Implicit Differentiation - Online Tutor, Practice Problems & Exam Prep

Finding The Implicit Derivative

At this point in the course, we should be quite familiar with how we can take derivatives of equations that look like this, where we have y=f(x). Now it turns out you're going to see problems in this course where you are not going to have this form of your equation. What's going to happen is y is going to end up being some variable inside the equation, rather than being on one side, but you're still going to be asked to find the derivative of y with respect to x. Now how would we do something like this? Well, it turns out there is a strategy, and that strategy is called implicit differentiation or the implicit derivative.

And I'm going to walk you through the steps in this video because I think you're going to find this process is actually very familiar to things we've already seen up to this point. So let's get started. Now when you need to differentiate a function where y is just some variable inside the equation rather than being on one side of the equal sign, what we need to do is take the derivative with respect to x of each term on both sides, or the derivative of whatever variable you are with respect to. In this case, we're dealing with x. Now for this implicit derivative here, if we want to find the derivative of y with respect to x, I'm going to make this, what we just talked about, our first step.

So I'm going to take the derivative of this entire equation right here, meaning we're going to have the derivative of x2 plus the derivative of y2. Keep in mind, this is all with respect to x, and then it's all going to be equal to the derivative of 49. Now when we see this derivative here, there are some steps that I know how to do already, because I know that the derivative of x2, using the power rule, is just going to be 2x. And I also know that the derivative of 49 or any constant is just 0. So really the big question of this equation here is how do we take the derivative of y2 with respect to x?

That seems kind of curious. Right? Well, it turns out there is a process for this. Now what you want to do is use the chain rule. And the chain rule is something that we've discussed, where you have the derivative of the outside function multiplied by the derivative of the inside function, assuming you have a function within a function.

So let's say, for example, we have y=f(x), and we want to take the derivative of this function, it'd be the derivative of the outside multiplied by the derivative of the inside. Now that strategy, we can also use with taking the derivative of one variable with respect to another. So what you need to do is think about what the outside function is and what the inside function is. Well, I know that the derivative of y2, if I just use the power rule, would just be 2y. But I also need to multiply this by the derivative of the inside function, which would just be y.

And the derivative of y with respect to x, well, that's going to be dy/dx right here. So this would be our derivative of y2 with respect to x. So really, it's just one extra step, you use the power rule, but then you also have to find this inside derivative right here, because you need to use the chain rule as well. So this right here would be our derivative. And notice something, once we do this, we can actually clearly see where the dy/dx is, and that's what we were trying to solve for in this problem.

So what do I need to do here? Well, at this point, it just becomes a basic algebra problem. So looking at the situation that I have, I can first take this 2x and subtract it on both sides of the equation. Now that's going to get the 2x to cancel on the left side, meaning that all I'm going to have on the left side is that 2y times dy/dx is equal to negative two x. Now keep in mind, I'm trying to isolate this dy/dx on one side of the equation.

So what I can see here is that I can cancel the twos on both sides of the equation because dividing 2 on both sides will get those to cancel. And then from here, I just need to divide this y out. So dividing the y on the left side will get those y's to cancel right about there, and then I'm just going to have my dy/dx by itself. So that the dy/dx is equal to this expression on the right side, which is negative x over y. And this right here would be the solution to our problem, and that is the implicit derivative right there.

Now something else that I want to mention here is that notice how we got y inside of our final equation when we found dy/dx. Sometimes you're going to be asked to get the entire thing in terms of x only, so you'll need to find a way to eliminate this y. But the way that we can actually do that is by going up to our original equation that we had and solving it for y first. So if I solve this for y, which you can do just using some algebra here, you should get that y×y=49-x2. You just need to move some things around and then square root both sides to get this y by itself.

So you can take this expression for y right here and plug it in for y in the relationship that you got. If we do this, we'll get that dy/dx is equal to negative x divided by this term for y right here, which is the square root of 49-x2. And notice something here. Notice the result that we got. This result is the exact same thing that we would have gotten if we just had y solved for on one side of the equation already.

It's the same result for both of these, so why exactly did we even need to do this process for implicit differentiation? Well, that's because you're going to see situations where you actually have to use this strategy that we talked about in this video. You're not always going to have these nice equations where you can just get y on one side easily. Sometimes you'll have y raised to a really high power, or you'll have multiple y terms in the equation, in which case using this implicit differentiation strategy is going to be much more efficient than just having y on one side or solving for y and finding the derivative with respect to x. So this is the process for finding the implicit derivative.

Hope you found this video helpful.

Find $dy/dx$ for the equation below using implicit differentiation.

$5x^2+y^3=12$

• A. $dy/dx=-\frac{10x}{3y^2}$
• B. $dy/dx=\frac{10x}{3y^2}$
• C. $dy/dx=\left(12-5x^2\right)^{\left(\frac13\right)}$
• D. $dy/dx=-\left(12-5x^2\right)^{\left(\frac13\right)}$

Find $dy/dx$ for the equation below using implicit differentiation.

$3\sqrt{y}=x-y$

• A. $dy/dx=3\sqrt{y}-x$
• B. $dy/dx=\frac{2\sqrt{y}-3}{4\sqrt{y}}$
• C. $dy/dx=x-3\sqrt{y}$
• D. $dy/dx=\frac{3}{3\sqrt{y}-2xy}$

Find $dy/dx$ for the equation below using implicit differentiation.

$xy=\sin\left(y\right)$

• A. $dy/dx=\frac{\sin\left(y\right)}{x}$
• B. $dy/dx=\frac{x}{\cos\left(y\right)}$
• C. $dy/dx=\frac{y}{\cos\left(y\right)-x}$
• D. $dy/dx=\cos\left(y\right)$

Finding The Implicit Derivative Example 1

Let's give this problem a try. So here we are given this equation, 2x-1y=36. And the first step here asks us to find y' using implicit differentiation. Now something I'm going to tell you before we get into this problem is this is going to be a more complicated case that you'll see with implicit derivatives. But if you can confidently solve this problem that we're about to go over, I think that you're going to be pretty solid on this concept of implicit derivatives.

So let's get right into things to see how we can solve the more complicated problems we see in this course. So what I'm going to do first is rewrite this equation. You'll have 2x, and that's going to be minus 1y is equal to 36. That's the equation we're dealing with. And this first part asks us to find dy/dx, using the implicit differentiation.

Now the first thing I'm going to do is take the derivative of this entire equation right here. So doing that means I can distribute this derivative into every single term that I see. So we're going to have the derivative with respect to x of 2x minus the derivative with respect to x of 1y is equal to the derivative with respect to x of 36. Now from here, I just need to deal with each one of these terms separately. Now the derivative of 2x, well, you need to recognize that 2x-1 is the same thing as 2 times x to the minus 1.

We can always take whatever is in the denominator and bring it to the numerator as long as we make it a negative exponent. Now the reason I did this is because notice the power rule becomes a lot more simple when we take this derivative. So we get negative two times x to the negative two power, which that whole thing is going to end up becoming -2x2. So that's going to be this first term. Now next, we need to subtract off this next term, the derivative of 1/y.

Now one over y can be written as y-1. Again, we can use this negative power trick here when we're dealing with these fractions. And, again, we can use the power rule right here, which will give us that this whole derivative ends up becoming -1y2. But this is not the final derivative because notice we're taking the derivative of y with respect to x. Here we had all x's, so it was fine.

But since we're taking the derivative of a different variable here, we need to multiply this by dydx to show that we did the chain rule as well. And dy/dx is just the same thing as y'. I'm going to write it like this just because it's less to write, and I think it's going to make the algebra a little bit easier later. So we get this as our result for y', then we have the derivative of 36, which 36 is just a constant. So this whole thing goes to 0.

So this is going to be our derivative right here. Now notice I'm trying to find y'. So what I'm going to do is try to isolate y' on one side using some algebra. Now if you cancel the negative signs right there, I'm also going to add 2x2 on both sides of this equation. This is just some algebra that I'm doing.

Now what this is going to do is get the 2x2 to cancel on this left side, meaning that all we're going to have on the left side is 1y2y'. We have 1y2y'. That's going to be equal to 2x2. Now that's what we get on this other side of the equation. Now from here, I can go ahead and multiply both sides of the equation by y2.

So if I multiply this side by y2 and that side by y2, notice the y2 will cancel right there. So all we're going to be left with on the left side is y'. And then on the right side, we're going to get 2y2x2. And this right here would be our result for y'. So that's how we can find the implicit derivative, and that would be our answer for step a.

Now let's move on to step b. Step b asked us to solve the equation explicitly for y and differentiate to get y' in terms of x. Now this is, in my opinion, the trickiest part of this problem, because we have to take this entire equation right here, and what we have to do is we have to rewrite it so y is on one side of the equation. There's a lot of fractions here, so the algebra is actually going to get pretty tedious. So we have this equation, 2x-1y=36.

Now the first thing I'm going to do is subtract this 2x} on both sides of the equation in a pretty straightforward way that will get the 2x to cancel on this left side here, leaving me with just negative 1y. That's all going to be equal to 36 minus 2x. Now from here, I can multiply both sides of this equation by negative one. Let's go and get the negative here to cancel, leaving me with 1y. And that's going to be equal to we're going to have to flip these 2 since we're multiplying everything by negative one.

We're distributing the negative one completely in here, giving us 2x-36. So that's what this equation is. Now from here, I need to take the reciprocal on both sides of the equation because the reciprocal of 1y is just going to give us y. But then I need to take the multiplicative inverse here as well, which is going to give me 12x-36. Now I'm not just going to write it like this to my final result because I think I can get this to look a little more simple than what we have.

The way I can do that is by multiplying 36 by x and then dividing by x as well. I can do this because those x's would just cancel, but notice this gives me a common denominator. So we'll get that 12-36x, subtracting these two values, and then that whole thing will be divided by x, the common denominator, in which case I can just flip this entire fraction and bring it up to the top, giving me that y is equal to x2-36x. So this right here would be our answer for y. I noticed that this is a much more simple looking answer than that.

So this is going to be the equation that we go with. So that's what we found for y. Now the next question becomes, how exactly do we use this equation that we just found to solve part b? Well, part b asks us to find y', the derivative. So if this is what we got for y, what I need to do is find the derivative of this whole thing.

And to do that, I can use the quotient rule. The quotient rule says that the derivative of the top of your fraction minus the bottom of your fraction is equal to the derivative of the top times the bottom minus the top times the derivative of the bottom, all divided by the bottom square. So since we can see this is the top of our fraction, this is the bottom of our fraction, that's what we can do to find this derivative. So I'm going to take the derivative of x, this top piece here, which is just going to be 1. Then I can multiply that by the bottom here, which is 2 minus 36 x.

Now what I can do from here is subtract off the top of this fraction, which is x, multiplied by the derivative of the bottom of this fraction, which the derivative of 2, that's just going to be 0, and then the derivative of negative 36 x is just going to be negative 36. So that that's the derivative of negative 36 x right there, and this is what we end up getting. Now all of this is going to be divided by the bottom piece squared, which is going to be 2-36x2. So this is what we get. Now from here, I can simplify things.

Because I can multiply this one out through here, so we're going to get 2 minus 36 x. And I can cancel the negative signs, leaving me with positive 36 x. And all of that is going to be divided by 2-36x2. That's going to be y'. Now from here, notice that the negative 36 and positive 36 x on top will cancel.

So this whole thing is just going to become 2 all divided by 2-36x2. So this would be our result for y'. So this is what happens when we take the derivative of y with respect to x, and we have y already solved for on one side of the equation. So this is the result we got here. That would be our solution for part b.

Now this is our solution for part b. Notice that it has x in the equation and only x. This is our solution for part a, and notice it has both x's and y's. But that's what happens when you take the implicit derivative versus taking the derivative where y is already on one side. But notice what part c asks us to do.

Part c asks us to check that our solutions are consistent by substituting the expression for y derived in this part right here and plugging it into this equation we got to see if these two solutions are going to match. So let's go ahead and do that. Now at this point, I can clear this quotient rule because we should know how to do this already, and I can go ahead and clear this as well. Now what we're trying to do is see if the equation that we just got is accurate by checking it with what we did for the implicit derivative. So let's figure this out.

Well, what I can do is take this entire expression for y right here and plug it in for y in this equation. That's going to give me that y

Here’s what students ask on this topic:

What is implicit differentiation and when is it used?

Implicit differentiation is a technique used to find the derivative of a function when y is not isolated on one side of the equation. It is particularly useful for complex equations where isolating y is difficult or impossible. To perform implicit differentiation, you take the derivative of each term in the equation with respect to x, using the chain rule to differentiate y terms, treating y as a function of x. This method allows you to find dy/dx even when y is embedded within the equation.

How do you apply the chain rule in implicit differentiation?

In implicit differentiation, the chain rule is applied when differentiating terms involving y. For example, if you have y², you treat y as a function of x. The derivative of y² with respect to x is found by first taking the derivative of the outer function (y²) with respect to y, which is 2y, and then multiplying by the derivative of the inner function (y) with respect to x, which is dy/dx. Thus, the derivative of y² with respect to x is 2y(dy/dx).

Can you provide an example of implicit differentiation?

Sure! Let's differentiate the equation x² + y² = 49 implicitly. First, take the derivative of each term with respect to x: d/dx(x²) + d/dx(y²) = d/dx(49). This gives us 2x + 2y(dy/dx) = 0. To solve for dy/dx, isolate it: 2y(dy/dx) = -2x, then divide both sides by 2y: dy/dx = -x/y. This is the implicit derivative of the given equation.

How do you express the final result of implicit differentiation solely in terms of x?

To express the final result of implicit differentiation solely in terms of x, you need to solve the original equation for y and substitute this expression into the derivative. For example, if you have x² + y² = 49, solve for y: y = √(49 - x²). Substitute this into the implicit derivative dy/dx = -x/y to get dy/dx = -x/√(49 - x²).

Why is implicit differentiation important in calculus?

Implicit differentiation is important in calculus because it allows us to find the derivative of functions where y is not easily isolated. This technique is essential for dealing with complex equations, such as those involving multiple y terms or higher powers of y. It provides a systematic way to differentiate these equations and is widely used in various fields, including physics, engineering, and economics, where such complex relationships frequently occur.