Learn about Implicit Differentiation Video

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Learn about Implicit Differentiation

Learn about Implicit Differentiation Welcome to the video on implicit differentiation. So, let’s say that I have—well let’s just explain the difference between implicit and explicit first. So, if I had a function that was; let’s say y =—I don’t know, x2 + 2x + 3. In this situation, y; the variable y is defined explicitly in terms of x. Well how do I know that? Well, if you give me an x, I can just input it into this expression right there and I explicitly get y, right? It just pops out of the equation. But on the other hand, what if I had something like this; what if I had x2 + y2 = 100. Well now, if I give you an x, you still can figure out y but it’s just not as easy, right? You would—if I said x =—I don’t know—x = 9. You would put 9 in there, you would get 81 + y2 = 100. You would subtract that from; so you get y2 = what? 19. I don’t know. These numbers are weird. And then you get y = ± √19. All right. So, you still could solve for y but you have to—but this expression right here or this equation defined y implicitly. And you know that’s the different—you know. When people use the word explicit or implicit in everyday language, it’s the same thing when you say—you know, he explicitly said this; that means—you know. I am hungry. I explicitly stated that I am hungry. But you could implicitly state that you’re hungry. You could say, “Oh well, you know, I haven’t eaten for a long time and you know—boy, it would be nice to maybe eat something.” And that would be kind of an implicit expression of your hunger because someone would have to kind of do a little bit work to come to the conclusion that you are hungry, instead of you telling them. So with that out of the way, let’s actually do some differentiation. So what everything you’ve learned about differentiation right now has been explicit differentiation. So example, for this function, we know how to take the explicit derivative here. Well, it’s explicitly defined function, so we just take the derivative. We apply the derivative operator. We take the rate of change of both sides of this equation with respect to x. So you could do it like this. We’re taking the derivative operator. I know this is getting messy. We’re taking the derivative operator on that side and we’re also taking the derivative operator on this side and all that. I didn’t plan to write that there, that’s why it’s— So what did happen? When the left side, the rate of change at y changes with respect to x, well, we just write that. The rate of change of y for every small change in y, what does this change in x? That equals and then we know how to this type of derivative. This is just a simple quadratic. So that equals 2x + 2 + 0, right? So there we go. We just did some explicit differentiation. So, how do we do the same thing here? Well, we’re going to do the same thing. We’re going to apply the derivative operator to both sides of this equation. So let me do that. I’ll do that in a different color. Let me erase all of this too because this is kind of obvious things that you might already know. Let me erase all of that. Okay. So I’ll take the derivative operator to both sides of this equation. So, this is no different than what we did up here. And as you learn in algebra, anything you do to one side of equation, you have to do it to the other side of function. Otherwise, it would no longer be an equation. Take the derivative operator because we’re taking the derivative—the respect to x. We could have taken it with respect to y. If x2 + y2 =—if you take the derivative of both sides, right? What is the derivative? Well 100 with respect to x right? So this equals the derivative with respect to x. Let me put in brackets. I was going to do parentheses but I think bracket—of x2 plus the derivative with respect to x of y2. And what is the derivative with respect to x of a 100. For every change in x, how much does this change? Well, it doesn’t change at all. It’s going to stay a 100, right? So this is zero. The derivative of a constant really respect to any changing variable is going to be zero. So that was just zero. And what’s the derivative of x2 with respect to x? Well, we did that up here, right? When we took this derivative, we just took the derivative each of these terms, the derivative of x2 with respect to x. Well, that is just 2x. So, we get 2x +. Now this term here is where you might get a little bit confused. I am taking the derivative of y2 with respect to x, the derivative of y2 with respect to x. So let me take you on this side. You know you should not do this every time that you do one of these implicit differentiation problems but I am doing this because I want you to understand what we are doing. So let’s say I just defined g = y2, right? So let’s say that this was g. And I want to take the derivative of g with respect to x. So I apply the derivative operator on both sides of this. So let me do that. So the derivative—I’ll do a d of x of that side. d of x on that side. That is a d. There you go. So, what we get is the derivative of g with respect to x is equal to the derivative with respect to x of y2, right? And so far, we haven’t really done much. I’ve just rewrote things. But what does the chain rule tell us? The chain rule tells us that the derivative—let me stay in the same color. I was switching colors. The derivative of any variable with respect to x or with respect to another variable is equal to the derivative of g with respect to y times the derivative of y with respect to x. And you could view this if you don’t have the full intuition of the chain or—yeah there’s a couple of ways to view it. You know, all of these dg’s and dy’s that well you see throughout integral and differential calculus, all these really means are really, really, really, really small change in g, a really, really, really small change in y, a really, really, really small change in x. So, you can almost view this—so you know, if you have a—when you multiply this out. This really, really small change in y is the same as this really, really small change in y. So if these were—theses are essentially fractions, so you could cancel these out and that is why you would get the derivative of the change in g with respect to x or the derivative of g with respect to x. So this is what the chain rule tells us. So, given that this is what the chain rule tells us, how does that help us figuring out this? Well, we know that the derivative of g with respect to x is equal to the derivative of g with respect to y. But we already know that g = y2. I just made that definition. That’s where we started from. So what’s the derivative of g with respect to y? Well that’s just 2y, right? Just using our derivative rules or polynomial rules which hopefully you believe actually work because there are several videos where we prove it. And then times the derivative of y with respect to x, which we don’t know and that is actually kind of what we’re going to try to solve for. So, the derivative of g with respect to x is 2y times the derivative of y with respect x. Well g is just the same thing as y2, right? So the derivative of—this is g and you don’t—you know normally when you do this, you don’t have to go to the whole problem of substituting for g. I just wanted to give you that intuition. So what’s this derivative right here? Well, it’s going to be—the derivative of this expression with respect to y from the chain rule. So that is 2y times the derivative of y with respect to x. The derivative of y with respect to x is equal to zero. When you do it mechanically, it’s actually quite simple, you know. You just say, “Oh, well let me pretend like y is like an x, take a derivative and I multiply it times the derivative of y with respect to y.” Mechanically, it’s very simply but I really want you to understand that intuition that this is coming out of the chain rule. Well then actually the chain rule itself is not this—you know bizarre concept. It’s really—you’re just multiplying fractions. So let me clear up some space here and we can finish this problem and then I’ll do a problem where you have to do implicit differentiation and even in this one that I just gave. It’s really useful to do implicit differentiation. So, let me—I could clear this guy out up here. Clear that out. I could clear all this stuff out. There we go. Clear all of that out. We’re ready to do some more work. So, now we could solve this for the derivative of y with respect to x. So we get. Let’s subtract 2x from both sides, so we get 2y—oh sorry. Oh yeah, 2y dy/dx =—I’m just subtracting 2x from both sides—-2x and then what do we get? We get dy over dx =—divide both sides by 2y. The 2’s cancel out, you get minus x over y. And this is interesting just as it is, right? Because what was this? This was the equation for a circle with radius 10, right? You know r2. And if you know your trigonometry and all of that, this tells you that the slope of a circle at any point is equal to the negative of the x over the y. Well, you know, we know from trigonometry that the x is the cosine and the y is the sine and so that tells you something interesting about how the derivative is really to the tangent and all of that but anyway, I won’t go into that, we’re just learning implicit differentiation here. But it’s sometimes fun to just think about things. So, how is this different from everything else we’ve learned so far? Well here, we have the derivative—the derivative is actually a function of both the x and the y value. And if we feel really uncomfortable having this y here; if we don’t like that, what we could do is we can substitute. So let’s solve for y here. We get y2 = 100 - x2 and then you get y = ± √100 - x2. And so you just substitute that in here. So you get—another way of writing the derivative y with respect to x, sometimes you’ll see y prime. I am just switching notation just so you’re familiar with it—is equal to minus x over the ± √100 - x2. And to some degree, since we have plus or minus down here, this minus sign, it kind of does not matter, right? Because when you have some minus here, you have the plus or this plus is minus, so we can kind of ignore this. That is something interesting for you to think about as well. But anyway—and this was interesting because our traditional way of solving derivatives at—we have never seen a plus or minus here, so that might be a little bit interesting to you. So now, let’s do a problem that is actually a bit of a trick problem. It’s not a trick problem but it’s actually something that I think—when I first saw it either. When I was practicing for the AP Calculus exam, we actually saw it on the AP Calculus exam. So, it’s something that you might encounter but it’s really neat. Let’s say I have y = xx. So at first you say, “Oh, this looks easy. Maybe I could use a parallel.” But no chain rule but have 2 x’s here and I mean you might want to just pause and think about this for a little bit because this actually isn’t that straightforward a problem. And even though right here, y is explicitly defined in terms of x, right? If I give you x—if I say x is 3, well then y is 33, y is 27. So this is an explicit function. I explicitly defined y in terms of x. But it’s very hard to differentiate explicitly. And here is a trick. Let’s take the natural—or actually you could take any logarithm but let’s the natural log of both sides of this equation. So you get the natural log of y is equal to the natural log of xx. And why did I do that? It is because now, I can use what we know about logarithm properties. We know that log ab = b log a, right? You just want the exponent properties videos if you forgot that. So let’s rewrite this like using this property and I am going to switch colors. So we get the natural lny = x ln x. And now we can implicitly differentiate this. So what’s—so let’s take the derivative with respect to x of both sides of this equation. And I’ll switch colors again. Nope, that’s not what I wanted to do. d/dx—and I am writing this just because—just so you know what I am doing—= d/dx [x ln x]. So the derivative of this expression with respect to x is equal to the derivative of this expression with respect to y times the derivative of y with respect to x. So what’s the derivative of this with respect to y? What’s 1 over y? 1 over y times a derivative; just the ―times the derivative of y with respect to y. dy/dx =—and remember, just mechanically if you want to know—you know make sure you’re doing it right, you just take the derivative with respect to y and you multiply it times dy/dx. That is just the chain rule—equals and I would take the derivative of this. Well now, we can just use the product rule, the derivative of the first, right? It’s this times this. So what is the derivative of x? It’s 1. So it’s 1 × ln x +—the derivative of the second term—1 over x ×—the first term—x. Let me clear that up there. I am already overtime but I’ve become a YouTubeb partner so my videos can be longer but I don’t want to make them too long because you’ll get bored all right? So let’s simplify this. We get 1 over y dy/dx = ln x +—well that just cancels out and becomes 1, right? And so we have the derivative of y with respect to x is equal to y—I am just multiplying both sides of this equation times y. y × ln x + 1. Now, you might say you’re done but some people just don’t like it the change of y with respect to x. They don’t like it being defined in terms of y. So let’s substitute. We know that y = xx, so that equals xx × ln x + 1. So, another way of saying it is that y prime or the change of y with respect to x or the slope at any point of this curve is equal to xx ln x + xx. I just distribute the xx but that is pretty neat. This is kind of a—you know a five start problem that you might see in your books and you might seen it in a math competition and people will be really impress. They’d be even more impressed if you kind of recognize this pattern and you are able to find the antiderivative of it and recognize that it is xx. Anyway, I’ve talked for too long. I will see you in the next video.

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Learn about Implicit Differentiation

Welcome to the video on implicit differentiation. So, let’s say that I have—well let’s just explain the difference between implicit and explicit first. So, if I had a function that was; let’s say y =—I don’t know, x2 + 2x + 3. In this situation, y; the variable y is defined explicitly in terms of x. Well how do I know that? Well, if you give me an x, I can just input it into this expression right there and I explicitly get y, right? It just pops out of the equation.

But on the other hand, what if I had something like this; what if I had x2 + y2 = 100. Well now, if I give you an x, you still can figure out y but it’s just not as easy, right? You would—if I said x =—I don’t know—x = 9. You would put 9 in there, you would get 81 + y2 = 100. You would subtract that from; so you get y2 = what? 19. I don’t know. These numbers are weird. And then you get y = ± √19. All right.

So, you still could solve for y but you have to—but this expression right here or this equation defined y implicitly. And you know that’s the different—you know. When people use the word explicit or implicit in everyday language, it’s the same thing when you say—you know, he explicitly said this; that means—you know. I am hungry. I explicitly stated that I am hungry. But you could implicitly state that you’re hungry. You could say, “Oh well, you know, I haven’t eaten for a long time and you know—boy, it would be nice to maybe eat something.” And that would be kind of an implicit expression of your hunger because someone would have to kind of do a little bit work to come to the conclusion that you are hungry, instead of you telling them.

So with that out of the way, let’s actually do some differentiation. So what everything you’ve learned about differentiation right now has been explicit differentiation. So example, for this function, we know how to take the explicit derivative here. Well, it’s explicitly defined function, so we just take the derivative. We apply the derivative operator. We take the rate of change of both sides of this equation with respect to x. So you could do it like this. We’re taking the derivative operator. I know this is getting messy. We’re taking the derivative operator on that side and we’re also taking the derivative operator on this side and all that. I didn’t plan to write that there, that’s why it’s—

So what did happen? When the left side, the rate of change at y changes with respect to x, well, we just write that. The rate of change of y for every small change in y, what does this change in x? That equals and then we know how to this type of derivative. This is just a simple quadratic. So that equals 2x + 2 + 0, right? So there we go. We just did some explicit differentiation.

So, how do we do the same thing here? Well, we’re going to do the same thing. We’re going to apply the derivative operator to both sides of this equation. So let me do that. I’ll do that in a different color. Let me erase all of this too because this is kind of obvious things that you might already know. Let me erase all of that. Okay. So I’ll take the derivative operator to both sides of this equation. So, this is no different than what we did up here. And as you learn in algebra, anything you do to one side of equation, you have to do it to the other side of function. Otherwise, it would no longer be an equation.

Take the derivative operator because we’re taking the derivative—the respect to x. We could have taken it with respect to y. If x2 + y2 =—if you take the derivative of both sides, right? What is the derivative? Well 100 with respect to x right? So this equals the derivative with respect to x. Let me put in brackets. I was going to do parentheses but I think bracket—of x2 plus the derivative with respect to x of y2.

And what is the derivative with respect to x of a 100. For every change in x, how much does this change? Well, it doesn’t change at all. It’s going to stay a 100, right? So this is zero. The derivative of a constant really respect to any changing variable is going to be zero. So that was just zero.

And what’s the derivative of x2 with respect to x? Well, we did that up here, right? When we took this derivative, we just took the derivative each of these terms, the derivative of x2 with respect to x. Well, that is just 2x. So, we get 2x +. Now this term here is where you might get a little bit confused. I am taking the derivative of y2 with respect to x, the derivative of y2 with respect to x.

So let me take you on this side. You know you should not do this every time that you do one of these implicit differentiation problems but I am doing this because I want you to understand what we are doing.

So let’s say I just defined g = y2, right? So let’s say that this was g. And I want to take the derivative of g with respect to x. So I apply the derivative operator on both sides of this. So let me do that. So the derivative—I’ll do a d of x of that side. d of x on that side. That is a d. There you go.

So, what we get is the derivative of g with respect to x is equal to the derivative with respect to x of y2, right? And so far, we haven’t really done much. I’ve just rewrote things. But what does the chain rule tell us? The chain rule tells us that the derivative—let me stay in the same color. I was switching colors. The derivative of any variable with respect to x or with respect to another variable is equal to the derivative of g with respect to y times the derivative of y with respect to x. And you could view this if you don’t have the full intuition of the chain or—yeah there’s a couple of ways to view it.

You know, all of these dg’s and dy’s that well you see throughout integral and differential calculus, all these really means are really, really, really, really small change in g, a really, really, really small change in y, a really, really, really small change in x. So, you can almost view this—so you know, if you have a—when you multiply this out. This really, really small change in y is the same as this really, really small change in y. So if these were—theses are essentially fractions, so you could cancel these out and that is why you would get the derivative of the change in g with respect to x or the derivative of g with respect to x.

So this is what the chain rule tells us. So, given that this is what the chain rule tells us, how does that help us figuring out this? Well, we know that the derivative of g with respect to x is equal to the derivative of g with respect to y. But we already know that g = y2. I just made that definition. That’s where we started from. So what’s the derivative of g with respect to y? Well that’s just 2y, right? Just using our derivative rules or polynomial rules which hopefully you believe actually work because there are several videos where we prove it. And then times the derivative of y with respect to x, which we don’t know and that is actually kind of what we’re going to try to solve for.

So, the derivative of g with respect to x is 2y times the derivative of y with respect x. Well g is just the same thing as y2, right? So the derivative of—this is g and you don’t—you know normally when you do this, you don’t have to go to the whole problem of substituting for g. I just wanted to give you that intuition.

So what’s this derivative right here? Well, it’s going to be—the derivative of this expression with respect to y from the chain rule. So that is 2y times the derivative of y with respect to x. The derivative of y with respect to x is equal to zero. When you do it mechanically, it’s actually quite simple, you know. You just say, “Oh, well let me pretend like y is like an x, take a derivative and I multiply it times the derivative of y with respect to y.” Mechanically, it’s very simply but I really want you to understand that intuition that this is coming out of the chain rule. Well then actually the chain rule itself is not this—you know bizarre concept. It’s really—you’re just multiplying fractions.

So let me clear up some space here and we can finish this problem and then I’ll do a problem where you have to do implicit differentiation and even in this one that I just gave. It’s really useful to do implicit differentiation. So, let me—I could clear this guy out up here. Clear that out. I could clear all this stuff out. There we go. Clear all of that out. We’re ready to do some more work.

So, now we could solve this for the derivative of y with respect to x. So we get. Let’s subtract 2x from both sides, so we get 2y—oh sorry. Oh yeah, 2y dy/dx =—I’m just subtracting 2x from both sides—-2x and then what do we get? We get dy over dx =—divide both sides by 2y. The 2’s cancel out, you get minus x over y.

And this is interesting just as it is, right? Because what was this? This was the equation for a circle with radius 10, right? You know r2. And if you know your trigonometry and all of that, this tells you that the slope of a circle at any point is equal to the negative of the x over the y. Well, you know, we know from trigonometry that the x is the cosine and the y is the sine and so that tells you something interesting about how the derivative is really to the tangent and all of that but anyway, I won’t go into that, we’re just learning implicit differentiation here. But it’s sometimes fun to just think about things.

So, how is this different from everything else we’ve learned so far? Well here, we have the derivative—the derivative is actually a function of both the x and the y value. And if we feel really uncomfortable having this y here; if we don’t like that, what we could do is we can substitute. So let’s solve for y here.

We get y2 = 100 - x2 and then you get y = ± √100 - x2. And so you just substitute that in here. So you get—another way of writing the derivative y with respect to x, sometimes you’ll see y prime. I am just switching notation just so you’re familiar with it—is equal to minus x over the ± √100 - x2. And to some degree, since we have plus or minus down here, this minus sign, it kind of does not matter, right? Because when you have some minus here, you have the plus or this plus is minus, so we can kind of ignore this. That is something interesting for you to think about as well.

But anyway—and this was interesting because our traditional way of solving derivatives at—we have never seen a plus or minus here, so that might be a little bit interesting to you. So now, let’s do a problem that is actually a bit of a trick problem. It’s not a trick problem but it’s actually something that I think—when I first saw it either. When I was practicing for the AP Calculus exam, we actually saw it on the AP Calculus exam. So, it’s something that you might encounter but it’s really neat.

Let’s say I have y = xx. So at first you say, “Oh, this looks easy. Maybe I could use a parallel.” But no chain rule but have 2 x’s here and I mean you might want to just pause and think about this for a little bit because this actually isn’t that straightforward a problem. And even though right here, y is explicitly defined in terms of x, right? If I give you x—if I say x is 3, well then y is 33, y is 27. So this is an explicit function. I explicitly defined y in terms of x.

But it’s very hard to differentiate explicitly. And here is a trick. Let’s take the natural—or actually you could take any logarithm but let’s the natural log of both sides of this equation. So you get the natural log of y is equal to the natural log of xx. And why did I do that? It is because now, I can use what we know about logarithm properties. We know that log ab = b log a, right? You just want the exponent properties videos if you forgot that.

So let’s rewrite this like using this property and I am going to switch colors. So we get the natural lny = x ln x. And now we can implicitly differentiate this. So what’s—so let’s take the derivative with respect to x of both sides of this equation. And I’ll switch colors again. Nope, that’s not what I wanted to do.

d/dx—and I am writing this just because—just so you know what I am doing—= d/dx [x ln x]. So the derivative of this expression with respect to x is equal to the derivative of this expression with respect to y times the derivative of y with respect to x. So what’s the derivative of this with respect to y? What’s 1 over y? 1 over y times a derivative; just the ―times the derivative of y with respect to y. dy/dx =—and remember, just mechanically if you want to know—you know make sure you’re doing it right, you just take the derivative with respect to y and you multiply it times dy/dx. That is just the chain rule—equals and I would take the derivative of this.

Well now, we can just use the product rule, the derivative of the first, right? It’s this times this. So what is the derivative of x? It’s 1. So it’s 1 × ln x +—the derivative of the second term—1 over x ×—the first term—x. Let me clear that up there. I am already overtime but I’ve become a YouTubeb partner so my videos can be longer but I don’t want to make them too long because you’ll get bored all right?

So let’s simplify this. We get 1 over y dy/dx = ln x +—well that just cancels out and becomes 1, right? And so we have the derivative of y with respect to x is equal to y—I am just multiplying both sides of this equation times y. y × ln x + 1. Now, you might say you’re done but some people just don’t like it the change of y with respect to x. They don’t like it being defined in terms of y. So let’s substitute. We know that y = xx, so that equals xx × ln x + 1. So, another way of saying it is that y prime or the change of y with respect to x or the slope at any point of this curve is equal to xx ln x + xx. I just distribute the xx but that is pretty neat.

This is kind of a—you know a five start problem that you might see in your books and you might seen it in a math competition and people will be really impress. They’d be even more impressed if you kind of recognize this pattern and you are able to find the antiderivative of it and recognize that it is xx.

Anyway, I’ve talked for too long. I will see you in the next video.

Transcription by: Scribe4you Transcription Services