 Alright so we will start today with differentiation, when I say differentiation I would also talk about differentiability. Okay so differentiability first of all just like we talked about continuity the last class. So you are not sharing your screen? Okay one second. Can you see the screen now? So you are not sharing your screen? No sir. Yeah now I can. Okay so differentiability is basically testing whether a function has a derivative at a given point or not. So when a function is differentiable and when a function f of x has a derivative at a point So sir I am hearing two voices, I am hearing one of, there is a lag and there is your voice also, I am hearing two voices. Oh yeah last time I also had the same other thing, now is it proper? Okay fine. When a function has a derivative at x equal to a we say that the function f of x is differentiable at x equal to a. Now are you hearing two voices? Now what is the meaning of the fact that a function has a derivative at x equal to a? It means two things. One the function f of x must have a unique tangent at x equal to a and secondly the unique tangent must have a finite slope. Is that fine? Yes sir. Okay now under what circumstances a function may not have a unique tangent at a point x equal to a or it may not have a finite slope tangent at x equal to a. So let us understand when is a function not differentiable at x equal to a that would give us a clear idea as to where the function will be differentiable. Okay so let's take the scenario where f of x or conditions where f of x is not differentiable at x equal to a. The first scenario is when the function is discontinuous at some point when f of x is discontinuous at some point or at that point x equal to a it will not be differentiable at that point also. So last class we had talked about five situations where the function is discontinuous at x equal to a. So any one of those five conditions if it arises the function will be said to be not differentiable at x equal to a. Okay. Okay. Second scenario is when the function has a corner or a cusp or a key at x equal to a. So is that cusp? Yeah see USP cusp. What is corner? What is corner? Are you good? Like this. This is a corner. So at x equal to a this is a corner being formed. Okay cusp is something like this. This is a cusp. Okay this is a cusp. So here it is a corner. So here it is a cusp. And kink is something like this. You have a line and straight away there is a fall like this. So here at x equal to a there is a kink. So I don't get the difference between corner and cusp. Corner is when the two lines meet at the point they form a corner. Okay. So these are lines. These are curves. Oh. Okay. This is a line in a curve. Okay. So at all such points you would find that the function will not have a unique tangent at that point. For example if I talk about mod x function. Yes. When I have talked about mod x function, mod x function graph is a v. So it has clearly a corner at x equal to 0. X equal to 0. This is a function mod x at x equal to 0. Now if I try to make a tangent at 0 by choosing a point to the right of this function. So we all know the definition of mod x function. Mod x function is redefined as x when x is greater than equal to 0 and negative x when x is less than 0. And x is less than 0. Now if I want to find the derivative of the function by choosing a point slightly right to 0 let's say I choose a point 0 plus h. Okay. And I use the first principle that is limit h tending to 0 f of 0 plus h minus f of 0 by h. Let's see what do we get. f of 0 plus h is a positive quantity right. Yes. Now here 0 plus h means you are following the first definition that is x. So it will be 0 plus h means h only. Yeah. Okay. Right by h you get your answer as 1. Okay. So I am getting the slope of the tangent which I am obtaining by choosing a point to the right of 0 as 1. On the other hand if I find the slope of the tangent by choosing a point left to 0 what will I write? You get limit h tending to 0 minus f of 0 minus h minus f of 0 So upon h it will give you minus it will give you 1 again. By negative h. Yeah, yeah. f of 0 minus h is a negative quantity so I will follow this definition. So it will give you minus of minus minus 0 minus 0 upon minus h it will give you minus 1. Which will give you negative 1. So basically what I am trying to convince you over here if I try to make a tangent like this this tangent red one will have a slope of 1. On the other hand we have a minus 1. Right, but if I make a green tangent by choosing a point to the left of 0 then this point will have this line will have a slope of negative 1. Right. So this implies I cannot have unique tangent at that point. That means there is no unique tangent at that point. There is no unique tangent at x equal to 0. So the function is non-differentiable at x equal to 0 point. So this criteria is violated. This criteria is violated. Unique tangent criteria is violated. That means violated. And same will happen even when there is a cusp or a kink. Okay. Now even though if the function doesn't have a cusp or a corner or it is continuous at that point it may still be non-differentiable provided the slope of the tangent drawn at that point let's say this is my point x equal to a if this slope becomes infinite then also we say at this point x equal to a the function is not differentiable. Okay. Is that fine? So remember the two conditions when it has a discontinuity it will definitely be not differentiable when it is having a corner it will be not differentiable or when it has an infinite slope of the tangent at that point then also it will not be differentiable. Is that fine? Yes, sir. Now let me just give you a question. Okay, let's take this question. So we have a function which is defined as 4x square minus 4 by x minus 1 when x is not equal to 1 and it is 4 when x is equal to 1. Okay. Which of these statements are true? Which statements are true? Which statements are true? Statement number 1 is limit of f of x as x into 1 exists. Statement 2 f is continuous at x equal to 1 Statement 3 f is differentiable at x equal to 1. Now the options given to you are none of the statements are correct. Option B Option B 1 only is correct. Option C 1 and 2 both are correct. Option D 1 and 3 are correct and option E 1, 2 and 3 all are correct. I think it's none of the above. None of the above. Okay, let's verify this. First is does f of x exist? What is the limit of f of x as x tends to 1 minus 0 by 0 4 Okay, that's fine. So it becomes 4 x minus 1 by x plus 1 by x minus 1 as x is very close to 0 you can just cancel this off. When you put 1 you get 4 into 1 which is 8. Isn't it Isn't it more to be minus x minus 1 in the denominator? Minus x minus 1 is? When you when you redefine the function you will get Why does the function need to be redefined? Is there any mod over here? You don't. Other than generally you generally have to express it in terms of x greater than 1 and x less than 1 and x equal to 1 that makes it easier. Who told you this rule? Sir, I thought. We don't redefine the function. It is already in the redefined shape. I see. Now what is the limit of the function as x tends to 1 plus Same thing I will say. Limit exists. Again limit exists. So this statement is correct? So our statement is correct. But it's not continuous at x equal to 1. It is not continuous because the value at 1 doesn't match with 8. So hence only 1 is correct. Only 1 is correct. Is the idea clear? Now we will take another one. So there is a function which is having a graph like this. The graph is like this. From minus 5 it cuts the x axis goes up. So these all are straight lines. So there are straight lines. There are no curves. Yeah. This is minus 6. This is minus 3. This is minus 1. 0. This is 1. This is 2. This is 5. This is 7. Okay. Yeah. So the curve has been sketched from minus 6 to 7. Okay. Now answer the following question. The question is in the interval 1, 2, 2 f of x is give the equation of f of x. Equation. Yeah. By the way let me give you some values on the y coordinate as well. I have not given you. This is 3. Okay. This is again 3. This is 4. This is 4. Okay. So basically I am trying to find out the equation of the line this line. I am showing it with white color. So you can do point slope form. This line. Of course I can do a point slope form. Yeah. Or you can easily guess from here that it is basically a line which is having a slope of negative 1. Correct. Yeah. Passing through. Origin, correct? Yeah. Or is it passing through origin? This line, this small part is not the whole line is passing through origin. Yeah it is. No, this point, this point over here you can just use your 2 point form. This point here is 1, 3. This point here is 2, 4. Okay. So tell me an equation of a line passing through 1, 3 and 2, 4. You mean to y minus y1 equal to y2 minus y1 upon x2 minus x1 into x minus x1. Right. Y minus y1 is equal to y2 minus y1 by x2 minus x1 times x minus x1. Correct. So this is going to be 1 itself. So this will be y minus 3 is equal to x minus 1. So x minus 1. Just give me a way. Okay. 1, 3 is equal to 2, 3. y2 minus y1 by x2 minus x1. What slope is positive? This is minus 3 actually. So this is minus 3. So it will be Yeah, yeah. So this will be plus, this will be minus 4. Again, this is 2, minus 4. So there is a mistake in writing the points, coordinates over here. So this is 2, minus 4 and 1, minus 4. 2, minus 4 and this is 1, minus 3. Okay. Yeah, yeah. So if you give slope as minus 1. Exactly. So the slope here would be negative 1. So we will have y plus x equal to 2. y plus 3 is negative x minus 1. So y plus 3 is equal to negative x plus 1. So y is equal to negative x minus 4, correct? Yeah, yeah. So plus 4. Sorry, negative x minus 2. This will be negative x minus... Yeah, correct. So this will be the equation of the line connecting that point. Next question is over which of the following interval f dash x equals 0? So f dash x equals 0 in which interval? Or you can say in the intervals. f dash x equal to 0 means the line is parallel to the x axis. Parallel to the x axis. It means minus 3, minus 1 and 2, 5. Alright. So minus 3 to minus 1 and 2 to 5. Okay. Next question is Yes. How many points of discontinuity how many points of discontinuity does f dash x have? How many points of discontinuity does f dash x have? Sir, is it 0? No, no, no. The question is f dash x have. This is the graph of f of x. Read the question properly. How many points of discontinuity? Ask you, plot the graph of f dash x. How would the graph look like? Let's plot the graph of f dash x. So from the interval minus 6 to minus 3 if you look at the function from minus 6 to minus 3 it is a line. What is the slope of that line? Yeah, minus 2 of the 1 2. It is having a rise of 3 and it is having a run of 2, correct? Yeah, 3 by 2. So the answer is 3 by 2. So while I am plotting the graph of this function from minus 6 to let's say minus 3 I would plot it as just 3 by 2 line like this. Yeah. Because I have to plot the slope now. Okay. What will happen from minus 3 to minus 1? It will fall down to the x axis. So it will fall down like this? Yeah. Minus 3 to minus 1. Okay. Yeah. Exactly at this point the function you will see there is a break in the function. It's a break yeah. Correct? I cannot find the derivative at these points. Okay. What about from minus 1 to 0? Minus 1 to 0. It's 3 by It's having a slope of negative 3. Right. It has a slope of negative 3. So it will be like this, correct? Yeah. And this trend continues plus 1 also right? Yes. Because the line here continues to go like this straight till minus 1. So instead of stopping here I have to stop at 1. So till 2 right? No. Till here the line is different. Okay. But here the slope this slope is actually minus 3 but this slope is actually minus of 1. So it will go Yes. It will go at minus 1 position. Okay. Till it reaches 2. So this will be the Again from 2 to 5 2 to 5 again it will be on the x axis. Right. 2 to 5 it will be again on the x axis. 2 to 5 the line will be like this. 5 to 7 Okay. Okay. And from 5 to 7 it will have a slope of I don't know I have not written the points. Okay. I'll show you. Yeah minus 4. So rise is 4, run is 2. So it will have a slope of plus 2, correct? Yes. So plus 2 means it will go up. So let's say this is your point this is your 3 point right? Somewhere over here. This was 3 by 2. So it will be somewhere over here. 3 to 7. Okay. Now tell me how many points of discontinuities are there? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12. 12 points of discontinuities. Yeah. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12. 12 points of discontinuities. Yeah. 1, 2, 3, 4, 5, 6, 7, 8. Yeah. Yes sir. Yeah. Yes sir. Okay. So 1 is at top of minus 6. That will be 1 point. Okay. Once again. Here if you have to check the discontinuity, what do you do first? You see the value. Check whether there is a... Remember at corner points what did I tell you? Corner points you said that. The function is starting over here. Okay. So if I take the value of the function at minus 6 and compare it with the limit of the function to the right of minus 6 point. Are they equal? So here it is continuous. It is, no? Yeah. This value is positive. This value is whatever is 3 by 2. The same value is here also. Discontinuity will happen at this position. Discontinuity will happen at this position. Discontinuity will happen at this position. Discontinuity will happen at this position. Discontinuity will happen at this position. Discontinuity will happen at this position and discontinuity will happen at this position. Corner points as of now we cannot tell because you just have to see one important thing which I wanted to tell. If you want to check the differentiability at corner points let's say at minus 6 then we just check whether the derivative at minus 6 plus it should exist and should be finite. That's it. Okay. If you want to check at 7 then derivative at 7 minus must exist and must be finite. Must be finite. Right? Yeah. So here the function is exhibiting 5 points of discontinuity. So those two are taken as 1 point. Okay. Which two are taken as 1 point? I mean the ones which you have circled have taken as 1 point of discontinuity. This itself is a 1 point. X equal to 5 is 1 point only. X equal to 2 is 1 point. X equal to 1 is 1 point. X equal to minus 1 is another point. X equal to minus 3. These are the points, right? Yes. We don't count it as 2 points. There is one X value where it is discontinuous. The next part of the question is from minus 6 to 3 minus 3 what is the value of f dash x? From minus 6 to 3 what is 3 by 2? Minus 6 to minus 3 are 3 by 2. Is that fine? 3 by 2? Absolutely correct? Yeah. Okay. Next question. Which of the following statements about f dash x is false? Statement A. It consists of 6 horizontal segments. It consists of 6 horizontal segments. Okay. Option B. It has 4 jump discontinuities. 4 jump discontinuities. Option C. F dash x is discontinuous. F dash x is discontinuous. In the sets minus 3 minus 1 1, 2 and 5 Option D F dash x ranges from minus 3 to 2 and option E is in the interval minus 1 to 1 F dash x is negative 3. Which of the statement is false? You can refer to this graph also. Sir, I think it's B. It's B? Yeah. Because I have 5 jump discontinuities. Absolutely correct. So this statement is false so B becomes the right option. Is that clear now how to deal with these kind of problems? Yes, sir. Okay. Now, some properties with respect to differentiability. Properties of differentiability. First property is if your f of x function is differentiable at x equal to a then any constant multiplied to the function will also be differentiable at x equal to a. Okay. Second property is if f of x and g of x are differentiable at x equal to a then the sum the difference the product. They will also be differentiable at x equal to a. Okay. Even the ratio is also but provided g of a should not be 0. Sir. Can you put this session on YouTube? Yes, yes. Okay. Okay. Why what happened that makes you conscious. So little so happens. Don't worry. Fine. Yeah. And one of the most important property is if f of x is discontinuous at x equal to a then f of x is definitely not differentiable at x equal to a. Correct. Okay. But if f of x is continuous at x equal to a then it may or may not be differentiable at x equal to a. Are you getting this point? Yes, sir. We will now start talking about differentiation per se. I think in the class we had already talked about differentiation of implicit function differentiation using logarithmic function, correct? Yes. So I will quickly just take few examples and move on because I don't want to waste our time doing the same stuff. Hope you are aware of implicit differentiation. Yeah. Can I give you a problem on that? Okay. So let's take this question. So we have a function we have an equation like this which is an implicit one. Okay. Yes, sir. Find dy by dx. So do you remember the formula which we discussed the other day? Do I? Do you remember the formula that we discussed the other day? Make this as f of x, y equal to zero. Then dy by dx is directly given as minus dou f by dou x divided by dou f by dou y. Yeah. So you can directly use this formula here. So bring this minus one on this side even though it is not required but let's say I bring the minus one. Had there been some other function in x and y you should always bring it to one side. This is what we call as f. Correct? Yeah. So this is negative of partial derivative of this with respect to x. So what is the partial derivative of this with respect to x? It will be 6x. It will be minus 2y. And the rest of the thing will be zero. Yeah. Correct? Yeah. Divided by dou f by dou y. So what is the partial derivative with respect to y? This will be zero. Okay. Yeah. So if I drop the factor of 2 what will I get? Negative. 3x. Negative 3x. Negative y. So I can write it as 3x minus y by x minus 5y. Yeah. Yeah. And this is your implicit differentiation. So this is actually a part of your CBC curriculum as well. That's why I'm not spending a lot of time on this. Okay. So I'll move directly to parametric differentiation. I think this is also something which we had already done in the class. Yeah. So let's take an example on this as well. Let's say I have a question x is e to the power theta cos theta y is e to the power theta sin theta. Find d2y by dx square at theta is equal to pi by 2. d2y use logarithmic function also in this. Is it required? You may use I mean I'm not saying you cannot but is it required here? Because let's say if I have to find dy by dx I can use the formula dy by d theta divided by dx by d theta. Yeah. To differentiate this I don't need log actually because I can directly use product rule. So e to the power theta derivative of sin theta will be cos theta plus again e to the power theta sin theta. Minus. What about the denominator? Divided by e to the power theta sin theta minus e to the power theta cos theta. Correct. Now if you want you can get rid of theta from everywhere so it becomes cos theta plus sin theta by cos theta minus sin theta. Okay. And this term if you divide the numerator and denominator by cos of theta what do you get? 1 plus tan theta upon 1 minus tan theta. Excellent. Okay. It's tan of 5 by 4 plus theta. 5 by 4 plus theta. Now having got dy by dx as tan of 5 by 4 plus theta what would be your d2y by dx square term? D2y by dx what? Derivative of dy by dx correct? 6 square theta. What will I write? 6 square theta. 6 square theta or 6 square pi by 4 plus theta. That's it. Theta by d theta by dx Absolutely. Please don't forget here your function is in terms of theta but you are differentiating with respect to x so you have to use the chain rule over it. Yeah. Now so secant square pi by 4 plus theta and d theta by dx would be reciprocal of dx by d theta so reciprocal of this term so that will be 1 by e to the power theta 1 by e to the power theta cos theta minus theta. Now you have to evaluate this when theta is pi by 2. So secant square here if you put pi by 2 you get 3 pi by 4 into 1 by e to the power pi by 2 this will become 0 minus 1 correct? Yeah. Pi by 2. What is secant square 3 pi by 4? 6 square 5 by 4 6 square pi by 4 2 divided by minus e to the power pi by 2 that will become your answer or negative 2 by e to the power pi by 2 or you can say negative 2 e to the power minus 1 is that fine? Yes, sir. Okay. So next concept that I am going to talk about is slightly new to you that is the concept of derivative of inverse of a function. Derivative of inverse of a function. I think we had already discussed the concept of inverse of a function in our function chapter. Yeah. So if f of x is the function its inverse is represented by f of x minus 1. Yeah. Now we also know from the graph that these two functions are actually mirror images of each other about y equal to x. Y equal to x line, yeah. They are mirror images of each other about y equal to x line. Okay. So I will give you a scenario. Let me see how do you analyze that. So let's say I have given you a sketch of a function something like this. Okay. And let's say this point here is 3, 8. Let me call this point as point A. Okay. So point A is 3, 8. Correct. Yeah. Let's say at this point I sketch a tangent. Okay. And I say that the slope of this tangent is 5. Yes. Yes sir. Okay. The function is y equal to f of x. I want to know from you what is f dash 8. Sorry. What is f f inverse dash 8. Let me write in a proper way. So what did be 5? f inverse dash. So what did be 5? Are you sure? Yeah. Sir, it's wrong. It's over half a minute. Sir, it will be the slope of the line which is reflected about the yx above 1 equal to x line. Okay. First question. How would be the graph of f inverse x? Okay. This is, this line is your y equal to x line. Okay. So it will be inverted about that. Right. Yeah. So if I am not wrong will it look like this or not? Correct. Yes. If this point A is 3, 8 what is the mirror image of this point let's say about the y equal to x line. What will be this point B? Can you tell me that Shreya? Yeah. Y equal to x line. It's 90. If 3, 8 is reflected about y equal to x then what is the coordinate of B point? So B is the reflection of A about y equal to x. Won't it be 8, 3? See x equal to y means x and y are interchanging their positions, isn't it? Whatever is on the x axis will go on the y. Whatever is on the y will go on the x. So basically you can treat this as a mirror which is reflecting the points from y axis to x axis and x axis to y axis. Correct? Yes. Now when I ask you the question find the derivative of f inverse at 8. What did I mean by that? You have to find the slope of the element I want to find the slope of the tangent at this point. Correct? Yes or no? Yes sir. Now how is the slope related? Let me just give you a quick analysis over here. Let's say I want to treat the slope of this as the ratio of these two quantities. Okay? So if dy by dx is your 5 let's say dy by dx is 5 that is rise by run is your 5 Correct? Yes. In this case what will happen? In this case you would realize the same dy will now come over here and the same dx will now come over here. So for this it will be the dx by dy ratio. So if dy by dx is 5 then what will be dx by dy? 1 by 5. Absolutely correct. So the slope here will be 1 by 5. Are you getting this point? Yes sir. This I tried to show you geometrically. Okay? Yeah. Mathematically if I have to explain this then I would use a very important theorem which is this f of f inverse x that's always x. But only if it's 0. And same is true for this. That means it is a function which actually returns the same value as you input to it. Yes sir. Is that fine? These two properties will be very helpful even when we are dealing with questions involving inverse of functions. Is that fine? Yes sir. Now see. Let's say I take the second equation for the previous question itself I was given that f dash 3 was 5 and we wanted and we wanted f inverse dash at 8 and it was given that the curve was passing through 3, 8 wasn't this the information given to us in the previous question? Yes sir. So slope of the function at 3 was given as 5 the value of the function at 3 was given as 8 and we were asked to find out the slope of the tangent drawn to the inverse of the function at 8 point. Yes or no? Yes sir. The expression f inverse f of x is equal to x. Let's differentiate throughout with respect to x. Let's differentiate with respect to x. What do we get? We get f dash of inverse f of x into f dash x is equal to 1. Yes. Just try to recall this. You would be knowing this. If you want to find the derivative of this with respect to x. I told you in the class also it's f dash g of x into g dash x. That's what we call as the chain rule right? Yes sir. This itself is your chain rule. Now here I need f inverse dash at 8. Can I write it like this? Okay. I brought this term to the denominator. Yes. If you want this term to be 8 what should be your x value? Look here. Absolutely. If you want f inverse dash at 8 you have to put x value as 3. 5. 5. So your answer is going to be 1 by 5. Is that correct? Okay. Can you conclude that the inverse of a function will have the reciprocal of the slope? The derivative of the inverse of a function will be inverse of the slope. Is that right? See derivative at which point that is also important right? See here you are finding the slope at 3 to be 5. At 3 the slope is 5 but here the slope at 8 is 1 by 5. Right. It's because 8 is the very point that you get when your x is 3. So basically the curve was passing through 3, 8. So yes if these two reverse that means f dash f dash at 3 would be reciprocal of the inverse dash at 8. This you can say if the curve is passing through 3, 8. Okay. So if you want to generalize you can generalize it like this. Let's say a, b lies on f of x. Lies on f of x. Okay. Then I can say that f inverse dash calculated at b will be 1 by f of f dash calculated at a. Are you getting this one? Yes sir. This you can generalize. Okay. Let's take another question based on this. If your function f of x is defined as x2 plus x minus 2 and let g of x be the inverse of f of x. Okay. Evaluate g dash 0. Evaluate g dash 0. Of course yes. Let's say it is a non calculator based question. See here the idea is if you want to find g dash 0 it will be reciprocal of f dash that value of x which will give you the value of this as 0. So basically tell me what should be your a says that this becomes 0. Can you guess that the value will actually be 1? Because if you apply your factor theorem you know that the remainder theorem you know that the 1 plus satisfying this equation. Okay. So your answer will be nothing but 1 by f dash 1. So if this is your f of x what will be your f dash 1? f dash 1 is f dash 1 will be 3x square plus 1 at 1. 3x square. 4 right? So your answer is going to be 4. So this is going to be your answer. Understood how it works? Yes. Let me see if I can give you another one. Okay let's take this question. If f of x is given by 2x cube minus 3x and x is the inverse of f of x then h dash minus 1 will be equal to 2x cube. Good carefully. Yeah. We cannot get 2 answers. So what do you do first? I found the derivative of f of x. It has to be 1 by derivative of f at such a value a where this gives you the value as negative 1. Yeah. So 2a cube minus 3a plus 1 equal to 0 again I can see a equal to 1 is a possible value. Yes sir. f dash 1. So what is f dash 1 then? f dash 1 will be nothing but 6x square minus 3 at x equal to 1 will be 3. So your answer will be what? 3. 1 upon 3. 1 by 3. Is that fine? Yes sir. Now couple of things that differentiation actually helps us in is first it also helps in evaluating limits in many of the cases by the use of a very important rule called the Lopital rule. Lopital rule. Okay so are you aware of this rule? Yeah. So can we do some questions where we can evaluate the limits by using Lopital rule? To differentiate and then apply the limits that you eliminate the 0 by 0 form. Okay so I will just review it for you. Lopital rule is applied to problems of this nature where the indeterminate form involved is 0 by 0 or infinity by infinity. And how do we evaluate the limits in such cases? We actually differentiate the numerator function and denominator function separately. Put this value a to see if it gives me a finite value. Correct? Yeah. If indeterminacy persists the process is repeated. What process? The process of differentiating the numerator and denominator and putting the value of X equal to it. So that process will repeat. Okay. So I will give you a simple example to understand this. Let's say I want to evaluate limit extending to negative 2 not negative 2 let's say I take infinity X cube minus 4 X square plus 7 divided by 3 minus this. Right? I think the answer for this is already known to you by the fact that it's since the degree of the numerator and denominator is same it's just 1 by minus 2 right? Yes. Let's say I want to solve this by using Lopital. First I would ask myself can I apply Lopital over here that means is it 0 by 0 or infinity by infinity form? Yes. The answer is yes. So what we do is we differentiate the numerator we write 3 X square minus 8 X divided by minus 6 minus 6 X square again it's infinity by infinity so we will continue the process 6 X minus 8 and then again since it is infinity by infinity we will continue the process and ultimately we get the answer as minus 1 by 2. Now here the word applied to is important because unless until it is in these two forms you cannot apply it but nevertheless you can actually use to solve any indeterminate form question for example let's say I want to solve this question limit X to the power 1 by X extending to infinity which indeterminate form is exhibited over here can you tell me? Infinity raised to 0. Absolutely this is infinity raised to the power 0 form I can still use Lopital to solve this question. Let's say I call this as L take log on both the sides the base of E so can I say taking log on this will become 1 by X ln X 1 by X ln X so it is nothing but it is nothing but ln of X by X so do you see now it is of infinity by infinity form so now again you differentiate now we can apply Lopital on this so if you differentiate your numerator what do you get? 1 by X if you differentiate your denominator you get 1 and add in 3t this value is going to be 0 so it is a finite value so I will stop over here so ln of L is 0 so what is your L? again you differentiate L is 1 oh different we will stop over here okay understood? yes sir I will give you one more yeah what is limit of X sin 1 by X as X tends to infinity X tends to infinity X in determinant form why not? because it will be X into sin 0 which is 0 so 0 into infinity is an indeterminate form this is another indeterminate form so can I apply Lopital in this present form of the problem? no sir it doesn't work so first of all what do I do? first of all it can be your product rule hey there is no product rule first of all I have to bring it to 0 by 0 form or infinity by infinity form there is no product rule applied and all in Lopital the differentiation of numerator and denominator separately please remember when you are doing F dash A by G dash A you are not applying any kind of a quotient rule you are differentiating it separately and this you are differentiating it separately but you cannot start the process of differentiation unless until you have brought it to 0 by 0 form or infinity by infinity form so how did you know you have to make it like that? what else option do you have? you could have made X divided by cosec of 1 by X that is fine but cosec will involve a bit of lengthy differentiation process this was the most simplest I could think of so the idea is you have to manipulate with your function and bring it to 0 by 0 or infinity by infinity form even before you start differentiating the numerator and denominator yes sir not tell you what is the derivative of sine of 1 by X with respect to X cos cos 1 by X into 1 by 2 root X 1 by 2 root X minus 1 by 2 root X where root 2 X comes and then denominator will be minus 1 by X correct when I get cancelled so X divided by 8 will be this term and this term gets cancelled and this will actually lead the answer to 1 because this quantity is very very small but close to 0 and cos 0 is going to be 1 okay in reality if I just show you the graph of X sine 1 by X you will understand how it actually works the graph of X sine 1 by X you see a graph like this it comes from 1 comes down over here oscillates very vigorously and again even dies at 1 so this line is actually Y equal to 1 line so Y equal to 1 is asymptotic to this it is helpful to know this graph because it is very widely seen in AP calculus is the idea clear okay so let's take another one evaluate plus H minus 1 by H H tending to 0 which form of indeterminancy is here 10 which 0 by 0 can I apply Lofikar directly on this no I can't why can't I I just said no but I can't hear you my question is we can't can't we apply Lofikar on this because it's 0 by 0 okay so what will be the derivative of the numerator a separate differentiation we have to do so it's 1 by E plus H into yeah 1 by E plus H derivative of E plus H E is constant yeah so it's 1 what's the derivative of minus of 1 0 what's the derivative of H with respect to H 1 so now when you put H as 0 in this it is 1 by E is it a finite answer so this becomes my answer oh right I started using those standard limits no we are applying Lofikar over here don't worry evaluate the limit 1 plus tan 5 by 4 minus x raised to the power of c to x where x is sending to 5 by 4 yeah so you use logs right right so let's say this is L so you first make it as H tending to 0 you take minus 5 by 4 equal to H first want to do that I have taken logs first okay then what do I do next so I can write this as limit extending to 5 by 4 ln of 1 plus tan 5 by 4 minus x divided by cos of 2x right now which form of indeterminancy is here 0 by 0 0 by 0 0 by 0 can I apply Lofikar over here I thought shouldn't equal to H shouldn't you do that because it's not ending to 0 doesn't matter can do that you want to do this x is 5 by 4 plus H and then you make H as 0 you can do it because it doesn't make a difference to your answers so let's say okay let me do that first so this will become H tending to 0 ln of 1 plus this will become minus H so I'll write it down as negative tan yeah negative tan raised to cos divided by this will become cos of 2x so cos of 5 by 2 plus 2H what is that negative sin never mind do it no worries now once it has been reduced to 0 by 0 form can we apply LH rule yes we can 1 upon tell me what is the derivative of ln of 1 minus tan H 1 upon 1 minus tan H into minus x square H what happens what happens to the derivative of minus sin to H it becomes cos minus cos to H the negative gets cancelled into 2 right yeah yeah into 2 yeah yeah sir now when you put H as 0 it becomes 1 minus 0 this will become 1 by 2 so what's your answer half but this is ln of the limit so what is limit itself 2 e to the power e to the power half ln L is half ln is e to the power half yeah yes sir okay now we will quickly talk about the mean value theorems under mean value theorems we are going to study two theorems one is called the rose theorem and other is called the langrange's mean value theorem sometimes this is only referred to as mean value theorem they may not use the word langrange's okay I have seen in your ncrt syllabus also curriculum they avoid using the word langrange's they just refer this as mean value theorem okay so both of these theorems essentially talk of the same thing and let me start with rose theorem rose theorem says if there is a function which is continuous in close interval a to b and it is differentiable in open interval a to b and if f of a is equal to f of b then there exists let me write k over a then there exists at least once c belonging to the open interval a to b where the derivative of the function becomes 0 so this whole theorem is actually known by the name of rose theorem which was given by michael roll in 16th century now what does this theorem actually mean geometrically is something which is very obvious so mr roll meant that if you have a function like this i am just taking a simple example if you have a function which is continuous from a to b let's say this is your a let's say this is your b and let's say it doesn't have any kind of corner cusp that means it is differentiable also in the open interval a to b and f of a and f of b have the same values so let's say this is your f of a and it is same as f of b then it says that there will exist at least one point c where you would realize that the slope of the tangent becomes 0 at this point this is having a slope of 0 so f dash c becomes 0 that means it will become parallel to the x axis which is 0 now why at least one c because you can have a situation like this also let's say this is a okay there are like 3 points yeah so here we can have 3 points one is this point let's say alpha let's say beta let's say gamma where the slope of the tangent will start becoming 0 so it's very obvious thing that he stated okay let me take a question on this demonstrate rho's theorem for the function x sin x on the interval 0 to pi the graph is just showing that it's x equal to 0 without drawing the graph also you can do there is nothing like you have to always draw the graph but you have to use a calculator for this definitely how to solve equations on calculators probably know how to use a calculator for graphs not solving equations no no sir I just try the graphs so please try that out also today because many a times you would need to use your calculator to solve equations which you cannot directly solve now here if I have to demonstrate rho's theorem the first thing that I need to show that f of x which is your x sin x is continuous function in 0 to pi how how it is continuous in 0 to pi it's very simple because you know the property of continuity that if there are two functions if x is continuous x is continuous and the product is also continuous absolutely correct f of x is also differentiable in the open interval 0 to pi 0 to pi and again why is that so because x is differentiable from 0 to pi sin x is differentiable from 0 to pi perfect now is f of 0 equal to f of pi yeah yes because 0 will make it 0 and pi will make sin pi 0 both are same both are equal to 0 now rho's theorem says that the derivative of the function at c will vanish for at least at least one c belonging to the interval 0 to pi that's what it's trying to say so we have to demonstrate that this actually happens this is the step that we need to show right so we need to show this or we need to demonstrate this okay so first of all what is the derivative of the function x it's x cos x plus sin x correct and let's say I am finding at c so that means I am putting x as c so it is c cos c plus sin c correct and if I am putting it 0 I need to know my c first okay now there are various ways to solve it one is in your calculator other is if you see this something like this c is equal to minus just minus tan c correct I am sorry that means I am trying to see where does minus of tan x meet the line y equal to x correct so let me take you to g o g bra y is equal to negative tan x and y is equal to x correct so you would see here that in the interval 0 to 5 this is the point where they meet which point this is the point a a is 2.03 radians it will always give you radians okay yeah so I can say that the value of c over here is going to be 2.03 radians so c here comes out to be 2.03 radians which definitely belong to the interval 0 to pi and hence a rose theorem is verified or demonstrated please don't write rose theorem is proved because rose theorem is already proven you are just verifying it is that clear how it works yes sir okay let me pick up another example in fact in your Aadi Sharma when you open you will find a lot of questions on rose theorem okay so without wasting much time can I go to lang ranges mean value theorem okay so lang ranges mean value theorem which basically a generalization of what rose theorem actually meant okay so lang ranges said that if your function is continuous in or on the closed interval a to b and if your function is differentiable in open interval a to b then there exists at least one c belonging to the open interval a to b where the derivative of the function at c is nothing but the slope of the secant line connecting a and b the slope of the secant line connecting the points a and b let me show you from a graph so lang ranges was very smart person so what he did he actually took the same concept as what rose said okay let's say this is your function which is continuous in the closed interval a to b differentiable in the open interval a to b then he said that there will exist at least one c let's say this is that point c where the slope of the tangent will become same as the slope of the line connecting let's say this point a with b so these two will actually become parallel in fact in this given diagram you can see there would be two such points one would be like this as well yeah three like that so you can have let's say alpha you can have beta whose slope will be parallel to the secant line connecting a and b okay it's like saying that when you are travelling on a car the average speed of your car will match with the instantaneous speed at some value of your time in the journey okay so this is nothing but your average velocity this is nothing but your instantaneous velocity is that fine just one moment I will get a problem on this so we will take a question verify row's theorem for x-a to the power n x-b to the power of n so is this the continuous function let's say I give you the interval from a to b is it continuous is it differentiable yes so what's the derivative of this function true x-a raised to m into nx-b raised to n-1 into d by dx of x-b which is 1 into 1 plus x-a raised to m and x-b okay yes so what is f of b-f of a by b-a can I say it will be 0 only so that means c-a to the power m-1 c-b to the power n-1 m c-b plus n c-a will be equal to 0 yes now assuming that these two assuming that c is not equal to a and c is not equal to b this will be equal to 0 then what I'm trying to say I'm trying to say that c lies between open interval a to b c cannot be a or c cannot be b open interval it cannot take the values at the extreme positions so this becomes m plus n c is equal to m-b plus n-a so c is nothing but m-b plus n-a by m plus n I should have mentioned over here m and n are natural numbers do you see this is actually like a section formula yes so your point c has to be somewhere between a and b hence mean value theorem is very nice do this verify verify mean value theorem for this function on the interval 0 to pi so Tuesday I will be sending you a test of onr yeah everything will be included so is it a continuous function in 0 to pi because it made up of some of two continuous functions right is the differentiable function also because it is made up of some of two differentiable functions so all I need to show that there exists a c which will satisfy this criteria by pi minus f of 0 by pi minus 0 where your c should belong to 0 to pi open interval what will be f dash c cost to x is to cost to x correct cost to x f of pi would be what 0 f of 0 will also be 0 okay so this will be 0 itself so that means cost c plus cost to c is 0 if you use your transformation formula you can say it is to cost 3c by 2 into cost c by 2 is equal to 0 so this implies 3c by 2 could be values like pi by 2 okay 3 pi by 2 pi pi by 2 etc and even c by 2 could be values like pi by 2, 3 pi by 2 pi pi by 2 etc from this top one I can get a value of 3c by 2 let's say equal to pi by 2 then c is equal to pi by 3 rest other values will either be pi or exceed the interval and here the minimum value is pi so we got this value as our one of the values which belongs to the interval 0 to pi which satisfies our main value theorem is it clear how these main value yes sir okay great so I'll give you some questions in general there is a parabola y is equal to x square minus a tex plus 10 okay x value isn't equal to 4 question is find g inverse sorry g dash c where g is the inverse of this function this is first question second question is f of u is sin u and u is x square minus 9 find f o g dash at 3 first one should give you the options for the first question should I give you the options okay minus half minus one sixth one sixth half 11 by 2 okay so g dash of 3 will be 1 by f dash of a where this a is such a value which will actually give you this as 3 correct okay so here a square minus a tex plus 7 is equal to 0 this is factorizable as this yeah so which one will you choose will you choose 1 or 7 you choose 1 because x is less than absolutely correct since x is less than 4 will only choose 1 will not choose 7 correct so my answer will be 1 by f dash 1 right so what is f dash 1 f dash 1 will be dy by dx at 1 that is going to be 2x minus 8 at 1 that is going to be minus 6 so your answer is minus 1 by 6 what did you say minus half okay next one second one 0 okay how do I get this first sign you so basically one second I think I forgot to tell what is your g in this case correct g is actually your u in this case give me one moment this is your g actually so what is f of g so what is the derivative of this minus 9 into what should I do that cos of x cos of x is 9 into 32x correct yeah so what is your answer then cos 0 into 6 which is 0 which is 6 okay yes now the last part of this chapter is something which is very easy this is called estimating a derivative at a point estimating a derivative numerically now you would have learnt about the first principles right when you find the derivative of a function at a what do we do we take the limiting case of f of a plus h minus f of a by h correct okay now there is something called symmetric difference quotient symmetric difference quotient which actually helps us to evaluate f dash a in a better way like this f dash a is given as limit h tending to 0 or you can say it is approximately equal to let me remove this approximately equal to f of a plus h minus f of a minus h by 2h okay indirectly it's nothing but it's an average of it's an average of the progressive derivative expression how do you do that how do you convert how do you make it f of a plus h minus f of a minus h by 2h see what happens when we are finding the derivative at a let's say okay what do we normally do we normally take a point slightly right to the function let's say a plus h correct yeah okay and let's say this point is a comma f of a then we draw a secant line like this yeah correct we find the slope of the secant line as y2 minus y1 by x2 minus x1 correct no yeah now what instead of doing this if I take a point to the left of a let's say a minus h then this point will be a minus h comma f of a minus h correct and if I take the limiting case of this secant line let me show you in white color then what do I write I will write f of a plus h minus f of a minus h that is y2 minus y1 by x2 minus x1 what is x2 minus x1 2h this is what I have written over here okay because many times question comes to evaluate the approximate value of derivative at a given point let me take an example to show you this let's say this is a differentiable function it follows the following values so at 1.0 its value is 8 find f dash 1.5 find f dash 1.5 so what I will do is this value is approximately equal to f of 1.6 minus f of 1.4 divided by 1.6 minus 1.4 so I am using this formula f dash a is approximately f of a plus h minus f of a minus h by a plus h minus a minus h here what is my a 1.5 here what is my h h is 0.1 okay so from this I get so this value can easily evaluate from the table given to you minus 14 by 0.2 that is 8 by 0.2 which is 4.5 okay this is how we get the estimate value of the derivative at a point by using symmetric difference quotient so here we will end up the session right now okay and I will share with you the assignment