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 \( y \) is equal to our function \( f(x) \). 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 \( x^2 \) plus the derivative of \( y^2 \). 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 \( x^2 \), 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 \( y^2 \) 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 \( y \) is equal to some function, 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 \( y^2 \), 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 \( y^2 \) 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 \( 2x \). 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 \( dy/dx \) is equal to this expression on the right side, which is negative \( x/y \). 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 \( y \) is equal to the square root of \( 49 - x^2 \). 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 \( 49 - x^2 \). 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 is equal to 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 y', which is 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 - 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 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 simpler 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 negative two over x squared. So that's going to be this first term. Now next, we need to subtract off this next term, the derivative of 1y.

Now one over y can be written as y to the negative one power. 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 negative one over y squared. 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 dy/dx 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 2 over x squared 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 2 over x squared to cancel on this left side, meaning that all we're going to have on the left side is yy2 y'. We have y2 times y'. That's going to be equal to 2 over x squared. 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 y squared.

So if I multiply this side by y squared and that side by y squared, notice the y squareds 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 2 times y squared divided by x squared. 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 is equal to 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