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Felicia S.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

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So now that we talked about partial derivatives of the function, we have a sense of how to differentiate functions more than one variable. So if you recall, there's a very important differentiation rule her functions of one variable and that's the chain rule. So in one variable, the chain rule goes like this. Suppose that why is a function of you and you is a function of X Then if I want to take the derivative of why with respect X so go from the dependent variable all the way down to this independent variable skipping this intermediate variable then what do I do? Well, I first take the derivative of why, with respect to you, then multiply by the derivative of you with respect Toe X like that or, in other words, f prime of g of x. Oh, I'm sorry. What I mean is the derivative of f of G of X is well, it's the derivative of F evaluated G of X And then I multiplied by the derivative of the inside she prime effects. So this is sort of the live minutes notation. This is more the traditional function notation. But that's what the change Will says in one variable. So what about the two variable case? Well, the problem here is that we meet might mean different things by the two variable case. So case one goes like this Z is a function of X and Y X is a function of tea And why isn't function of team? This is sort of like the chamberlain one variable. Except now we have two intermediate variables that depend on this independent variable T and Z is the dependent variable. That depends on all of them. So what is the derivative of Z with respect to D So I have a dependent variable that really just depends on the independent variable T. But I have these intermediate variables. Well, it looks like this. So first I take the partial Rosie with respect to X, and I notice I have to use partial here because there's some ambiguity about what I mean by the derivative of Z. I could take the derivative and see with respect to X or why so here I take it with respect to X and then multiply by dx DT. And I'm using the regular d here because there is no ambiguity about what independent variable X depends on. It's just a function of T, so that's just like the one variable case. But I've ignored why? Well, what do I dio? I just had the chain role in multi variable. I just add all of the different possibilities. So I add taking the other choice. I take the partial of Z with respect why and then multiplied by Dwight D. T like that. So this is what the chain rule says in the two variable case. In case one where the intermediate variables X and Y just depend on one independent variable t. So I told you about Case one where the intermediate variables just dependent on one independent variable. But now let's talk about case, too. Case to it's more interesting. So let's say again, Z is a function of the to intermediate variables X and y. So I'm saying intermediate because X and we're not going to be independent. X and Y are gonna actually themselves pinned on to independent variables like this. You can see where this could get sort of endlessly complicated as we go to more and more variables. But this will be the case will consider Z is a function of the two independent variables or the to intermediate variables X and Y. And then those intermediate variables each are functions of the independent variables s antique. So now I have a choice. If I want to take the derivative of C with respect to an independent variable, I have to say what I mean. Do I mean s or do I mean t So in that case, when I have the ambiguity I need to use the partial derivative, it's a partial derivative of Z with respect Toe s. Well, I just have to consider the contribution from X and the contribution from why so have partial Z partial X, partial X partial X s plus Parcells e partial. Why partial? Why Partial s so notice. I'm taking the derivative with respect to s. And so I need to consider both contributions from the intermediate variables X and y So your ex And here's why. But X and y themselves e need to differentiate with respect to X. And if I wanted to take the partial with respect to T, then I have partial Z. I mean, you could probably guess it at this point partial X partial T this time and then plus partial Z Partial. Why? Partial? Why? Partial team? So what about Mawr Variables? Well, I'm not going to actually write down, you know, any more explicit formula, but I'm gonna tell you a general principle, and I hope that you kind of think about this and maybe even write down the case for three variables, But for more variables, you just add the appropriate terms. And what I mean by that, if you had another intermediate variable se z, well, then you would just add on the contribution from that variable. And this is one of those things where you know, you just think about it and wrap your head around it and, you know, it's very natural. You just sort of keep adding thes, uh, partial derivative factors for all of your intermediate variables and for each one of your independent variables. So there's one thing in Calcutta one. I'm not saying this is the only thing, but it's definitely something that was a little bit unjustified when you talked about it. And what that is is implicit differentiation. And now, with the multi variable chain role, we're actually in a position to give a justification for implicit differentiation. So the idea with implicit differentiation is that you didn't have wise a function of X you had. Why was an implicitly defined function of X so you can actually write a set up for implicit differentiation like this? You have an equation that looks like this. I get everything on one side now notice. I have not solved. Why as a function of X, because maybe I'm not able to do that. Maybe remember, like X squared. Plus y squared equals one. Why was not an explicit function of X? Why could only be defined as an implicit function of X? So here, if I have an equation like this and I want to find de y dx, what can I dio? Well, let's say that why is a function of X? Let's just pretend that why is a function of X, then we're actually in a set up, and we'll say X is also a function of X like that. Now, of course, X is equal to itself. So this is a little silly to right, but I just want to set it up into set up of the multi variable change role. I have these two intermediate variables x and Y that depend on the independent variable X. So what is the multi variable change will tell me. It tells me that the partial derivative with respect to X of X y times, dx, dx So that's the derivative. This is like partial f partial x times. Uh, D x t x So another. I mean, um I mean, this is a little weird to write dx dx, but I'm saying X depends on itself. So it's just X. So it's just one. And then I add the partial with respect. Why times do you i d x Okay, because the intermediate variable y depends on this independent variable X, and then that's going to be zero. But this is just one, of course, the X over d x and I can actually just solve for do I D x do I D x. It's just minus. Subtract this over f X over death for me, so this not Onley justifies implicit differentiation. It actually makes it a little bit simpler because we can use this idea of partial derivatives, whereas if you recall implicit differentiation had a lot of algebraic manipulation. But now it's so easy. Now we just get everything on one side, take the partial with respect to X, then take the partial with respect to why just take the question. That's D Y. D X.

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