 So, I was trying to give you that condition. So, anyway this we have proved that differentiability implies it is continuous. So, that is because the right hand side goes to 0. So, here is a theorem which I was referring, which is a sufficient condition for differentiability, which is a very nice one and very useful one. It is only says that if this is happening, then the function is differentiable. It does not say the converse is always true. So, the earlier condition was if and only if, right? It is an equivalent way of defining differentiability. But this condition says suppose f is defined in a domain D such that the following hold both the partial derivatives exist in a neighborhood of the point. So, something more is coming. Not only the partial derivatives exist at that point, we require they exist in a neighborhood of that point. Why we need that? Because we want to talk about f x and f y are continuous at the point a b. The partial derivatives as functions of two variables. I want to talk about the continuity of these functions of two variables. For it to define the continuity of a function of two variables, the function should be defined in a neighborhood. Otherwise, you cannot talk about the continuity of that function. So, for that reason the first condition is required because I want to state that f x and f y are both continuous at a b. So, they should be first of all defined in a neighborhood of a b. So, first condition that partial derivatives exist in a neighborhood and are continuous at that point. That is good enough to say that the function is differentiable at the point a b. So, this is called the increment theorem. So, not only we want partial derivatives to exist at that point, we want them to exist in a neighborhood and to be continuous at that point. That is good enough. That is sufficient enough to ensure that the function is differentiable. The proof is there in the slides, but will not go through the proof. That is not slightly long. So, will not go through the proof. If you are interested, you can read that. So, here is the same example that we showed. We showed that this example of a function which is differentiable at the point x 0 y 0. You can calculate the partial derivatives. Partial derivative at 0 0 we saw they exist and are both 0. Actually, in a neighborhood of 0 0 also they exist. For example, what is the partial derivative of this with respect to x at a point x not equal to 0. At x is not equal to 0, the function is defined by this x square plus y square into this. So, I can apply the product rule to calculate the derivative in a neighborhood. That means, y is not 0. So, this function is as it is, where y is treated as a constant. So, x square plus y square sign of 1 over x square plus y square at a point not equal to 0 0. That is the definition. So, if I put y equal to a constant that will give me derivative with respect to x at a point other than 0 0. So, product rule applies x square. So, it will be 2 x into sign of this plus x square into derivative of sign that is cos of we can calculate that derivative. So, if you calculate that you will see the cos of 1 over something is coming and you can check that function is not continuous. So, derivative at a point in a neighborhood exist and is not continuous. The function is differentiable. So, this is an example to illustrate that the increment theorem, the derivative exist and are continuous at that point is only a sufficient condition that is not necessary for the function to be differentiable. So, one can check. So, this kind of derivative will come and this function is not continuous. You can easily check this function is not continuous at the point 0 0. So, partial derivatives are not continuous. Function is differentiable. So, that increment theorem is only a sufficient condition if something happens, if the partial derivative exist in a neighborhood and are continuous at that point, then the function is differentiable. The converse need not hold. Function can be differentiable without the partial derivative being continuous at that point. So, there is an example. So, good enough. Now, here is something corresponding to one variable chain rule. In one variable, we had say, if you have a function f in C d and you have a function g in R, say that f composite, this is g composite f is defined. So, that will be a function from A B to R. You first apply f and then apply g. So, that is a composite function. And we had the chain rule. So, what was the chain rule in one variable? You take a point here, say x0. This will go to a point f of x0. Call that as y0. g is a function defined there. So, that takes this point to g of y0. And the composite function is defined at the point x0. So, I want to say, when is composite function derivative at the point x0 equal to? The composite function is defined on A B. x0 is a point. We want to know what is the derivative in one variable. So, what does the chain rule say? It says it is you need to have the derivative of f at the point x0. f at the point into g dash of, what is the chain rule? g dash of f of x0, that is y0. So, that is an increment that comes. Now in several variables, you can have a function from r2 to r. And then you can have a function from here to here. So, here is a function f and here is a function g. So, what will be composite function? g composite f will be a function from r2 to r. Or you can also have a function say f from r2 to r2 and then a function g here. So, g composite f will again be a function from r2 to r. So, there are many possibilities that become apparent. So, you can have a function from here to r2 itself. So, you can have r2 or you can have from r2. There are many possibilities. For example, you can have a function from r2 to r2 and g from r2 to r2. So, various compositions are possible. And we want to know, in the first one, your f is a function of two variables composed with one variable. So, you can ask what is the meaning of differentiability of the composite function. So, in each one of them, one can ask what is the notion of differentiability. So, for that, there are chain rules. So, I will just state some of them. So, let us look at a function w, which is function of x t, y t. That means what? There is a function g, which is a function of, f is a function of two variables, f of x t, y t. So, it is a function of two variables. But from where does x t, y t come? It depends only on t. So, there is a function of one variable. You can call it as g, r2, r2. So, t goes to x t, y t. Are you following what I am saying? So, it can be, for example, you are observing the path of a particle in a plane. At time point t, what is the position in the plane? The position of the point in the plane is given by the x coordinate x t and the y coordinate y t. So, a curve in a plane is described by a function of one variable taking values in the plane. And that is combined with a function with f. So, f does something to that. So, it is a function. So, it says that this function will be differentiable at a point t0. And the derivative is, there is a composite of f of x t, y t. So, f is a function of two variables. So, there will be increment in the direction x, increment in the direction of y. So, f x, partial derivative in the direction of x, at that point, chain rule one variable, how does the one variable change? x dash of x t0 plus partial derivative of f in the y direction and the increment, the rate of change given by y variable. Is this clear? Let me just draw a picture probably. So, this is the last one actually. T goes to x t and y t and then something that goes to g of x t, y t. So, t goes to the composite function. Intuitively, one can write it as, w is a function of one variable. The composite is a function of one variable. So, we will talk about the derivative of a function of one variable at a point. So, d w by d t. Now, w is a composite function f of x t, y t. So, f in the direction what is the rate of change? Partial derivative of f with respect to x by chain rule d x by d t. The second variable also to be taken care of, so rate of change of f in the direction of y d y by d t. So, that is how you calculate the derivative of function of one variable, which is a composite of two functions. Now, there are more possibilities. For example, this g could itself be a function of two variable, t and s. Then you will have a x comma x of t comma s, y of t comma s. So, then instead of d x by d t, there is a partial derivative of that in the direction of and so on. So, let me just show you those also. You just have to, it is not difficult. They are not that straight forward to prove. So, let me go to the proof of this. We will just assume these things. So, this is what the pictorial it says. w is a function of two variables. x is a function of one variable, x t, y is a function of a variable t. The combined thing is a function of one variable again. So, we have to go along this plus. So, partial derivative of f with respect to x, the increment d x by d t. So, that is along this branch plus over the other branch. For example, if you have something more, there are many chain rules. I think you should just look them and understand them. How to, here is another one. w is a function of two variables x and y, but x and y themselves are functions of two variables. Earlier it was x t, y t. Now, x is a function of two variables r and s. So, it gives you two different pictures. So, the composite function is a function of two variables. So, it will have now partial derivatives. Earlier the composite function was one variable. So, now function of two variables. How do you get the partial derivative now? So, two variables. So, partial derivatives, you want to know whether they exist or not. If they exist, what will be? From one branch, partial f is a function of two variables. So, partial derivative of this with respect to x. x is a function of two variables. So, which variable you are looking at? Here, we are looking at with respect to r. So, partial derivative of x with respect to r. Similarly, for the y, r. So, add up. That gives you partial derivative of the composite function with respect to r. You want partial derivative of the composite function with respect to s. So, partial derivative of w with respect to x. That is okay. Partial derivative of x with respect to s, it will come. So, that is the second one. So, depending on what is where, how many directions you are going, you will write the chain rules. So, get used to these chain rules. Just go through them once. Once you understand, you will know what you can have this kind of thing. One variable to two variable to one variable. So, one variable and then going to two variables. So, the composite is a function of one variable, but taking values in two variables. So, partial derivatives again will come. So, these are various kinds of possibilities. Why they are useful? We will see one application of these kind of things. Other possibilities. Let me not go through all of them. They are just see how many branches, how many variables, those many pictures will appear. Now, for example, let us just look at one example that f is a function of two variables f x y equal to x square plus y square. Let us see x. This x itself is a function of a variable t, x t is equal to e raised to the power t. y also is a function of t that is equal to t. So, what is the composite function? t goes to x t comma y t, f of x t comma y t. So, it is a function of one variable. Composite is a function of t goes to a value in r, t goes to x t comma y t that is in r 2, but f takes x t comma y t to real line. So, it is a function of one variable only. So, what will be the differentiability derivative at a point? For one variable, the notion of derivative only. So, what is the derivative? So, composite function if you call it as w, we want partial derivative of w with respect to t at some point say or at any point t. So, what will be the derivative? Partial derivative is of f with respect to x into x dash of t plus partial derivative of y f with respect to y into y dash of t. So, let us write partial derivative of f with respect to x, it is 2 x. So, first term will be 2 x into e raised to the power t plus 2 y, derivative is 1 plus 2 y. So, that is the, you can put the values in terms of t again x and y, you can put the values in terms of t. So, that is partial, that is the derivative. For another example, let us just, so that is our computation, f x y is same, function is same, but now x is a function of 2 variables s and t and y is a function of 2 variables. So, x s t is this, y s t is this. So, what is the composite function? w now, it starts with s comma t 2 variables, maps into x of s comma t comma y of s comma t. So, this is a function of r 2 to r 2 and then there is a function f of 2 variables taking into real variable. So, the composite function is a function on r 2, where the first thing r 2 to r 2 and second one r 2 to r. So, the composite function is a function of 2 variables taking real values. So, when is a function of 2 variable composite function? Partial derivatives I can find out. So, what are the variables s and t? Composite function with respect to s, how do I find out? Partial derivative of f with respect to x, so that is 2 x into partial derivative of x, x with respect to s, because I am looking at, so into 2 s plus partial derivative of f with respect to y, 2 y into partial derivative of y with respect to s, so that is 2 t. At these 2 terms, you get the partial derivative of the composite function with respect to s and similarly with respect to t. So, this is what you will get. So, that is how you compute. See, what functions are coming and how many are there? So, that many additions and multiplications. So, I think chain rules are quite clear. More examples that you can read from the slides. So, let us look at something more. So, let us take a function f of 2 variables domain D in r 2 taking values in r. Let us assume f is differentiable. We do not need to assume that. So, we will come to that later. So, let me draw a picture, what we are doing so that you understand. So, D is a domain. So, there is a domain and for every point in the domain, the function gives some value. So, z is equal to f of x y. So, this is x y in the domain. So, you get various points. So, as I had said, you will get some kind of a surface. So, this kind of a surface you will get. So, s, which is not visible, yes, reasonably visible, s is x y f of x y. So, that is the surface collection of all such points. Now, we looked at partial derivatives of this function, existence. So, what we did? For partial derivative, we fixed one of the variables and see whether how much the change rate of change is coming along the other variable. We want to generalize this slightly. Instead of rate of change along x axis or y axis, we want to go along any vector. So, let us fix a vector u belonging to r 2. So, what is the meaning of a fixing a vector? Let us say the norm of this vector is equal to 1. So, let us say this is the vector u. That is the direction in the plane. That is the direction in the plane, in the x y plane. Now, I want to see what is the rate of change of the function along this direction, along this direction. So, at this point, let us call this point as x naught and y naught. Let us fix a point also. So, how do I move in the domain, in that direction? So, to move in the direction, I will draw a line which is parallel to that vector passing through this point x naught y naught. Is it okay? We will see as the point moves in this line, we will get a value. So, what you will get? You will get some kind of a curve. As you move on this line L, L is a line through the point x naught y naught parallel to that vector u, in the direction of that vector. Is it okay? So, as the point moves on this line, the value of the function will be a point on the surface S. So, as you move along this line, that will give you a curve on the surface. Is it okay? Geometrically quite clear. So, what I want to know is, as you move along this curve at this point, is there some rate of change of the function? How do I find out the rate of change of the function? So, what I have to do? The value at this point is f at x naught y naught. I should see how much is the increment. So, what is the nearby point? So, if I take a point near by x y, what is that in terms of x naught y naught? What is this point x y in terms of x naught y naught? It is on a line which is parallel to that vector. So, what is this x? Can I say x is equal to something and y equal to something in terms of x naught y naught? What is that? Of course, the vector u, because it is parallel to u. So, what is this? What is x? What is y? So, let us, if you want to write what is the vector x y, that is t times the vector u plus x naught y naught. Is that okay for everybody? Any point on this? So, let me demystify it for you. Imagine you are moving on that line. You are moving on this line. At the point 0, at time point 0, you are at this point. You are moving on that line. You are starting your journey at the point t equal to 0 and you are at that point. Then what will happen? You are moving in the direction time t units. How much you would have moved? t times the vector u, depending on whether it is positive, positive or negative, backward or that is physics. But, if you still do not understand it, here is. So, this is the point x naught y naught. This is the point x y. So, what are the coordinates of this? This direction is u. Vector addition now, if this is. So, what will be a point? That is t times u. This is t times the vector u. This vector is x naught y naught. So, what is x y? So, vector addition, if you like. So, this x y is equal to x naught y naught plus t times the vector u. So, let us call it u 1, u 2, the components if u is equal to u 1, u 2. Is it okay? Vector addition only. Nothing more than that. So, I want rate of change. So, f at a nearby point x y. So, that is t u 1. Sorry, what is the nearby point? So, f at x naught plus t u 1 y naught plus t u 2. That is a point nearby minus the value f at x naught y naught divided by how much is the change? How much distance you have moved? That is t, because we have taken the vector to be the unit vector. So, t times u 1 square plus u 2 square square root, that is the distance. So, that we have taken it as 1. What is that? Is that okay? We have taken the norm to be equal to 1. So, along this, this point x y is t of u plus the vector x naught y naught. So, what is the value at this point? Nearby point. Value at this point. So, there is this. That is f at the nearby point f of x y. This is f of x naught y naught. So, the increment divided by the distance. Rate of change I want to find out. So, that is what precisely I am writing. f at a nearby point minus the value at that point divided by how much is the increment? I should take the limit of this as t goes to 0. If this exists, it may not exist. We are trying to find out what is the rate of change. If it exists, it is called the directional derivative of f along the vector u at x naught y naught. So, it is of the function f along that direction u at that point. So, let us denote it by directional derivative capital D of the function f along that direction u at that point. So, let us denote it by directional derivative capital D of the function f in the direction u at the point x naught y naught. So, that is the notation we will use. So, can you say what is the partial derivative with respect to x? What is u there? It is a direction only. So, partial derivative, what is the vector u? If you want partial derivative with respect to x, it is nothing but the vector u is 1 comma 0, unit vector in the direction 1 comma 0, y direction 0 comma 1. So, those are the partial derivatives. So, they may not exist. Even partial derivative we have seen may or may not exist. So, if they exist, so this is the notation for that.