dash t as well or this can be written as s dash whichever way you want to express. All these things mean the same. Okay. Now I gave you this table. Oh this is done. So let's go to the next. So let us go to the next. Yup. So now write down this table. Now table. Okay table is this. So that means if you have... So you remember these rules and these rules will be handy. Okay. Okay so table was this. What was the table? The table was that if you have a standard function y is equal to fx let's say. Then what will be y dash or what will be f dash x or first derivative or dy by dx or d by dx of fx. All are same thing. All are same thing. So hence let's say y was equals to constant or I'm not writing y again and again. So it will mean same. So let's say point number one. If y was equal to c, a constant c then dy by dx will be zero. Example. So let us also take examples parallely. Let us also take examples parallely so that you will be thorough. Okay so hence let's say y is equal to five. If y was five dy by dx will be zero. Dy by dx will be zero. Why because it's constant. So it's not changing. So what will be the rate of change? The rate, the change itself is zero. If y is constant that means for any given value of x the y value is same. So hence if you remember it will be a line parallel to. This is a y is equal to c, c graph. Isn't it y is equal to c, this is y, this is x. So the change in y is not happening. So rate of change will be zero. Clear? Point number two. When y is of algebraic form, x to the power n where n is constant, n is constant. Many people make mistake here. Please mind x is, n is constant, x is variable. Right? So hence if it is equal to let's say x to the power we just checked x to the power solid. Here we'll have to write the differentiation. So the differentiation will be nx, n minus one. This is a differentiation, yeah? So example if it is x to the power two we just saw differentiation dy by dx was two x. So two comes here and two minus one. Two minus one, two x. If x was x cube it will be three x square. If it is x to the power four it will be four x cube. If it is x to the power one by two that is root x it will be one by two x to the power minus half. Yeah, if it is x to the power 0.5 or that same thing. Yeah, x to the power 0.5. Let's say it is x to the power 0.9. So it will be 0.9 times x to the power minus 0.1. Like that. This is what it means by, yeah, this is algebraic function. Third is trigonometric now let us say y is equal to sine x. So this thing will be cos x. Differential will be cos x. If it is cos x it will be minus sine x. And if it is let's say tan x then it is secant square x. Okay, sixth is let's say e to the power x exponential function. Again if you see e is a constant Euler's constant and what is the value of e? E is 2.71 something, something, something. It's an irrational number. So Euler's constant value is 2.71 and it is raised to a variable. Then it's differentiation you will get same. Okay, seventh is ln x, ln x is one upon x. Okay, where ln is nothing but log 2 base e. Yeah, is also written as ln x or ln x. This is this table you must keep handy. There are many more standard this thing but then most of them can be derived using these ones. So these seven basic ones you remember so that when you solve problems, keep this thing, keep this list in front of you and then only you solve problems. This is about the standard differentiation. Now we will give you some rules. What are the rules? Rules of differentiation, please take down. And this list which I just gave above and the list which now the list which I'm going to give now, these two will be very, very handy in doing basic integration. Don't think that this is the entire differentiation here but yes, this is good enough for you to deal with all physics and chemistry related problems later on. And to add an extent, if you're interested in taking up calculus a little early also this will give you a head start. Now, so rules of differentiation. So we did this, I'm just doing it again because it becomes very difficult to grasp it in few classes. So hence, don't worry about the repetitions as well. So rules of differentiation, please take down. And the best idea normally is that you copy them in a A4 sheet or something and keep it with you all the time or you can always fix it in front of your table. So don't hesitate to look at these rules while solving problems. You should not mug it up. And what happens is if you do it multiple number of times anyways you'll be able to recall the rules. First rule one, if y is equal to k times fx, k is a constant. Where k is constant or rather let me do it in a table form again but let me do this for the first case. So dy by dx is, dy by dx is k times f dash x. k times, okay, you can just pull out k. Example, y is equal to, so example could be y is equal to let's say 5x square. So dy by dx will be 5 times d by dx of x square which we saw is nothing but 2x. So hence it is 5 into 2x is equal to 10x. Okay, so problem for you. Do this question number one. Find dy by dx is equal to what? If y is equal to half sine x. See the table and quickly solve this. What will you answer quickly in chat box? Start answering. Yeah, quickly, all of you. Show mix says half cos x, good. So it's nothing great. So you can look at the table. Table says, table says, table says, sine x differentiation is cos x and the constant you have to remove and you can find half cos x. Any problem, I don't think any one of you would be having any problems in this. Let's go to the next rule. Next rule is sum rule, sum rule, right? So if y is equal to fx plus minus gx, that means if y is expressed as some of two functions, fx plus minus gx, then y prime or dy by dx, dy by dx differentiation is, I think, but f dash x plus minus g dash x, okay? Example, example. If y is equal to x square plus x cube, okay? So dy by dx will be, dy by dx is twice of x plus twice of x square, clear? Questions for you. Y is equal to log x plus three e to the power x. Find a dy by dx, quickly, quickly. Keep the table in front of you all the time. So I hope you have taken down the basic differentiation table and now you, yeah. Oh, for that matter, okay. Let me just simplify it a bit. It is not log, you can take it as, Ruhi, it is not log, it is ln. So let me just, I'll have to, yes, I'll have to. This is not log, this is ln, lnx. Yeah, if it is lnx, then only you can write that. If it is log x, then I will do something extra. One upon tan x, how come tan x, tan x is not there? Yes, all of you, quickly. I saw only, only two answers, yes, right? So I hope now you are getting it now when we went forward and then, now it's done. So let's say rule number three. So answer is simply one upon x plus three e to the power x. Good, next rule. So we are moving a little faster. If my pace is higher, you can also tell me that. So third rule is we call it product rule, product rule. And the product rule is product rule, product rule is, let's say y is equal to, let's say y is equal to fx times gx. There are two product, a product of two functions. Okay, then dy by dx is given as, you differentiate the first function and then product it with the second function and then add to it first function, product with the differentiation of the second function. f dash x plus gx, f dash x into gx plus fx into g dash x. Okay, so hence, now, example, y is equal to, let's say sine x cos x, this is the function. This is the function. So here, clearly, what is fx? This is fx, fx and this is gx. This is gx, so hence, what will be dy by dx? dy by dx will be dy by dx is equal to, first function differentiation. Sine x differentiation is cos x, cos x and multiplied by cos x, right? Then plus fx is sine x itself and differentiation of cos x is minus sine x. Okay, so hence, it is nothing but cos square x minus sine square x. How is log differentiated? Very good question. So let's do question from Trippan is, how is log differentiated? No, so let's say y is equal to log of x. Okay, y is equal to log of x. Then, how can you express? This is what, this is nothing but. So I think you guys are familiar with, right? So can I not say this as log of x to base 10? Now, can I not change the base by doing this log of x to base e divided by log of 10 to base e? If you remember the rules of logs, rules of logs are that if y is equal to, also not y is equal to, let's say log of log of, this is a rule of log of a to base b can be written as log of a to base c divided by log of b to base c. And you have to say c is greater than zero and c is not equal to one. That is so you can do that. Okay, so hence I can represent log x like that, now what is it? This is simply, so you can find, this is a constant, isn't it? This is a constant, so it is one upon log of 10 to base e or one upon log of 10 to base e, I'll do it in the next case and this is ln x, isn't it? Now, what is this? It's nothing but log of, or basically ln 10. One upon log of, one upon log of 10 to base e is nothing but, or not ln 10. One upon ln 10 is, I'm sorry, let me just remove this, yeah. So one is, what is that? It is clearly log of 10 to, sorry, log of e to base 10, so log e into ln x. Now, if you differentiate, this is constant, so you keep this constant and then this is log, so log e by x will be the answer. 10 upon x, no, not 10, log 10 upon x, okay. Is that clear to everyone? Yeah, so now next, move on to product tool is done, so problem for you. So here is, you remember this formula and then do this, problem for you, problem, problem for you. Problem for you is, find out y is equal to e x ln x. Find this, e x times ln x. Find the d y by d x of e x times ln x, quickly. Yeah, what is the answer, quick guys. You keep the table in front of you and then only you do it, don't guess. Alter is saying 10 upon x, okay, no. No. So differentiate the first function, then multiply it with the second one, then activate first function and differentiation of the second function. Kirtna says e x ln x plus one upon x, absolutely right. Good job, Kirtna. No, yeah. Good, e x one upon x plus ln x one upon x, no. No, no, no, no, no, no, something wrong. See, how do you solve this? Okay, so it is d y by d x will be first function. So differentiate this, so d y by d x of e x into ln x plus e x into d y by d x of ln x. ln x, so hence it is e x times ln x plus e x times one upon x, so hence it is e x times ln x plus one upon x. Perfect, so all of you understood, anyone is having any doubt, please, you know, you can text me or you can also put it in WhatsApp, whichever way you want to do it. So let's move on to the quotient rule. This is the other day we also did this rule and this rule number four. Quotient rule says that if y is equal to f x upon g x such that g x is not equal to zero, g x is not equal to zero, then, then d y by d x, d y by d x is given as what? You remember this, you know, what do you call mnemonic? It's like low d high minus high d low. So what does it mean? I'll tell you later, but you remember, low d high minus high d low, just to remember. What does it mean? So you first write the second function, so let's say if this is high, this is low, okay? So hence low, so write the lower function, g x. Then d, differentiate the first function. So d by d x of f x. Then minus, minus the high d low. So high function is f x times d by d x of g x divided by, but there's one more thing and this is g x whole square. So low d high minus high d low whole divided by the lower function square, okay? Remember this rule. Mind you, this holds only when g x is not equal to zero. So hence, okay, so hence now, now you can derive your own standard function for, or differentiation for example, problem for you. So find out if y is equal to cotangent x, cot x, then find out d y by d x. Hint, use quotient rule. Okay, for that matter, this is called quotient. Agitya, which one should I explain once again? The quotient rule. The quotient rule says y is equal to f x by g x if it is quotient of two functions, if y is f x upon g x. Then d y by d x is given as g x times first function derivative, the top function derivatives or d by d x of f x minus f x times the lower function derivative divided by g x whole square. So hence to remember this, many times people forget which one should be first, which one should be second. For that, we have devised the mnemonic. Low d high minus high d low, divide by the lower function square. So where high is the numerator, low is the denominator. So low d high means first write the lower function, then differentiate the first function, then minus the high function, and then differentiate the lower function. Didn't know how to solve this question. Okay, nevermind. So hence cot x, you have to express first of all in terms of f x by g x. So what can I write y is equal to? Cot can be written as cos x upon sine x. Yeah, and cot is not defined as x equals to zero. So we can, you know, we can eliminate x cannot be equal to zero or multiples of, even multiples of pi. Yeah, so that is the thing, right? So or let's say sine x cannot be zero, I'm writing, just in case you don't know. So sine x cannot be zero, right? Now, in that case, what can I write? So dy by dx, dy by dx will be low d high. So the lowest function first, so sine x into differentiate the first function. So differentiation of first function cos x will give you minus sine x from the table which I had given. Okay, minus, minus, minus what? High d low, so high function was cos x and differentiation of the low function. That is sine cos again, right? And divided by sine square x, the square of this thing. So this is equal to nothing, but if you see, what is this? Yeah, I don't have much space here, but anyways, let me write. So this is equal to, if you see, this is equal to, in the top, you'll get sine square x, sine square x plus cos square x, divided by sine square x. So for the positive space, I'm writing it here. Yeah, so this is nothing but, this is equal to, so you can, yeah, so this is equal to minus one upon sine square x, which is nothing but minus cos square x. Okay, so all of you have got it correctly. I hope you understood Aditya, yeah? Great, so let's move ahead. Let's move ahead, just give me a second so that I can pick up a few questions for you just a minute. Yeah, so now next is the last rule which we studied the other day was the chain rule. So let me go to a new, okay. Now what we are going to study is, this is rule number five, chain rule, chain rule. Okay, now imagine, let's take an example first, then you will understand what this chain rule is helpful into. So let's say I have to differentiate y is equal to sine square x, let's say this is a function. Y is equal to sine, sine square x. Now if you go to the table, which I just gave you, the seven standard differentiation of functions of standard, sorry, differentiation of seven standard functions I gave you in the first table, there is no mention of, there's no mention of function which is raised to a power here. Okay, then how to solve such? So I know if y is equal to sine x, if y is equal to sine x, I know dy by dx is, dy by dx is, dy by dx is two sine x, sorry, cos x, cos x, right, that table says. Now many people do a mistake here, they say that, okay, why can't we use the x to the power n rule? So normally when I give this problem to them, they will do dy by dx, dy by dx as two sine x. I have seen people making this mistake, where they think that, okay, it is nothing but sine x to the power two, sine x to the power two, and hence we can apply this rule, but this is wrong. Because here it is algebraic function, this is x, here it is sine x, these two functions are different. Though the power is constant, but x is not equal to sine x. So you can't differentiate like that, so this is not allowed. So hence what do I do? So hence I try to reduce it to some standard form. Sir, can we can solve the using problem? Yes, you can definitely use the product tool, Shomik, no doubts about it, but we can, but I'm illustrating the change rule, and there will be many cases where the product tool will also be not applicable. We'll see those examples as well. In this case, I have taken a very simple example to illustrate you. Product tool can be used, definitely, yeah, can be used. But now, how do I, I'm just highlighting what all errors people make. So it is not, you cannot differentiate it using x to the power n rule, that's not allowed. So always remember, x to the power n is always when x is an algebraic function, and some constant is used as the power. So that's not allowed. Once that is understood, how do we solve it? So hence the idea is to reduce it into some standard form. So let us start with this. So I'm writing this as y is equal to, let us say, v square, where v is equal to sine x, right? Now I get two functions. So I can say y is a function of v, yeah? And v is a function of, let's say, x. Right, so you can see a chain, right? Y is a function of, so you see the chain. Y is a function of v, and v is a function of x. This is a chain. So hence, how do I, so can I, is this standard form? Can you see, this is standard form? Yes, these are standard forms. So what can I do? I can do y is equal to y, dy by dv is defined. dy by dv is clearly two times v by standard rule, because y was a function of v and v square, so two v. But the question was to find out dy by dx, and not dy by dv. So this is what, this is what was to be found out. So dy by dv is two v, and what is dv by dx? I can do that also, this is nothing but cos x, right? So hence, dy by dv, if you see, dy by dx can be written as dy, sorry, dy, dy by dv into dv by dx. That's what is called chain rule. So it looks like dv is getting cancelled. Can you see that? But it's not getting cancelled. This is what the chain rule looks like. So dy by dv into dv by dx. Now do I know both of them? Yes, I know what is dy by dv, what was dy by dv? Simply write two v, and dv by dx was cos x. And what was v? Let's substitute it back. v was nothing but my sine x from here, sine x. So I can write two sine x, cos x. Okay, right? So this is what is the differentiation. Ruhi, this is not, you made the same mistake. So dy by dx of sine x is not two sine x. It is two sine x cos x. If you do the product rule, you'll get the same thing. How, let's say, product rule can be written as y is equal to sine x, sine x into sine x. This is fx, let's say, and this is gx. So dy by dx, dy by dx will be equal to differentiate the first function. So sine x becomes cos x. And let the second function be as it is. Then add to it the first function and the differentiated form of the second function, which is cos x. So hence it is two sine x cos x. Is it clear to everyone? Any doubt, please ask. Yeah, are you guys thorough with all the rules now? So we discussed seven basic standard functions whose differentiation now you know, and we also now know five rules of differentiation. Okay, so hence I will give you, for the fifth rule, let me give you one question and you guys solve this. So question is problem. Problem for chain rule. So let's say I have to differentiate, sine of three x square plus four x. Find out dy by dx. dy by dv is two v, yes, two sine x. But you have to do, then dv by dx also will be, yeah, dy by d sine x is two sine x, yeah. If you meant, yeah. If you meant like that, that's correct. dy by d sine x is two sine x, that's correct. Now solve this. Find out y is sine three x square plus four x, let's say. Again, this is non-standard form. If you see, I didn't give you anything of, I have given you y is equal to sine x. But here, x itself is a function of x, right? The argument within the sine function is a function of x. It's not plain and simple x. Then how will you solve it? Use chain rule. Yes, folks, quickly. Don't waste so much of time. So I'm not saying, oh, she deleted. Cos of three x square plus four x times six x plus four, yes, correct. Six x plus four, yes. Yes. There will not be any 10x. Six x plus, six x plus four, trippin. Cos three x square plus four x into six x plus four. Okay, how? Let's say, how do we solve it? So chain rule will be applied. So clearly, if you see, I know the standard form of sine of some x or v or t, but not of three x square plus four x and all. So how do I do it? So hence, let us say, let us reduce it to standard form. Let us say that it is sine of v, where v is equal to three x square plus four x. Okay, now if you see, here is a chain rule plus the sum rule, right? So there is a sum of two functions. So hence, if you see, what is dy by dv? Is cos of v and dv by dx will be nothing but differentiate this and then add the differentiation of four x to it. So it is nothing but six x plus four, right? So dy by dx will be equal to dy by, sorry, dy by dv is dy by dv. Into dv by dx, which is equal to cos v. Now what is v? v was simply this three x square plus four x into six x plus. Is it fine guys? Yeah, so try another one. So let me give you, now the question is, y is equal to under root sine squared x plus, no, not plus minus cos square x. Do this, so you have to find out. Again, you have to do chain rule. I think you will be facing difficulty in expressing in this thing. So let me, you will not be able to type in, I believe. But you can write root of whatever, root of, or in terms of substitution also you can write. Root of v times something, something, whatever you're saying. Yeah, many love guys. Yeah, what happened? Okay, it seems people have, yeah, you can express also, no worries. Could you guys solve it? Yes or no also is good. Anyone who couldn't solve it? Interact guys, you can respond. Hello? Okay, still doing two sine x cos x upon root v. Okay, where I'm believing, I'm assuming v is sine square x minus cos square x. So, okay, two sine x. Sine two x by root v. Sine two x by root v. I think there is some, okay, let me check. Two sine x cos x, yeah, looks like. But you have to just check whether the sine negative or not is there. Sine has to be checked. Anyways, how do we do it? So, let us solve solution. So, let's say y is equal to root v. Why did I check or why did I take under root stuff as v? Because I know if y is equal to root x or x to the power one by two, I know dy by dx. Dy by dx is half x, x to the power minus half. Sorry, this is not x, this is half. Yeah, this is half. So, half x to the power minus half, one by two root x. Yeah, so if y is x to the power half, dy by dx is one by two root x. Yeah, but if y is root v, then dy by dv will be, dv will be one by two root v. Simply replace x by v, you will get this. dy by dv is two root v. Now, v is equal to sine square x minus cos square x. So again, this is the sum of two functions, cos and minus cos, minus cos square, but I don't know what is differentiation of sine square using standard form. So again, I can write v is equal to, v is equal to let's say u square minus z square. So where u is sine x and z is z is cos x. Okay, I can always do that. Now I have to find out what dv by dx. So dv by dx will be, or yeah, so first of all, let us find out dv by dx. So dv by dx is cos x and dz by dx is minus sine x. Now, I have to find out dv by dx. Why am I finding dv by dx? Because let us write the chain rule first. So dy by dx will be equal to dy by dv and it is equal to, and then multiplied by dv by dx. This is what I have to do. dy by dv into dv by dx, right? Now, dy by dv is clear, I know, but I don't know dv by dx, so let us find out dv by dx. So dv by dx, dv by dx is equal to d by dx of u square minus z square. Now, how do I find out dy dx of u square? So do I know, is it standard form? So I know d by dx of, d by dx of u square minus d by dx of z square by some rule, right? But then again, I don't know, this u square doesn't look like to be a function of, you know, direct function, because if x were u, then I could differentiate it, but here it is not x, it is not u, it is x. So I can't differentiate, like, you can't say to you. You can't say to you, why? Because for that, x must be u, right? So what do I do? So let us make it u. So if I do that, can I write this as d of u square in by du, and then here you write du by dx. Now can you see, it looks like du is getting canceled and du square and dx are there, right? You can do that. And then minus dz square by dz into dz by dx. Now I'm getting this thing, yeah? So now what is du square du is to u. And du by dx, we just found out above. du by dx is cos x. So cos x, I write minus dz square by dz is two z into dz by dx is minus sine x. So it is two u, what was u? U was, where was u? U is sine x, sine x. So I can write this as two sine x cos x minus two z. z was, z was cos x. So it is two cos x and this minus and this minus will make it plus. So two cos x sine x. So hence I get what, four sine x cos x. So what was this dv by dx? If you remember, this is dv by dx. Let me explain again because I think most of you would not have got it. So dv by dx is dv by dx of u square minus dz square or z square where u was sine x and z was cos x. Yeah, so if you remember, we are coming from here, v is equal to u square minus z square. I have to find out dv by dx, but it is a function of u and z. So how do I do it? So I represent this as dv by dx is d by dx of u square minus z square. Now u square, I don't know how to differentiate directly with x because it has to be a function of u which is not there. So I made it a function of u. So du square times, du square by du times du by dx. So hence this item is, this item is sorted out. And then d by dz of dx of z square is dz square by dz into dz by dx. So this also is sorted out. Now I know d square by du is a standard form. It will be two u and then du by dx, if u was sine x, u was sine x, so du by dx was cos x. So I simply substituted here and then minus two z, minus sine x, I kept on substituting. And then finally I got four sine x cos x. Now it's time to do, find out dy by dx or dy by dx is dy by dv, which was one by two root v, if you see. And then dv by dx is four sine x cos x. So two and this goes, so hence it is two sine x cos x upon root v, which was under root, sine, I believe sine square x under root, sine square x minus cos square x. Yes, so many of you have got it correctly. Okay, so I hope you understood it. Let me give you another problem. This is full, so let me go to the next, you go. Okay, try this. Find out y is equal to root sine x, simple one I'm giving you first. Do it. So Sommick says cos x upon two root sine x, okay? Cool. So you guys are already doing it or did I really teach so good? Anyways, so y is root sine x, so how to find out dy by dx? So those who have completed this can go on to this, solve this, y is equal to, let me give you. Yeah, try to find out sine square, sine square two x plus three. So I am now solving this one. So dy by dx is, let's say, what I have to find out, right? So let us say y is root v, where v is sine x, okay? So dy by dx is equal to dy by dv into dv by dx, right? dy by dv, what will be it? dy by dv, dy by dv will be nothing but half, times v minus half, that is nothing but one by two root v. And dv by dx is very clear, dv by dx is cos x, right? So hence this will give you one by two root v into cos x. So hence nothing but cos x upon two root sine x, is the answer. Find out the second one, y is sine square, four sine two x plus three cos two x plus three, yeah, correct. How to do it? If you see y is equal to, let us say, v square, okay? Again, this is, you know, how to substitute this? For example, there is a power here, right? So I don't know what is sine square differentiation. So I start with that substitution only. So first, if you see, if I have to, you know, drill down, this is nothing but first you see, there is an algebraic function. Algebraic function, why? Because something to the power, something, right? Algebraic function. And then followed by a trigonometric function. And then followed by an algebraic function again, right? What is that? So it is a trigonometric function of an algebraic function and, you know, then it's algebraic function of a trigonometric function of an algebraic function. So if it is like composite function, if you remember. So we said, you know, y is a function of, let's say, y is a function of v, which is an algebraic function and is given as v square, right? And then v is a function of x or not x, let's say, u. Let us say u and this is a trigonometric function, sine of u. And then finally, then finally, u is a function of, let's say, hx, which is nothing but 2x plus, 2x plus 3. So hence, this is a composite. So hence, y is f of g of h of x, g of h of x. We discussed composite functions in the class. So hence it is. So hence, what will dy by dx be? dy by dx will be dy by dv, dv by du, and finally, du by dx, chain rule, chain rule. So dy by dv is clearly 2v, so 2v, so dy by dv is 2v. Sine u is cos u and du by dx is simply 2v. So hence, and you substitute back, so it is 2v. 2v is, what was v? 2 sine 2x plus 3, that was v into cos of u. So cos of u is, cos of u was 2x plus 3, and then finally 2, so this 2 and this 2 will become, this will make 4. So hence, answer is, answer is 4 sine 2x plus 3 into cos of 2x plus 3, okay? So I hope now you guys can, I'll tell you a quicker method of doing it, yeah? So you don't need to substitute again and again. So what you can do is, let us take the same problem. Quicker method is, you can do the substitution in your head itself. So sorry, so you can do a substitution in your head and then you don't need to do this full length, this thing. So what was the question, question was, y is equal to sine square, sine square 2x plus 3. So the first thing which comes to your mind is, y is equal to square of something. So hence, in your mind, this should go something, something square, yeah? So something squares derivative is 2 times something, is it? So hence, dy by dx or let's say, I'm not writing x now, x now, I'm writing y dash, y dash is 2 times something, 2 times something, 2 times something, write like that. This should go in your mind, not here on the notebook. So 2 times something it is, okay? Now, ask this question, did you differentiate with respect to x? No, you didn't differentiate with respect to x, you differentiated with respect to something here, isn't it? Then only you got 2 times something. So that means you have to differentiate once more. So what was something here? Sine of 2x plus 3, right? So sine of 2x plus 3 was something, right? Now differentiate sine of 2x plus 3. So now consider sine of 2x plus 3 as the something and differentiated, differentiated. Now I am differentiating sine of 2x plus 3, it will be cos of something, cos of something, right? So you differentiated with respect to, it was sine of something, right? So you just redefine something initially, again, once again, I'm repeating, initially treated as something square. So differentiating with respect to something, it will be 2 times something, but did we differentiate with respect to x? No, we differentiated with respect to something. So hence something needs to be differentiated once more, this time possibly with respect to x. But let's check, what was something? Something was sine of 2x plus 3. Now I have to differentiate sine of, again, I can now redefine something else 2x plus 3. So it is, in your head you say sine of something. Sine of something. So you differentiate sine of something. So sine of something, differentiation with, maybe cos of something, but did we differentiate with respect to x? No. We differentiated with respect to something. Now what was something here? Here something was 2x plus 3, okay? Now differentiate with respect to x. Can I differentiate with respect to x? Yes, I can. So what is the differentiation? 2. So hence you get what? 2 times sine of 2x plus 3, then cos of 2x plus 3 into 2 in one step itself. Yeah, so similarly all other chain rule problems can be solved like that. For example, I gave you y is equal to root sine x, isn't it? So I can say that it is root of something. Now root of something, differentiation is one by two or root something, right? So did I differentiate with respect to something, x? No, I differentiated with respect to something. So hence I got this. Now what was something? Something was sine x, something was sine x. Can I differentiate with respect to x? Yes, I can and this is cos x. So hence answer is cos x upon 2 root sine x. This is chain rule. This is how you have to, this is how you have to think. Yeah, so obviously when you do lots of practice, then so the continuation is here. So hence I'm marking this as star mark here. If you wish I can share this notes with you so that you guys can go through. So I think we are okay with the basic differentiation. Now you need to practice a lot. Next class onwards I'll take up maximum minima and I'll also explain the concept of integration. So that in the month of November, you should be good at basic differentiation integration and maximum minima. And this will help you in your 11th grade stuff also. So you can start picking up higher level mathematics, physics also when Rohit also deals with higher level kinematics. This is going to help you there in solving problems. I'll also give you some application problems in your kinematics itself so that you are familiar with the rules of differentiation and how to apply them there. Okay, so the only, please remember you have not learned calculus right now. You have just learned a few tools of calculus. So there is much more thing in it. And there are lots of other aspects to calculus which probably will not be able to go in this standard because you need some more understanding of higher mathematics. But you are good for let's say another couple of years or using application of calculus, basically in physics and chemistry and to an extent in mathematics as well. So this should be it. And what we'll do is I'll be sharing you and probably I remember last class also I shared a few questions to solve. I hope you guys have done it next class when we meet then you please get this solution notebook with you so that I can have a look. Any queries, any question guys? So I think you are good now. So you'll be able to do most of differentiation problems. Any question please, please shoot. Otherwise we'll call it a day and then we'll meet the next class. Any questions and any other thing you need just let me know so that we can work on that. Yes, you can have an interaction or you can save if you need anything. Yeah, okay. So thanks a lot for your time and all the best from tomorrow onwards, school starts. So do solve the problems which I'll be giving to you. And on Friday, we are on Saturday, sorry. We will be having, I think you have a full day school that day, right? So we will be having a class after your school. Any other doubt guys? You can always touch base with me, okay? So thanks a lot for your time. Have a nice day. Bye-bye.