 Let's actually go to sort of like the original setting. Can we find the tangent line of a circle? Well, this could be done from a completely trigonometric perspective, but we're going to use calculus here because it's a whole lot easier. Let's take the equation x squared plus y squared equals 25. Let's find the derivative dy over dx implicitly and use that to find the equation of the tangent line here at the point 3 comma 4. You'll notice 3, 4 is a point of course on this circle here. So, if we follow the strategy we did previously, we're going to take the derivative of the left-hand side, x squared plus y squared, and we're going to take the derivative of the right-hand side. In both situations, we're taking the derivative with respect to x, and we're using this fundamental truth of equations. If we do the same thing to the left-hand side as the right-hand side, that is if we apply a function to the left side and the right-hand side. Because it's a function, the output is independent. I should say the output will be consistent even if you take different representations of the same input. So, whatever you do to the left-hand side, if we do that same function to the right-hand side, equality will be still preserved. That's this important principle of taking functions of equations here. Now, on the right-hand side, since we're taking the derivative, we might abbreviate this as, of course, x squared plus y squared prime. Same thing on the right-hand side. We might abbreviate using this prime notation, where it's understood that we're taking the derivative with respect to x. The right-hand side can be pretty easy, right? The derivative of constant should be 0, so it's just going to disappear to be a 0. What about the left-hand side? Well, the usual properties of derivatives apply right here. If you have a derivative of a sum, you can take the sum of the derivatives. You can take them separately. So, we have to take the derivative of x squared with respect to x. We have to take the derivative of y squared with respect to x right here. Now, in the usual sense, when you take the derivative of x squared with respect to x, this is going to give you a 2x. But the thing that's a little bit different here when you do this implicit differentiation is if you take the derivative of y squared with respect to x, how do you compute that thing? Well, you use the chain rule. Instead of saying the derivative with respect to x, I'm going to take the derivative with respect to y and make sure you also take the derivative of y with respect to x. The inner derivative comes into play here. Therefore, the derivative of y squared with respect to y, that's going to be a 2y. But don't forget the inner derivative, y dy over dx. You're going to end up with this y prime. The most common mistake with students to calculate this with implicit differentiations, they forget that inner derivative. Much like with that elliptic curve we saw on the previous screen, we could solve for the y prime minus 2x from both sides. We then get a 2y prime. 2y y prime, excuse me, is equal to negative 2x. Divide both sides by the 2y. 2y. The 2s within cancel out. And we get that dy over dx is equal to negative x over y. And then let's plug in the points here. So we take our general formula, y minus y1 is equal to m times x minus x1. The x and y coordinates remember we're three common four. So we're going to get y minus four is equal to something times x minus three. What's the slope here? We have to evaluate this slope dy dx at the point three four, which by the formula we see above here, we're going to get a slope of negative three fours. So we put that in for the slope there. Then if you distribute in combined like terms, that'll then, this equation would then translate to become in slope intercept form, the equation you see right here. So again, we can find the slope of this tangent line if we compute the derivative implicitly. So the last thing to compare here is what if we did it the old fashioned way? Because if we take the equation x squared plus y squared equals 25, we could solve for it explicitly, right? We could get that y squared is equal to 25 minus x squared. We could take the square root of both sides, y equals the square root of 25 minus x squared. But we have to be cautious here. There's a plus or minus square root. Considering the above picture, we see that the point three comma four is on the upper semicircle here. So when you take the plus or minus, the plus is gonna be the top one, the minus will be the lower one. So we can get away with that and be like, oh, okay, I just need the plus one, no big deal. You take the derivative now. So if you wanna take the derivative dy dx, you're gonna have to take the derivative of the right hand side, the square root of 25 minus x squared prime. In which case you end up with, by the chain rule, right? Because we have a function inside of a function, you have this 25 minus x squared. It sits in five of the square root function, which we really should think of this as the one half power function. That's much easier for the calculus here. So by the chain rule, you're gonna get a one half times 25 minus x squared to the negative one half power. And then you have to multiply by the inner derivative of the derivative of 25 minus x squared. You'll notice the chain rule comes into play. That's a very important observation here. For which case then the 25 minus x squared to the negative one half power becomes a square root of 25 minus x squared in the denominator. You also have this coefficient of two. And when you take the derivative of 25 minus x squared, you're gonna get the 25, well, it's derivative zero. It's a constant. Then the derivative of the negative x squared becomes a negative two x. We see that the twos cancel out and then we end up with y prime equals this negative x over the square root of 25 minus x squared. I wanna point out to you something very important here. The square root of 25 minus x squared, that's y. We could simplify this formula as negative x over y. That seems awfully familiar. This was the derivative we found implicitly. But when you calculate the derivative implicitly, you didn't have to go through this complicated chain rule calculation with the square root. We had much simpler when we took the derivative of these things separately. So I guess what I'm trying to say is if you're trying, if you have an implicit equation between two variables x and y and you wanna calculate the derivative, there's never really an advantage for solving for y and then calculating the derivative. The derivative will not be simpler or better than if you found it implicitly. One might actually make the argument that the implicit form is simpler in its appearance. But the, so the thing is explicit differentiation never is gonna prove to be a better tool for derivatives. But on the other hand, implicit differentiation potentially avoids some of the cumbersome algebraic techniques necessary to solve for the equation explicitly. So really, I hope you take this as an advertisement that implicit differentiation is awesome and we should encourage ourselves to use it as we start studying these implicit relationships between functions and maybe you're interested in the tangent lines of their associated curves.