Calculus: Introduction to Implicit Differentiation

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Implicit Differentiation
Implicit Differentiation Basics
An Introduction to Implicit Differentiation Page [1 of 3]
Well, you should feel great at this point. Maybe you're a little bit tired, but you really should feel great. Because you are now empowered to basically take the derivative of almost any function that someone can throw at you. You know, that is really cool.
Now, no matter what the function is--someone could give it to you. It could have sines. It could have cosines, natural logs. It could have exponentials. It could have x squared. It could be all sorts of crazy things. But what we know is that if someone gives you the function--let me draw it to you in terms of a picture. If someone gives you some kind of crazy function--or not crazy function--you are now empowered to find the derivative--to figure out the slope of the tangent line at any of these points, no matter where the point is.
Maybe it's over here. There's a tangent line. Maybe it's over here. There's the tangent line. You can actually find the slopes of these lines by computing the derivative. You can actually take the derivative of really complicated-looking objects. But can you take the derivative of things that aren't functions?
What does that mean exactly--taking the derivative of things that aren't functions? Well, what's a function again? Well, you might want to go back and review the little function discussion we had a while back. Well, remember that a function is nothing more than a little machine where you input an x and you output a particular y. So, for something not to be a function--and in math terms, we call that a relation--what that means is that for certain x's, it might not be clear what the value of y is, because there might be a lot of different answers.
And in fact, let me show you what a non-function would look like just graphically. I'm going to draw you a really, really, really non-function. Look at this. Woof. It's pretty, but it's not a function, because--and maybe your remembered this from your days in high school--it fails the vertical line test. Do you remember the vertical line test?
The vertical line test is if whenever you draw a vertical line, if you only pierce your graph at one point, then the thing is always a function. Because for any particular x, there's a unique y. So, you put in an input, spits out an output. Look at this. This thing dramatically fails the vertical line test. I put it down here, and notice that, in fact, this red curve crosses this vertical line at two points. So, if that x value were, let's say, three, if I say to you, "What's the value of the curve at three?" You'd say, "I don't know. It's negative one or it's positive two." It's not a uniquely determined thing. So, this is not a function. It's a relation.
The thing about this relation, though, is that it's very smooth. Could you imagine running your hand along that? It would be so nice, and it would feel so good. In particular, notice that it has all sorts of tangent lines. Tangent lines are very pretty, and they nicely just track along. Here, look how pretty this goes along here. Down here, you can see how beautiful the tangent lines just go up against it, very lovingly.
Certainly these tangent lines have slopes. And so, a natural question to ask is how can we find the slopes of the tangent lines of things that aren't functions? Well, we don't know how to take derivatives of things that aren't functions. We only know how to take derivatives of functions themselves. So, this inspires us to think about a way of actually finding derivatives of things that aren't functions.
By the way, let me just show you a very pretty and very important example of a non-function. This one is a very exotic one, artistic one. But this is probably the most important example of a non-function--the circle. The circle is not a function, because you can easily see, it fails the vertical line test. However, a circle, which is so symmetric and so smooth, has tangent lines all around it, and it would be nice to be able to find the slope of those tangent lines.
So there's a nice example, and there's a very pretty example. You can certainly see the issue. The issue is how do you find derivatives of objects that are more exotic than functions? Well, the answer is I don't know. So, let's see if we can bootstrap our way up to that by first considering functions again.
So, let's now go back--you know, when you're starting something new, and you don't know what to do, of course, the thing to do is stop, regroup and study what you know as carefully as you can, understand it really, really well, and then push forward.
Now, what I'd like to do is consider an example of just a simple function that it's not a big deal to take the derivative of. I'm going to use the y-notation here, and I'm going to introduce some new language. So, this is going to be a little bit of a language lecture first. So, suppose I have y equals--and I'll put a simple function in here just for fun--three x squared minus four x plus one.
And now, suppose the question is to find dy/dx. First of all, remember what dy/dx means. Dy dx is the derivative of y with respect to x. Let me write that out, because I really want you to think about that for a second. Recall what that means. Dy dx, which is what we used to call--well, we sometimes call--f prime of x. But if we use the y-notation, we call it dy/dx. This is the derivative of y with respect--oo, oo, oo--just made it--to x.
So, that's what this symbol means. It's just the derivative. And there's the fancy phrasiology. The derivative of y with respect to x. Why is it, "with respect to x"? Well, because I have y as a function that depends upon x. So, if I want to see how it's changing--the relative changing--it's changing with respect to x. So, this is the derivative of y with respect to x.
Now, in fact, if you notice the way I wrote this, there's actually a little vocabulary quiz here. I can actually ask a grammar question. What kind of object is this thing? Well, dy/dx is actually an object. It's a noun. So, in fact, dy/dx is a noun. Here's the little grammar lesson for us right now. Dy dx is a noun. It's an object.
What is it? It's the derivative of y with respect to x. So, whenever I write dy/dx, I want you to think of that as an object. Okay, now, how do we traditionally do this problem? Well, traditionally, we just spit out the answer. I mean, that's how good we're getting at these things. What's the answer--answer? And we can just spit. Go ahead, spit--heh, heh--but not on the monitor. So, here's the answer. Dy dx equals six x minus four plus zero. Just spit out the answer. Not a big deal. We just take the derivative.
Now, there I am using the fact that, well, this is a function. Y is a function that depends upon x. I just took the derivative. What I'd like to do is try to enlarge this process to show detail and do this in slow motion. Instead of just spitting it out, I want to actually work through it in a couple of little steps. I'm still going to get the same answer, but now, I'm going to elongate that process by introducing some intermediate steps.
Now, you're saying to yourself, "What kind of intermediate steps do we have? This is such an easy function, how can he possibly prolong the solution to find the derivative to this?" Well, don't doubt me. I can prolong anything. In particular, I'd like to prolong the solution to take the derivative of this. And here's how I'm going to do it. I'm going to add in one little intermediate step.
And again, this intermediate step is going to seem really foolish, but what else is new. This is going to be, again, three x squared minus four x plus one. And the question again is find dy/dx. That, remember, is the noun--an object--the derivative of y with respect to x.
Now, how am I going to do that? Now, before, we just looked at this and said dy/dx equals six x minus four. What I'd like to do now is I'd like to add a little intermediate step--a step where I tell you a command what to do--differentiate. So, what I'd like to do is solve this problem in a slightly different way--in a way where we never actually use the fact that this was a function. You see, if I could figure out a way of finding the derivative of this without ever using the fact that this is a function, then that method, whatever it may be, might be extendable to look at things that aren't functions.
And so, here's the new notation. I'm going to write d/dx to be a command. I want you to think about this as a verb. This is an action, and the action is to take the derivative with respect to x. So, I want you to think about this as a command. Take the derivative with respect to x. So, that's a command. In particular, I want you to think about this thing d/dx as a verb. It's an action.
Remember that dy/dx is an object. It's the derivative of y with respect to x. This is a command. I'm telling you that I want you to do something. Now, let me show you how I'd actually write this symbolically. So, what I would write is the following. I would say okay, let's start with this entire thing that we have there. I'm just going to write that thing out. So, there is that thing.
And what I want us to do now is take the entire expression--equal sign and all. And since I'm trying to find the derivative of y with respect to x, I'm going to take that whole thing, and I'm going to tell you right now I want you to differentiate the whole thing. So, I'm going to write d/dx. So, that's a command, which says, "Take this whole expression, equal sign and all, and I want you to differentiate the whole thing with respect to x."
That's exactly what we did, but now I'm actually writing out the extra steps. This says take the derivative of this. Well, how do you do that? Well, we'll take the derivative of the left-hand side, and then we'll take the derivative of the right-hand side. So, let's actually write that out. If I take the derivative of the left-hand side, I would say take the derivative of that. And that would equal what you get when you take the derivative of this.
Notice how I'm using the language here. This thing is an operation. This thing is a command. This thing is an action. Take the derivative. It's not a derivative. It's the commandment to tell you to take the derivative of that.
Okay, now, what would that entail? Well, what that would entail is the following. In fact, let me just take this thing, and I'm just going to put it over there so you can see it. So, what do we have? So, I have to first look at the left-hand side and then look at the right-hand side. Well, let's look at the left-hand side first. The left-hand side is the command. "Take the derivative with respect to x of y."
Well, what does that equal? Well, let's think about that. That's the derivative of y with respect to x. And what is that? What is the derivative of y with respect to x? The answer is, it's dy/dx. So, if you actually perform the action that's being asked of us here, what we get is dy/dx. And why? Because remember that we've already said that dy/dx is the derivative of y with respect to x. And this commandment is asking us to take the derivative of y with respect to x. So, we get dy/dx.
Look how nicely this notation meshes, you see? This is a noun. And if you take this object and perform this action on it, you basically get dy/dx. So, the notation is sort of nice and consistent, even though there's sort of different meanings here. This is a command to do something to here, and this is what you get out.
Okay, so the left-hand side of this expression--the expression that's right on the side of me here, but let me just post it back up for you here--so, this left-hand side is just dy/dx. That's just dy/dx. And what about this? How do you take the derivative with respect to x of that? Well, that's something we already know how to do. We just take the derivative. That would be six x minus four.
So, in fact, the left-hand side, we see dy/dx equals--and the right-hand side, when we take the derivative of that side, which is actually--it's what we did before. We get six x minus four plus zero. And so, look. That's the answer we get. And you'll notice that's the exact same answer we got when we took the derivative just directly. We had the function. We just took the derivative.
Now, I put in a whole bunch of intermediate steps. In particular, I told you to take this entire thing and differentiate it with respect to x. So, I told you take that action--that verb--d/dx. And when you did that, what you got was that the left-hand side--the derivative of y with respect to x is dy/dx--and then, the right-hand side, the derivative of that quantity. All that stuff there was just six x minus four.
So, the bottom line is I actually got the same answer. But, you'll notice that never, ever did I use the fact that we had a function at our disposal. I just differentiated one side and differentiated the other side. I didn't care if there were x's or y's in there either with them. And if fact, this method that I'm outlining will actually be extendable to looking at things that are not functions.
So, we still haven't quite answered the question, but now we've built the language in order to resolve it. So, up next, we're going to actually attempt to find out derivatives of things that aren't even functions. Okay. Stay with us.