Video transcript
What I want to showyou in this video is that implicitdifferentiation will give you the same result as,I guess we can say, explicitdifferentiation when you can differentiate explicitly. So let's say that Ihave the relationship x times the square rootof y is equal to 1. This one is actuallypretty straightforward to define explicitly interms of x, to solve for y. So if we divide both sidesby x, we get square root of y is equal to 1/x. And then if you square bothsides, you get y is equal to 1 over x squared, whichis the same thing as x to the negative 2 power. And so if you wantthe derivative of y with respect to x, thisis pretty straightforward. This is just an applicationof the chain rule. So we get dy dx isequal to negative 2 x to the negative 2 minus 1--x to the negative 3 power. So that's prettystraightforward. But what I want to see is ifwe get the same exact result when we differentiateimplicitly. So let's apply ourderivative operator to both sides of this equation. And so let me makeit clear what we're doing-- x times the square rootof y and 1 right over there. When you apply thederivative operator to the expression onthe left-hand side, well, actually, we're goingto apply both the product rule and the chain rule. The product ruletells us-- so we have the product oftwo functions of x. You could view it that way. So this, the productrule tells us this is going to be thederivative with respect to x of x times thesquare root of y plus x, not taking its derivative,times the derivative with respect to x ofthe square root of y. Let me make itclear, this bracket. And on the right-handside, right over here, the derivative with respectto x of this constant, that's just goingto be equal to 0. So what does this simplify to? Well, the derivative withrespect to x of x is just 1. This simplifies to1, so we're just going to be left with the squareroot of y right over here. So this is going to simplifyto a square root of y. And what does thisover here simplify to? Well the derivative with respectto x of the square root of y, here we want toapply the chain rule. So let me make it clear. So we have plus this x pluswhatever business this is. And I'm going todo this in blue. Well, it's going to be thederivative of the square root of something with respectto that something. Well, the derivativeof the square root of something with respectto that something, or the derivative ofsomething to the 1/2 with respect to that something,is going to be 1/2 times that something to thenegative 1/2 power. Once again, this rightover here is the derivative of the square root ofy with respect to y. We've seen this multiple times. If I were to say the derivativeof the square root of x with respect to x, you wouldget 1/2 x to the negative 1/2. Now I'm just doing it with y's. But we're not done yet. Remember, ourderivative operator wasn't to say with respect to y. It's with respect to x. So this only gets uswith respect to y. We need to apply theentire chain rule. We have to multiply thattimes the derivative of y with respect to x in orderto get the real derivative of this expressionwith respect to x. So let's multiplytimes the derivative of y with respect to x. We don't know what that is. That's actually whatwe're trying to solve for. But to use thechain rule, we just have say it's the derivativeof the square root of y with respect to y timesthe derivative of y with respect to x. This is the derivative ofthis thing with respect to x. So we get this onthe left-hand side. On the right-handside, we just have a 0. And now, once again,we can attempt to solve for the derivativeof y with respect to x. And maybe theeasiest first step is to subtract the squareroot of y from both sides of this equation. And actually, let me moveall 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 alot of real estate, but hopefully thishelps a little bit. And actually, Idon't even like that. Let me leave itwhere it was before. So then, if we subtractthe square root of y from both sides-- andI'll try to simplify as I go-- we get this thing,which I can rewrite as x times-- well, it's just goingto be x in the numerator divided by 2 times the square rootof y. y to the negative 1/2 is just the square rootof y in the denominator. And 1/2, I just put the2 in the denominator there-- times dy dxtimes the derivative of y with respect to xis going to be equal to the negativesquare root of y. I just subtracted the squareroot of y from both sides. And actually, this issomething that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here, justto continue our simplification solving for dy dx. Well, to solvefor dy dx, we just have to divide both sides byx over 2 times the square root of y. So we're left with dy dx--or dividing both sides by this is the samething as multiplying by the reciprocal ofthis-- is equal to 2 times the square root of yover x-- over my yellow x-- times the negativesquare root of y. Well, what's thisgoing to simplify to? This is going to beequal to y times-- the square root of y times thesquare root of y is just y. The negative times the2, you get negative 2. So you get negative 2y over xis equal to the derivative of y with respect to x. Now you might besaying, look, we just figured out thederivative implicitly, and it looks very different thanwhat we did right over here. When we just used the powerrule, we got negative 2 x to the negative third power. The key here is to realizethat this thing right over here we could solve explicitlyin terms of-- we could solve for y. So we could just makethis substitution back here to see that theseare the exact same thing. So if we make the substitutiony is equal to 1 over x squared, you would get dy dx, thederivative of y with respect to x, is equal to our negative2 times 1 over x squared, and then all of that over x,which is equal to negative 2 over x to the third, which isexactly what we have over here, negative 2 x to thenegative third power.
