DIFFERENTIABILITY I

Added: 2026-05-17

[00:00:03]of our mind.

[00:00:06]So the independent is also >> All right, guys. Um it's again and I want to use this video to quickly break down differentiability in its simplest form. Like this video is meant to make this topic nothing to you.

[00:00:43]Do you understand? because after yesterday's class, majority of you guys started complaining that um the concept is quite confusing and I guess it's because I was blending the idea of derivative and differentiability together which looks quite confusing to some of you guys. Now here's the thing. In part one, you were taught differentiation, total differentiation.

[00:01:12]I mean you have y to be equal to f ofx in part one. And then you were told that derivative of this guy is any frime of x such that this guy is limit as a particular h t a small value that is approaching zero of um the limit of sorry of the function of um x + h minus a function of x all over h. Yeah, this was what you guys were taught in part one and they called this first principle.

[00:01:48]Now the idea of differentiability is not farfetched from this idea and we just have to build from here to make it much more sensible to you. Now this is first principle.

[00:02:03]Do you understand?

[00:02:05]Now the idea here is that it is actually possible for a function to be differentiable at a single point.

[00:02:13]Do you understand? So let's call the point point x not. Now what happens if x is equal to x not that is when do you see a function is differentiable at a point x not like this. Now if function f is differentiable at point x not if two things hold which is the left hand derivative and the right hand derivative being or existing and both being equal. However, here's how you define differentiability at a point x not. So you define it on x note like this frime at x not which is equal to the limit as h ts to z of f at x + h minus f at x all over h. You know we discussed all of this in class. Now this is how you define the derivative at point x not. Do you understand? Now you should note this which I observe most of you didn't really note and that point is that if frime at x not exist if frime at x not exist then keep this in mind.

[00:03:38]Then we say then we say that the function then we say that the function is differentiable.

[00:03:51]Please follow through. We say that the function is differentiable. I repeat if this particular condition or this definition exists then we say that the function is differentiable at x not or another way to write it or put it is that f has f has a derivative at x not. f has a derivative at x not. So most of you guys just like to accept this idea and then you find this big to use that why we say is differentiable why why why are we not saying it has a derivative at x well the moment this exist then you see that the guy is differentiable at that point do you understand it's differentiable at that point now let's also discuss the two conditions I talked about for for you to consider or observe before you conclude that a function f of x or f at x not h or f of x generally has a derivative at x not and what are the two conditions is the idea of right right and left hand limit h right and left hand uh limit sorry yeah right and left hand derivative This concept you were given in class right and left hand derivative and what was right and left hand derivative let's start from the right that's frime at x not plus being equal to limits as h tends to z from the right h tending to zero from the right of f f at x not + h - f at x all over h. This is the right derivative. Right hand derivative. Now for the left hand derivative, we denote that by frime minus at x. And this guy's limit as h tends to zero from the left of f at x plus h minus f at x not all over all over h. So note these two things.

[00:06:19]Now the condition attached is that of course these guys these two these two guys must exist.

[00:06:28]The two of them must exist. However, a function note a function f ofx. So, let's keep notes there. A function f ofx has a derivative.

[00:06:46]A function f ofx has a derivative at x = x.

[00:06:53]If and only if if and only if the left and the right hand derivative are equal. If and only if the left and the right hand derivative are equal like this.

[00:07:13]So now see don't confuse yourself. You have a function f of x and then they're saying confirm whether this guy is differentiable at a point x not. How do you confirm that? Simple as everything.

[00:07:31]You pick the function and then you evaluate the limit as h is tending to zero from the left of this particular definition.

[00:07:41]Then when you're done doing that, you also evaluate the limit as h is tending to zero from the right plus of you get f of what we have blah blah blah. Now you want to firstly confirm if these two guys exist. Then after confirming that you also want to confirm if they are equal. If they are not equal it means that the function is not differentiable at that point.

[00:08:08]Or what are you talking about? Example, if I ask you is the function f ofx which is the modulus of x is this guy differentiable at x = x before this concept or before we start talking about the idea of differentiability.

[00:08:26]I'm sure you must have been thinking what do they mean by is it differentiable at x equal to zh is just differentiation. Why can't you just differentiate directly and then just write the answer? No, they want you to investigate whether this guy is differentiable at this point, which means that you have to be critical with the idea. And how do you confirm if this guy is differentiable at this point?

[00:08:51]Simple. You evaluate the left hand and the right hand derivative and you show them if they are equal or not. Good news, I'm going to be solving this question in this video. However, before before I go into the question, let's consider the definition of differentiability itself. So everything I've discussed so far is just derivative. What you did in part one derivative derivative derivative. So and don't also forget I established here that if frime at x not ex exist then we see that the function is differentiable.

[00:09:27]So you have the right to say the function is differentiable at x=0. I don't be scared to make the statement differentiable even when you are solving at a point. Do you understand? Now I told you in class don't forget derivative deals with the point differentiability is talking about an open interval and I explained the the entire idea of open interval which is open um neighborhood. Let's let's let's talk about it. Let's talk about it.

[00:09:55]Differentiability.

[00:09:58]Now a function f is differentiable on a closed interval a comma b if it is differentiable in an open interval and if the following limits exist and what are the limits this particular limit.

[00:10:13]Now this is limit as h is tending to 0 + now it means that h at this particular point a is here you have an open interval and b is here like this. This is an interval. Now anything you have in this interval is approaching a from the right from the right. So it means that it means that whatever you have approaching a is surely greater than a and anything in this interval is approaching b from the left of b. So which means that it is less than b.

[00:10:47]That's the idea my friend. That's the idea. Simple as ABC and that is why it is limit as h ts to 0 plus of the function of a + h. Now we are saying that a function is differentiable in a closed interval. If it is differentiable in an open interval to just show you how critical differentiability deals with open neighborhood open neighborhood open neighborboard open neighborhood values approaching values approaching and not not the um boundary point values approaching not the boundary point.

[00:11:26]So values approaching a and b and not the values of a and b themselves. So we are saying oh um x is approaching a h from the from the right and it's approaching b from the left h. So how do you confirm differentiability in an interval? Good.

[00:11:48]All you just need to confirm is the limit as h is as h is tending to zero from the right h that is values approaching a from the right and you have this definition and then values approaching b from the left this limit two. So if these two guys exist then you see that the function is differentiable in that interval.

[00:12:11]Hence we say a function is differentiable. Hence we say a function is differentiable in a closed interval.

[00:12:19]If it is differentiable on an open interval a comma b like this and its right hand derivative and left hand derivative exist at a and b respectively simple as ABC. So it means that when you're talking about derivative derivative is just a point. Oh yeah, is this is a function differentiable or does it have a derivative at this point?

[00:12:42]How do you confirm if a function has a derivative at the point? You confirm the left hand derivative and then you confirm the right hand derivative. If the two of them exist and they are equal, the function has a derivative at that point. But differentiability is talking about open interval. Open interval. So what do you need to check?

[00:13:01]You need to check what is happening with the limits approaching a from the right.

[00:13:06]Limit of values approaching a from the right and then uh what is happening when the values are approaching b from the left. If you confirm that those limits exist then you are saying the function is differentiable in that interval.

[00:13:21]Let's solve some questions to deepen your knowledge.

[00:13:26]Now this question says investigate the continuity and differentiability of f ofx which is the modulus of x at x =0. So we let's quickly solve that because of our time. This video is already entering 40 minutes. Now um in order to investigate the continuity of this particular function it will be very much bright if we can define it out explicitly. Now um modulus function we've defined that as many times as possible in class. The modulus of x is defined on positive x if x is greater than zero. It is defined on zero if x is equal to z right and it is defined as negativex if x is less than zero. Now there are two ways to actually show this guy. If you want to be sharp and smart, you can just find the derivative of this guy directly from here.

[00:14:27]And of course that's um + one if x is greater than zero. That's 0. If x is equal to z and that's -1 if x is less than z. Now if you check the two points, I mean this is obviously derivative at a point right. So if you check point Z which is the only point we have I told you derivative gives you a point you talk about differentiability as when you're dealing with an interval however if it is if it has a derivative at that point then you see that it's differentiable at that point you understand now if you check what is happening when x is approaching z from the left that is when x is less than zero and you check what is happening when x is approaching zero from the right that is when x is greater than zero from our definition.

[00:15:21]The value of the limit of this function as x is approaching zero from the right is + one. And as x is approaching z from the left we have minus one. Which tells you that the left hand limit h the left hand limit at 0 at 0 and the right hand limit are not equal.

[00:15:45]They are not equal. So that tells you just one thing and that's the fact that this guy this particular function does not have a derivative at point x=0 or another way to put it is that it is not differentiable at x=0. However if I ask you is it continuous of course there are three cases to check for continuity.

[00:16:07]Number one, you check for existence of limits at number two, you check if the function is defined at that point. And number three, you check if the first and the second condition are equal. So if you check for the limit, you check for the limit as x ts minus of the modulus of x. Of course, that's the same thing as finding the limit as x ts 0 of - x, which is 0. Right?

[00:16:33]Also if you find the limit as x ts to 0 from the right of the modulus of x you are finding the limit as x tends to z of x itself and that's also zero. So the two values are equal they exist and they equal. So that means that the limit of this particular function does exist at zero. The second case or the second condition to check is if the function is defined at 0 and of course f of 0 f of 0 is also zero the modulus of 0 is 0. So the limit exist the function is defined and limit and the function at point z are equal which implies that this particular function is actually continuous at point x=0.

[00:17:20]So that tells you that it is possible for a function to be continuous and not be differentiable. Do you understand? So we see continuity does not imply differentiability. However, if it is differentiable, then it is surely continuous. So that tells you that differentiability implies continu. I mean I gave you that theorem in class.

[00:17:40]Differentiability differentiability implies continuity.

[00:17:52]The converse is not true. The converse is not true. Do you understand?

[00:17:57]Differentiability implies continuity. In case you are hearing my resent, you're not my students. Well, you're missing bro. Just join us. No, no stress. Join us for detailed well detailed explanation to concept. Now, the way I solve this question is quite what I say fast. Let's go through the the route I taught you. I mean I gave you some some some limit formulas. So let's use those limits ideas to solve this question. So from the left I mean we want to see if the guy is differentiable at 0. So the limit as the limit as h is tending to zero from the left of um of f frime of x. Yeah like this which is same as the limit as h ts to 0 minus of f at 0 + h because the point this point is zero j 0.

[00:19:07]minus f at 0 all over h. That's the same thing as the limit as h is tending to zero from the left of um now if you if f of x is modulus of x then it means that f at h from the left will be minus h then -0 all over h at the end of the day you get you get -1 there right that's -1 and if you find what is happening from the right limit as h ts to zero from the right of the derivative that's um limit as h * 0 + um f of 0 + h - f at 0 all over h. Now from the right f of h is h is h itself. So this is h - 0 all over h which is + 1. And you agree with me that the two of them does exist. Yes. But they are not equal. And the condition for um a function to have a derivative at a point is that the right and the left hand derivative must be equal. Don't forget the right and the left hand derivative must be equal. Sorry. Yeah, these two things the two of them must be equal. The two of them must be equal. They must be equal.

[00:20:43]They must be equal. Don't let me I mean another way to express what I wrote there. Let me make it simpler for you is to just write it as f prime at zero.

[00:20:54]This is from the left and this is from the right. Simple as so. But the two of them are not equal in this case. I'm saying frime minus at 0 is not giving us the same value as frime plus at zero. So that means that the guy is not differentiable like you don't need to stress the guy does not have a derivative at zero or better put it is not differentiable at z simple as my friend. Now let's solve another question that looks like this but boy a little advanced. They will now solve questions that has intervals so that it will make more sense to you.

[00:21:37]All right. They want us to investigate the continuity and differentiability of this guy. Now listen listen very well.

[00:21:44]Um the initial function I gave you was defined as modulus. I mean this guy deals with signs. So you know something is coming from the left and from the right. But this time around we have critically x² which is a non- negative value. I'm sure you agree with me that x is non negative. And then you have um you have s which is a trick function. Um we all know that the sign of any x lies within + one and minus one like this. So it means that all we just need to check as far as this particular function is concerned here is if the limit as h is ting to zero from I'm sorry of f of um x not which is 0 + h minus f at 0 all over h if this guy exist that's all I mean look at the definition from here I don't need it to get everything mixed up just pay attention look at this definition I gave I said if this guy exist then we see that the function is differentiable at x= x. So you get that's the derivative of so we just need to check if this guy exists for that particular function. So whenever you need to be critical with signs then I mean the idea of left and right comes in left and right derivative. Now how do you check this?

[00:23:12]Simple as I've already explained or defined the derivative of this guy at 0 is defined on limit as um h ts to 0 of f at 0 + h like I defined. So that's f at h minus f at 0 all over h which is the same thing as limit as h ts to zero of um now if you put h into this guy I mean the function at point when when the guy is approaching zero not when it is exactly zero when it is exactly zero you have zero so that's why what we have here is h² sin and 1 / h then - 0 all over 0. Do you understand? All over Z all over Z.

[00:24:10]The only reason why this guy is zero is because they give us in the definition. They did not give us in definition. And we want to find f at 0.

[00:24:23]you know that will be the same thing as 0 2 sin 1 / 0 which is the same thing as 0 * infinity and 0 * infinity is an indeterminate so that's why they give us in the definition so that we won't be hooked at that point I just needed to chip that in but yeah let's go back now what we have here sorry it should be over h not zero all right now what we have Okay. Is then the same thing as um limits as h ts to 0 of h sin 1 / h. Do you understand?

[00:25:04]Because this h we divide um the h². Now the question is what is the limit as h ts to 0 of h sin 1 / h?

[00:25:17]This is where squeeze theorem comes in.

[00:25:19]Right? How do you do that? I mean we know that sin 1 / h lies between lies between one and -1 for any h. Do you understand? For any average. Now um with that said if you multiply all through by h you have minus h less or equal to h sin 1 / h and this is less or equal to h. Now if you evalate the limit of both ends which um as h ts 0 like this if you evalate the limit as h ts 0 of minus h you get zero which is literally the same thing as what you get if you evalate limit as h is t0 of h itself. Do you understand? They both zeros. So that means that this guy will give you zero.

[00:26:11]So it means that the derivative is zero. It exists, right? It exists. And because it exists, that tells us that this particular function is differentiable at x=0. This particular question they actually asked them last session if this was not the question. The question was very close to this. I will confirm check yet. So this function is differentiable at x=0.

[00:26:42]And of course because it is differentiable you can tell that it is continuous very continuous right why because differentiability implies continuity the converse is not true do you understand the converse is not the case so let's solve a question that deals with interval do you understand so that you understand differentiability well I'm tired but let's solve this question.

[00:27:12]This will be our last question for this video or more. The video is already entering 30 minutes. I sure hope you getting it. That's the most important thing. So they said we should discuss continuity and differentiability of this particular function. Of course in the intervals given I think we should start from the idea of continuity and of course continuity depends on existence of limit. Now at point x = 2 do you think this function is continuous? Let us check it together. Now if x is approaching 2 from the left from the left of course means that x is less than 2. Are you following? x is less than 2.

[00:27:58]Now when do you think x will be less than 2 from what we have? So it means that if two is here then x is approaching two like this.

[00:28:10]If it is from the right it means that x is approaching to like this. From the right it means that x is obviously greater than two. Now from what we have it will be approaching x from the left.

[00:28:22]I'm saying x will be less than two or less or equal to in this particular um first function.

[00:28:31]You can see that for the first one x lies between 0 and two. So yeah um the limit as x is ting to um two from the left of the function is the same thing as the limit as x ts to 2 from the left of um 2x - 3 which is the same thing. If you put two inside you have 4 - 3 and that is one. If you check what is happening from the right limit as x ts to two from the right of the function of course that's the same thing as finding the limit as x ts to 2 + of um x² - 3 of course like I said that's when x is greater than 2 and from what you can see you can see that x is obviously greater than 2. So if you put two inside you have 4 - 3 and that is one.

[00:29:27]And of course from what you can see the limit does does exist. Do you understand the limit does exist and if you put two into these functions directly you will get you get defined values one and that implies that tells us that the function is actually continuous. Do you understand? The function is continuous at this function is continuous.

[00:29:56]The function is continuous at x =0 sorry at x = 2 and the guy is continuous at x = 2. Now let's talk about the differentiability of the function which is the main reason why we are here.

[00:30:16]Now from what we were given you can obviously see that we have intervals closed interval of 0 and four I'm saying x is between this interval from 0 to 4 at first was splitted between 0 and 2 and then two and four like this. So if you check it very well x is in this closed closed interval. So let's check the differentiability of x in this closed interval. Don't forget from our definition of differentiability we say that a function is differentiable on a closed interval if it is differentiable on the open interval. What's the idea behind the open interval? We said it. We said open neighborhood. Do you understand? So yeah, let's check it.

[00:31:05]The first thing we need to check is the existence of the left that is this boundary. this boundary here the limit from that end and then limit from this end once the two of them exist then we are sure that the function is differentiable in that interval so I am saying um the function from the right of course x is at the right of zero any x you have in this interval is is at the right of zero so I'm saying if this is the real line and this is 0 and four here this guy is at the of 0 and it is at the left of four.

[00:31:44]So f prime at 0. Why? Because it is at the right of 0 uh is equal to the limit as is equal to the limit as h ts to 0 + or 0. You don't even need to. We've already indicated the sign there. then of um f at a our a z now z you know a is the boundaries denoted by a comma b that's 0 + h minus f at 0 all over h and if you evaluate that that's um as limit as h tends to zero Then over um from the left of course that's um that's 2x - 3. So that will be 2 h - 3. Then minus if you put zero inside this thing you get - 3 - 3 close that all over h which will give you limit as h ts to 0 of um 2 h - 3 + 3. That means - 3 and + 3 will cancel themselves out all over h will cancel h and the result here is true. So we've seen that the right and or the right limit under differentiability you come and see differentiability here and we said that a function is differentiable if these two limit exist that has a condition for differentiability my friend if these two limits exist this condition here.

[00:33:33]So for this particular question we've we've we've cons we've observed or we've discovered that this guy exists. Well let's check the other one which is when it is coming from the left when x is approaching four from the left. So we are saying frime from the left of four is equal to the limit as h ts to z. H must always t as far as differentiability is concerned of the function. Then b is four now. So that will be 4 + h - f at 4 all over h which is the same thing as the limits as h ts to 0 of now from the left you have x²us 3. Do you understand why? Because x is less than 4 here in this region. Yeah. like this x is less than four in this region less or equal and we we observing open interval so x is less that mean what we have there is the same thing as 4 + h all 2 I mean what we have there is x square - 3 so that will be 4 + h all square - 3 - 3 then minus if you put if you drop four inside this thing you will have 60 - 3 which is um 13 then -3 all over h which is the same thing as limit as h ts to zero over um if you open this thing up you have 16 + 8 h then plus h² - 16 you know That's - 3 - 13 all over h.

[00:35:33]That's the same thing as this. This one will cancel this one out. So you have limits as h ts to zero of 8 h + h² all over h. So h is common there. You have the limits as h is ting 0 or um 8 + h.

[00:35:57]Can you understand? H will cancel the common h there. So if zero enters you have eight. Now we have we have discovered that we have discovered that the left e the left hand derivative and the right hand derivative exists. And because the two of them exist differentiability um intervals closed interval and then definition based on open interval is on existence of both the left and the right. Do you understand? So these two guys exist. One give us two the other give us eight. Because two of them exist we are sure that the function is differentiable. If it was at a point then it means that the left and the right must give us the same answer. Do you understand? Which is derivative.

[00:36:55]But now we are dealing with boundaries um um intervals open intervals. So all you existence existence of the left and the right limits two limits once they exist then you're sure that the function is differentiable in that interval. So that's it my friend and um hopefully you understand this. There are still more questions to solve. I will solve them in the next class. Thank you so much. Do have a wonderful day.

[00:37:23]>> Subject.

[00:37:25]What lesson?