IMPLICIT FUNCTION DIFFERENTIATION//OMAKA's DelF FORMULA

[00:00:03]This is Greater Tomorrow Academy Enhancement School G match 41. You are welcome to our introduction to calculus class.

[00:00:12]In this video we are going to treat implicit function differentiation. Is that okay? Implicit function differentiation. What is an implicit function? How do you identify an implicit function? It's very simple.

[00:00:25]When a function depends on two or more variables, that function is an implicit function. Look at this on the board here. For a two variable functions, you can see f of x comma y. It means a function that depends on x and y. In our previous videos, the functions we've been differentiating with we've seen is actually an explicit function.

[00:00:49]Explicit.

[00:00:50]Where the function depends only on one variable. For example, you have f of x.

[00:00:56]I can say y is equal to f of x. It's very clear. You know that y depend on x and x only.

[00:01:03]But if I give you a function where you see in the function you have both x, y so involved in it. Are you getting? We say that the function depends on x and y in that case. And so we refer to such function as implicit function. We're going to see them in the next video anyway.

[00:01:22]How do we approach an implicit function problem? You want to differentiate it.

[00:01:25]What do you do? Now I will start by telling you that there is a general method most people use. And how do you do that? Whereby if the implicit function is given to you, let us maybe 3xy + x²y³, you know? And then they said you should differentiate this. Most people go by differentiating both the x and then the y. When we want to differentiate y, after differentiating any y, you multiply it with dy/dx.

[00:01:56]Then at the end of doing that, you simplify, factorize out your dy dx, and then make it subject of the formula.

[00:02:05]That is what majority of the people do.

[00:02:07]Here from Gmaths41, we are going to apply this principle that once an implicit function is given to you, no need of bothering yourself going to say, "Okay, anytime you differentiate Y, you introduce dy dx as a multiplier to read." It's going to really be a waste of time. We can use this formula straight to obtain our result just in one or two steps. Now, dy over dx is equal to minus del F over del X, then all over del F over del Y. Here, F represents the implicit function given to you in that case.

[00:02:41]And what does this del mean? This del F del Y del X, they are actually partial differential coefficients. Partial derivative sign.

[00:02:54]Is that okay? And so, we are going to apply this formula and get our result just as fast as anything.

[00:03:03]All right. In the next video, we're going to pick a question where we'll apply both methods. But, remember, you've been told that from Gmaths41, this is what we're going to make use of.

[00:03:13]So, let us solve this first example using the general method. This is the question given to us. Y is equal to 3x cubed minus 4x squared minus Y cubed.

[00:03:22]And then, we are to obtain dy dx. You can't get dy dx straight ahead here. No, by the general method. And remember that I told you in the introductory video, anytime you differentiate Y, what do you do? Multiply it by dy dx. So, let's go.

[00:03:34]If you differentiate this Y, you're going to get one.

[00:03:38]So, it's going to be one, you multiply it by dy dx.

[00:03:42]And then, that's equal to Come here and then differentiate 3x cubed. Use this power to multiply this, you get nine, then x raised to the power Subtract one from this power three. If you subtract one from power three, you get two. Then minus come here, use this power two to multiply this four, that is going to give us eight. And then X, again subtract one from this two. If you subtract one from this two, you get one.

[00:04:03]So, it's the same as X raised to power one, which is X. Come here then minus, differentiate this Y cubed, you're going to get three Y squared as usual.

[00:04:11]And then you're going to multiply this by dy dx because it is Y that you differentiate with respect to X there.

[00:04:17]At this point, we are going to collect all dy dx terms together. So, what do I do here? I have dy dx on the left hand side. So, I will move this dy dx on the right to this left hand side. So, 1 * dy dx will give me dy dx. And then if this minus 3 Y squared dy dx crosses, I'm going to have plus 3 Y squared dy over dx. Everything will be equal to the right hand side will be 9 X squared minus 8 X.

[00:04:41]Good. At this point, we are going to factorize dy dx. We want to get only one dy dx. So, we do what we call factorization from here. So, I will bring out dy dx, which is common to these two terms. Open bracket. If you factorize dy dx out from here, what will be left will be one. How do we know that? In factorization, I brought out dy dx. So, this thing that I brought out, I will use it to divide where I brought it out from. So, dy over dx divided by this dy dx is factorized, you get one left here. Then plus you have 3 Y squared dy dx divided by this dy dx here. Then you know that dy dx will now cancel out. You'll be left with 3 Y squared. You see this?

[00:05:21]Everything will be equal to 9 X squared minus 8 X. Finally, make dy dx subject of this formula. To make this dy dx to stand alone, it means you have to remove 1 + 3 Y squared from here. And that is multiplication, you know that. So, to remove this term from here, you need to divide both sides by this term. So, making dy dx subject of the formula, it will be equal to 9 X squared minus 8 X all over divided by the quotient of the dx, which is 1 + 3 y squared. And of course, this is the answer. This is the general method. In the next video, we're going to use the formula method to arrive at this same answer just in one or two steps.

[00:05:58]Now, let us use the formula method to obtain the result of this dy dx. Now, how do you handle the problem using the formula method? First, collect all the variables to one side, and then call it your f of x, y. To do that, now look at this y, I will take it to the right-hand side. Then I will answer that the function which depends on x and y now is equal to 3 x cubed minus 4 x squared minus y cubed. If this + y crosses, it's going to give us minus y.

[00:06:26]That is the first thing you do. Then remember the formula that dy dx is equal to minus del f del x over del f over del y. This simply means differentiate that function partially with respect to x.

[00:06:41]When you are differentiating with respect to x, y will be constant. That is the meaning of the partial derivative.

[00:06:47]Then, if you're done with that, go back again and differentiate that function with respect to what? y this time around. It means that x will be taken as constant there. Let's work around it and see. This will be equal to minus Now, let's go.

[00:06:59]I want to differentiate this function with respect to x now. I will come here.

[00:07:03]There is x here. So, if you differentiate 3 x cubed, use this three to multiply this and then subtract one from the power. You're going to get 9 x squared. Come here, you have minus Use this two to multiply this four and subtract one from this power. You're going to get 8 x. Then here, do you see any y here?

[00:07:21]Do you see any x there rather? Is there any x there? No. What you have here is y. And remember you are differentiating with respect to x. It means that y is constant. Since there is no x here, this is a constant. If you differentiate, you get zero.

[00:07:35]I'll put zero here. Then the same thing here, differentiate this, you get zero because there is no x there. The zeros there are not necessary.

[00:07:42]Okay, all over. This time around for the denominator, differentiate this function again with respect to Y, which means that X is going to be kept as what? A constant. Okay, let's come here. Do you see any Y here? No. So, this is a pure constant and if you differentiate it, you get zero.

[00:07:57]Do you see any Y multiplying this 4 X squared here? Not at all. I'm not seeing any Y. So, this is a pure constant. If you differentiate it, you still get zero. Once again, these things are not necessary here. So, if you're solving it, just forget about them because there is no Y there. Okay, this There is Y here, Y cubed. If you differentiate it, you're going to get 3 Y squared. You know, use power to multiply and then subtract one from the power. Then minus, if you differentiate this Y, what will you get? One. You write it there.

[00:08:22]You see, you're done. So, this will give us minus open bracket. This is 9 X squared minus This will give us 8 X all over Look at the denominator. This is minus 3 Y squared minus 1. Do you notice that minus minus is common? You can factorize the minus out. If I factorize this minus out, I'll have minus open bracket. What will be left? 3 Y squared plus 1. Take note. This is zero. Forget about them. Here, this is minus. This is minus. You factorize the minus out. So, in the bracket, you now have 3 Y squared plus 1. Look at it. This minus will cancel out as you know. So, that dy dx will now be equal to 9 X squared minus 8 X all over 3 Y squared plus 1 as you obtained in the previous video.

[00:09:11]We have the second example we're going to solve using the formula method. You see, you have to appreciate the formula method because it goes for all questions. The general method might just be so deadly in some kind of questions given to you.

[00:09:24]And that's why you want to appreciate the effort coming from GMAT 41 here you.

[00:09:28]Let us solve this question just in one or two steps using our formula method.

[00:09:33]And then, this is a question given to us. Remember the first thing you do is to collect all variables to one side and call it a function of X. X. You may choose to move all these things from here to the other side or you choose to move this 2y to the other side. Either ways, you still get the same answer. I will move this 2y to the left here so that my function of X {comma} Y will now be cos X + X squared Y squared minus sin Y. If this 2y crosses, you're going to get minus 2y. And so, I will now proceed to obtain my dy dx straight. Remember the formula. No need to write it down again.

[00:10:08]All I need to do, I'll put minus open brackets. The numerator is what? del F del X. That is, differentiate the function with respect to X partially.

[00:10:18]Meaning that Y is going to be kept constant. Are you ready? Let's do that together.

[00:10:24]This is cos X. So, I can differentiate with respect to X. You know that, right?

[00:10:27]If you differentiate cos X with respect to X, you get what? Minus sin X. Now, previous videos in standard derivative, we have already explained all of those.

[00:10:36]Come over here. This is X squared Y squared. There is X multiplying this Y squared. So, what do we do? You are going to keep the Y squared constant.

[00:10:45]You can see that. Then, differentiate the X squared. If you differentiate the X squared, you're going to get 2X.

[00:10:52]Take note.

[00:10:53]This Y squared will be kept what?

[00:10:54]Constant because there is X squared multiplying there's X variable. You differentiate just that X variable. Come over here. This minus sin Y. Please, do you see any X multiplying anything here?

[00:11:05]I'm not seeing any X. Therefore, if you differentiate sin Y, you're going to get zero because it is with respect to X.

[00:11:10]You know that.

[00:11:12]So, this one gives us zero. Forget about it. Minus 2y. There is no X multiplying this term. So, if you differentiate, it's a pure constant. You get zero. You are done with that. All over This time around, the denominator, you differentiate the function with respect to what? Y. Therefore, X will be kept as a constant. Come over here. This cos X.

[00:11:29]Differentiate. There's no Y multiplying anything there. So, it's zero. It goes off. Come over here. This is x squared y squared. Remember you're differentiating with respect to Y. There is Y multiplying this x squared here. So, you keep x squared constant and then differentiate this y squared, which will give you 2Y. Come over here. Minus This is sin Y. A Y term there. So, you differentiate sin Y, you get what? Cos Y. And then this is also a Y term.

[00:11:55]Differentiate minus 2Y. What will you get? You will get minus 2.

[00:11:59]You are done. So, just simplify and get the dy dx to be equal to If you open up this bracket here, this minus affecting this will give you positive sin x. The minus affecting this will give you minus y squared times 2x, which is minus 2x y squared. Then all over x squared times 2y will give us 2x squared y. And then minus cos y minus 2. This is your result.

[00:12:25]Now, let us play around this past question, okay?

[00:12:28]2014-2015 question, I think. Compute the differential coefficient of the equation x squared y plus y squared x plus 4x equal to 1 with respect to x at the point 1,1. Don't get confused hearing at the point 1,1, okay? It simply tells us that we should get dy dx. If you have gotten your dy dx, anywhere you see x, put the x coordinate 1. If you see y, put the y coordinate 1. Then simplify to get a number. In fact, we're going to meet that under application of differentiation when we're talking about gradients of a function, okay? Now, this question given to us, the function is an implicit function. We know that, right?

[00:13:03]Because it contains x and y. It depends on x and y. So, we write f(x,y) is equal to x squared y plus y squared x plus 4x.

[00:13:12]This one will cross to give us minus 1.

[00:13:14]So, let us get our dy dx using our formula. You know it is minus del x over del x all over del x over what? Del y.

[00:13:21]And you know what that means already.

[00:13:23]So, let's go. At this point, we are differentiating with respect to x for the numerator. This y is multiplying this x squared. So, we keep the y constant, differentiate x squared, you're going to get 2x.

[00:13:34]Good. Then here, plus again here, in differentiating with respect to x, you know that so this y squared will be kept constant. You differentiate this x, you get what? One.

[00:13:44]All right, then the next sign is plus again. This is 4x. Don't forget the numerator is with respect to x, so you can differentiate this. You'll get four.

[00:13:53]This is minus one, a constant. If you differentiate it, you get zero. It goes off. Then all over this time around for the denominator, we're going to differentiate this function with respect to what? Y. Which means that x be taken as what? A constant. Let's do it.

[00:14:06]x squared times y, we keep x squared constant and then differentiate this y.

[00:14:10]If you differentiate y, you get one.

[00:14:12]Then plus this time around, you have y squared x.

[00:14:16]So, x is multiplying it. You keep the x constant, differentiate this y squared alone. You'll get 2y. Then of course, look at here. Is there any y here? No.

[00:14:25]You differentiate 4x with respect to y, it goes to zero. This one also, zero.

[00:14:28]So, you're done with the differentiation. Now, I don't want to simplify anything any further because they gave us points (1,1). So, all I will do is go here and then substitute one for x and y. So, let's do that. Here, I will have minus instead of this, I'll put 1 * 2 * 1.

[00:14:46]Okay, then plus you come here. This is y squared. Y is one from the coordinate given. You square it times one. And then lastly, plus four.

[00:14:56]Everything will be divided by Come over here. This is x and x has the value of one, so you have one squared times one.

[00:15:03]Then plus This is x. X is one. You put one times two times the value of y, which is also one. And so, our dy/dx finally at that point (1,1) will be equal to If you solve this part here, 1 * 2 * 1 will give us two. Then here, 1 squared is 1 * 1. You get one here. This part is 4. Then here 1 ^ 2 * 1 will give us 1. 1 * 2 * 1 will give us 2. So, you see this is 2 + 1, which is 3 + 4. And then you get -7 as numerator. 1 + 2 is 3. So, this is the answer. Depending on the option, if the option is in mixed fraction, this will give you two whole number. That is -2 whole number 1/3. Is that okay? So, either of these is your result.