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# Showing explicit and implicit differentiation give same result

AP Calc: FUN‑3 (EU), FUN‑3.D (LO), FUN‑3.D.1 (EK)

## Video transcript

What I want to show you in this video is that implicit differentiation will give you the same result as, I guess we can say, explicit differentiation when you can differentiate explicitly. So let's say that I have the relationship x times the square root of y is equal to 1. This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1/x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. So we get dy dx is equal to negative 2 x to the negative 2 minus 1-- x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing-- x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, well, actually, we're going to apply both the product rule and the chain rule. The product rule tells us-- so we have the product of two functions of x. You could view it that way. So this, the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, times the derivative with respect to x of the square root of y. Let me make it clear, this bracket. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. This simplifies to 1, so we're just going to be left with the square root of y right over here. So this is going to simplify to a square root of y. And what does this over here simplify to? Well the derivative with respect to x of the square root of y, here we want to apply the chain rule. So let me make it clear. So we have plus this x plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of the square root of something with respect to that something. Well, the derivative of the square root of something with respect to that something, or the derivative of something to the 1/2 with respect to that something, is going to be 1/2 times that something to the negative 1/2 power. Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1/2 x to the negative 1/2. Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand side. On the right-hand side, we just have a 0. And now, once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over, so I have, once again, more room to work with. So let me cut it, actually. And then let me paste it. Let me move it over, right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then, if we subtract the square root of y from both sides-- and I'll try to simplify as I go-- we get this thing, which I can rewrite as x times-- well, it's just going to be x in the numerator divided by 2 times the square root of y. y to the negative 1/2 is just the square root of y in the denominator. And 1/2, I just put the 2 in the denominator there-- times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here, just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx-- or dividing both sides by this is the same thing as multiplying by the reciprocal of this-- is equal to 2 times the square root of y over x-- over my yellow x-- times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to y times-- the square root of y times the square root of y is just y. The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2 x to the negative third power. The key here is to realize that this thing right over here we could solve explicitly in terms of-- we could solve for y. So we could just make this substitution back here to see that these are the exact same thing. So if we make the substitution y is equal to 1 over x squared, you would get dy dx, the derivative of y with respect to x, is equal to our negative 2 times 1 over x squared, and then all of that over x, which is equal to negative 2 over x to the third, which is exactly what we have over here, negative 2 x to the negative third power.