John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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Implicit Differentiation

John Postovit
John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.


What are you going to do when you can't solve an equation for a single variable? Call a ghost-busters? That's probably before your time guys. Anyhow, if you can't solve for a single variable, you can still do implicit differentiation. It's a really cool technique.

Implicit differentiation lets you differentiate any kind of relation. We are going to start in this episode by showing you that it really works. After we show you that it really works, we're going to try a couple of practice problems. Then we're going to move on to doing a second derivative. So, let's try it.

No fooling, yeah it really works. Let's try one out. This problem is asking you to find the derivative with respect to x of that little equation, you notice it? It hasn't been solved for y like we usually do. Who cares? Implicit differentiation? Who cares? Watch what we do. We start of with y - x² equals, I'm going to leave a little gap here, gaps for a reason.

Remember the very first thing you learned in Algebra was whatever you do to one side, you've got to do it to the other? Well I'm going to treat a derivative as something that you have to do to both sides. Weird, but it works. My goal here is to get dy/dx so what I have to do is do the derivative with respect to x on both sides, like this. d/dx. I'm going to put this in a parenthesis because it has to be done to that entire side. If I'm doing the derivative of one side I have to do the derivative of the other side. So you've got d/dx of 5. Okay, now we can go to town.

Here is something though, it's a key component of it that's kind of brand new. We have to do the derivative of both of these and the derivative of x² is no problem. But the derivative of y with respect to x, it's not even the same variable. Now, what you do though is you pretend that this is x for the time being and you do its derivative. The derivative of y is 1. The derivative of a single variable variable is 1. But you don't really know how y relates to x. So effectively this is actually like the chain rule. It's like y is composed of a bunch of xs and there is a bunch of stuff in there and you have to do its derivative as well. But I don't know how y relates to x so I just have to leave that derivative unfinished like this.

So I've got to do the derivative of y with respect to x but I don't know how to do it yet so I'm just going to leave it like that. Now let's do the derivative of that, that's easy. -2x and on this side the derivative of any constant is 0. We are almost done. Recognize that? That's just another way of saying the derivative. I'm going to solve for this fraction and I'm finished.

Let's see what we have here. We are going to add 2x to both sides and we get dy/dx equals 2x. In the next slide I'm going to show you that this really mysterious technique really works. What I'm going to do is solve this for y and do it the way we normally did derivatives. So I'm going to find the derivative of that, I'm going to solve for y. You're probably wondering by now, well why don't you just solve for y? It's so easy to do. For this one you probably would.

Because the derivative then is really easy to do, it was easy to solve for a y. The derivative of that side would be 0 + 2x. We got the same answer. This of course is the same as dy/dx in Leibniz's notation. So we got the same thing. The cool thing about implicit differentiation is that sometimes you've got something where it's really hard to solve for y and you don't even have to bother.

So you're probably thinking, yeah Postivit says this is amazing and wonderful but when am I ever going to use it? When you're going to use this is when you deal with differential equations. We're going to do some simple differential equations later in this episode and with related rates. Related rates are a great topic, they are a big topic and big enough that we are going to tackle them with another episode. So let's try one.

This is one where you really wouldn't want to do this without implicit differentiation. Because to do this one regularly you'd have to solve it for y but that would be tough because you have a y³ and a y factor. That would be really ugly to solve for y. We're not even going to try. What we're going to do is set it up. So let me do this set up, we would have d/dx. d/dx (y³ + 2y + 4x²) equals d/dx of 7. Remember what I said a little while ago that we really are using the chain rule. It looked like it but we are. Let me show you why. Let's say I was able to solve this thing for y. I'm just going to make something up. This isn't really what y would be.

Let's say if I was able to solve for y it turned out to be y equals x to the four ninths power minus x. I know what y is now and I'm going to substitute these ys with that expression. So I would have dy/dx of, this y has to be replaced with x to the 4/9 minus x. There is a third power on it so this entire quantity here is the y³ plus that.

Let's say you've substituted this and you're saying, now I'm going to do the derivative the regular old fashioned way what you're used to. Let's say, I got to do the derivative of this but it's not just x in there. So I take the 3 up front, make the power one less but then I'd have to do the derivative of the stuff on the inside. That takes the chain rule.

Now if it was this you could just go ahead and do it. But it's not. We don't know what it is, it's just y and we don't want to even try to find out what y is. This is a case where the chain rule factor would be just the derivative of y with respect to x and you leave it that way. Now let's do the problem. So I'm going to write that out one more time. We have d/dx (y³ + 2y + 4x²) equals d/dx of 7. You might be wondering, there's two variables in there, why did I choose to put dx on the bottom? Just because I told you so. You could have done dy/dx in his equation. It's possible. But we won't in this case. Later on though we will be doing a derivative with respect to other variables.

So the derivative of this, if this was just plain old fashioned x, its derivative would be 3x². We're going to pretend it is so it's a derivative. This is the outer part of the chain rule. It gives us 3y² but this is the chain rule because this is not the same variable that we're deriving with respect to. So we have to have that times dy/dx factor. Now, if this was 2x, it's derivative would be 2. So we're going to pretend it is x. Its derivative is 2 but it's not x so I've got to backtrack and put in that dy/dx factor. And the derivative of that is just 8x and the other side, the derivative of a constant is 0.

Well, our goal is to find out what the derivative with respect to x is. My next step, think about this, we'll get two of these, what I'm going to have to do, solve for them. Since there it two of them, I'm going to have to factor them out. I'm going to take and subtract 8x from both sides. On this side, what I have left is two factors that have dy/dx in them. I'm going to factor that out, binomial factoring. Looks like this, we have dy/dx and if I factor it out of this expression I get 3y². And factor it out of that expression there I just get 2 and we're almost done.

It's dy/dx times that whole parenthesis. All I need to do is divide both sides by that whole parenthesis and you've got the answer. So the final answer for this one, is dy/dx equals -8x over this whole quantity, 3y² plus 2. There it is. If you want to follow along with solutions, bonus materials contain follow along that will have these solutions to the practice problems we do. There's one more elaboration that we can do on this.

This is the same problem with just one more step. You may be asked on the AP test to go and find the slope of an equation like that at a particular location. This one right here I previously solved, if we do the derivative of this one with implicit differentiation you get dy/dx equals -4x/y. You might want to try this just for a little bit of practice. These are relatively easy ones to do. Let's see what happens. You recognize this shape, it's an ellipse, you've been doing those since algebra 2. Let's make a little graph here, it looks something like that. If you were to go and do the slope at x equals 1, 1 is about here on the graph, x equals at that spot. You get a slope of somewhere around -1, but, x is 1 at that spot too.

Which is okay, it can happen because this thing is a relation it's not a function. And at that spot looks like you have a slope of +1 or something like that. So you can actually get two answers. Look what we got here, the slope actually requires you to use x and y, both x and y in it. What you need to do in order to go on is actually get both coordinates, the x and y coordinates for finding the slope now.

Before you all just needed the x coordinate because you had slope formulas that only involved x. This one also involves y so this location is 1 and if I put 1 into this formula, we will get y equals 3. Double check that for a second here, yeah, we're in business. So for that location, the slope is -4 of, the x location is 1 over 3. So the slope for this one is -4/3. If I were to work this one out its coordinate would be (1, -3). If I use the slope formula one more time, I get the second slope for when x equals 1 and this one turns out to be +4/3.

Second derivatives. Oh the agony! Well, these are kind of long but you may be called on to do one. So you want to make sure you've practised some of these. The second derivative is a derivative of a derivative, in other words the rate of change of a slope or the slope of a slope. The next slide is going to show you the steps. You might want to download these from the bonus materials so you can follow along.

So our goal is going to be to do the second derivative of this shape, x² + y² equals 36. That's a circle. Notice how this is written, it's kind of funny. The square is on the d here but it's on the variable down here. It's always that way so just don't worry about it.

Here are the steps, the first two are the ones that we've already practised, do implicit differentiation, and then solve for dy/dx. Next, you do implicit differentiation one more time and then you usually have to substitute for y'. And then sometimes you even have to go all the way back to the original equation and substitute that in.

So first thing I need to do is the first derivative of this, I'm going to write that over here. So we have to do d/dx of x² + y² and that has to equal d/dx of 36. Derivative of x² is 2x, derivative of y² now. Remember it's not the same variable, y and x are different variables, that's what counts. We're going to do the derivative as if it was x², that would give us 2y. But since it really isn't x, we need that chain rule factor of dy/dx. We're still doing the first implicit differentiation. On the other side, the derivative of 36, it's a constant, it's just 0. But don't count on the derivative of that side always being 0, it's just been that way for the couple of problems we've done.

We do the differentiation, we have to solve now for dy/dx, that won't take long. In fact, you could probably do this one in your head. dy/dx I would take 2x from both sides to do this and then I'd still have 2y dy/dx on this side. So divide both sides by 2y. These 2s divide away just giving us x/y, I'm going to take this negative sign and put it in front of the fraction. So dy/dx equals -x/y.

I'm going to do one more thing that's going to be a time saver when you do the second derivative. We're going to start and write this notation is kind of long so I'm going to start using Newtonian notation, I call it y'. So y' equals -x/y. There it is, you've done steps one and two. We're going to go on to doing step 3. Step 3 is to do the derivative of this one more time. So I'm going to be doing d/dx of y' and that has to equal d/dx of -x/y. Second derivative both sides. First derivative of y' is y''. We're using the chain rule here too. It doesn't look like it but we really are. The derivative of this would be 1 times dy/dx. Over here, just stop and look at that for a second, that can confuse you. We've got to do the derivative of -x/y. We're dividing variables.

Think back a couple of episodes. Which rule do you use when you're doing a derivative of variables being divided? Quotient rule. That can really surprise you. Not only is it quotient rule, the variable down here isn't the same one as the one we're doing the derivative on. So we're going to have a chain rule as part of it. This is a quotient and chain rule combination. Quotient rule comes first though.

So you recall the quotient rule requires you to do the derivative of the top part times the non derivative of the bottom part. This minus I'm going to take that out front so we don't have to deal with it until later. This we can take it out as a factor. Derivative of the top part, that's just 1 times the non derivative of the bottom part that's y. Now we have to do the next part in the quotient rule which is subtracting the non derivative of the top, non derivative of the numerator, times the derivative of the denominator. The denominator, it's not the same variable so its derivative is going to be 1 times dy/dx. And normally I wouldn't bother writing out the 1 but I want to do that because I really want to make it obvious that we are using the chain rule here. Instead of writing dy/dx I'm going to write y' to save us some writing time.

Last part of the quotient rule is to have the square of the denominator. That would be y². We are close to done. I'm going to go to the next screen to do the substitution so we have to take this y' and substitute it away. Same thing, I'm just going to write it down again. y'' was -y minus xy' over y².

Remember y' was -x/y, I've got to substitute that next. So this is the -y - x times -x/y all over y². Got a little bit of simplifying to do. This is going to require you to have common denominators. So over here this fraction is going to become negative of negative, that's positive, +x² over y. And this one right here, this y, that's y/1 so I'm going to have to multiply by y/y. And this y becomes y² over y and it's all over y² so now you can add these two fractions divide by the y² and we've got our final result.

The final result is going to be -y² + x² all over y³. Take a deep breath, calm yourselves. Go back over those steps. If you're methodical about it, these really aren't too bad at all.

Well, we've just done a lot of interesting stuff with implicit differentiation. We started out by explaining how it works, proving it works, doing a simple example, a little more complex example. We went and found some actual slopes for a couple of spots lying ellipse and finally we finished up with doing a second derivative. But your head still might be spending right now so take a break, relax a little bit and then come back for the related rates episode soon because you are going to be using these techniques with related rates.

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