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Well hello everyone, welcome to our second lecture on partial differentiation. We introduced

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partial differentiation in our last lecture and we looked at functions of two variables

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and we then looked at the geometrical interpretation of partial derivatives and looked at some different

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notation for the partial derivative. We then went on to calculate the partial derivatives

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of some standard functions before finishing off with looking at some chain rules for differentiating

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a function of two variables. Today we're going to carry on with partial differentiation

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and we're going to look at the product rule for a function of two variables, the quotient

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rule for a function of two variables and we'll finish off by looking at higher order partial

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derivatives. So but first of all let's remind ourselves of the product rule for a function

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of one variable. So one important thing to notice if you've got two functions multiplied

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together something like x squared sine x two functions of one variable then the product

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rule is not as many people right? It's not the derivative of the first function times

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the derivative of the second function. That's completely wrong. To differentiate a product

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you must use what's called the product rule and the product rule is given here. We've

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got the product of two functions of x, f and g. I haven't bothered writing the brackets

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x because we just accept that these are functions of x. So the product rule says that it's a

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derivative of the first function times the second function, f dashed times g plus the

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first function times the derivative of the second function that's f times g dashed. So

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that is the formula we must use when we're differentiating a function that's a product

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and there's only one variable involved. So we must use this function and you'll have seen

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that in the past. We're now going to extend differentiating products to functions of

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two variables. I'm going to follow this pattern when we're deriving a rule for the product

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of two functions, the derivative of the product of two functions of two variables. So here's

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our function z of x, y is equal to f of x, y times g of x, y. Now I'm going to draw the

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x, y notation. It just gets a bit crowded. We know that these are f and g are functions

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of the two variables x and y. So we're going to calculate that product. Now the rule stays

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exactly the same really. Here we only had one variable so we could write f dash to represent

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the derivative with respect to x, a. In this case now we've got two variables. So writing

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f dashed isn't really clear because we don't know whether it's the x we're differentiating

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with respect to or whether it's the y we're differentiating with respect to. So what we

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do in this case is we use the notation we saw in the last lecture, an alternative notation

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for the partial derivative with respect to x. We write x as a subscript. So this means

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find the partial derivative of f with respect to x multiplied by the function g. To that

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we add the function f multiplied by the partial derivative of g with respect to x. So it's

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very very similar to the formula for the product rule when we've got a function of one variable

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when we're multiplying two functions of one variable. It's very similar except that now

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we must explicitly write the variable that we're differentiating with respect to. And

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that's all that does. And we've got a similar definition for the partial derivative with

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respect to y when we've got a product. So it's f sub y times g plus f g sub y, where

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the f sub y and the g sub y represent the partial derivative of f with respect to y

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and the partial derivative of g with respect to y. So that's what we have to do. So let's

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have a look at an example. So our function said equal to f of xy if you like, said as

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a function of x and y. It's a function of two variables. We've got this function to start

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with x cubed plus four y squared and we're multiplying it by the second function e to

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the two xy. So this is clearly a product and we need the product rule for it. Clearly a

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product of a function of two variables and another function of two variables. So we need

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the product rule here and we're asked to find the partial with respect to x and the partial

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derivative with respect to y. So I'm going to define one function equal to f and the

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other function equal to g. It doesn't matter which. So I'm going to define the first function

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x cubed plus four y squared. I'm going to define that as f of xy. So that's going to

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be my f function. My g function or g of xy function is the second function, the exponential

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function e to the two xy. And if we remember from the previous slide, these are the formulae

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for the first order partial derivative with respect to x and the first order partial derivative

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with respect to y. So that means I've got f and I've got g to get the partial of z with

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respect to x. And first of all, I've got to find the partial derivative of f with respect

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to x and the partial derivative of g with respect to x. That's before I can use the

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formula. And I've got to do a similar thing before I can get the partial with respect

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to y. I've got to find the derivative of f with respect to y and the partial derivative

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of g with respect to y. So let's get started on that. So first of all, we'll do partial

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derivative z by dx. And we defined f as being x cubed plus four y squared and g equal to

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e to the two xy. So I'll differentiate f with respect to x. Well, if I do that, differentiate

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x cubed with respect to x, I get three x squared. Now, remember, we treat y as a constant

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when we differentiate with respect to x. So y and any function of y, like y squared

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or four y squared is treated as a constant. And when you differentiate with respect to

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x, and if you differentiate a constant, you get zero. So that term will vanish when we

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differentiate it. So we just get three x squared. Now, our second function g, we've got to

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differentiate that with respect to x as well. So if we look at it, though, we've got an

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exponential function applied to the function two xy. So this is a function of a function,

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it's a composite function. So we need the chain rule. So g is equal to e to the two

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xy. I'm going to write it like this x two xy. It's maybe clearer for some people to

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see that we need the chain rule. An exponential function applied to the function two xy. So

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the chain rule, what do we do? We differentiate the outside function, and the exponential

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doesn't change when differentiated. So that just stays as an exponential, leave the bracket

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as it is, and differentiate the term inside the bracket with respect to x. Okay, so we're

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going to multiply by this quantity. So we're going to do partial d by dx of two xy. So

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we could, like we did in the last lecture, you know, we've got d by dx of two xy. So y and any

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function of y is a constant. So let's take the two y outside, and we've just got to find

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the partial with respect to x of x. So I'll just write that in. So that's partial d by dx

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of x, which is just one. So we've got two y times one, which is equal to two y. And you know,

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we're taking the exponential outside. So we've got the exponential times two y, and we can rearrange

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it as two y e to the two xy. So that's where that derivative comes from. Okay, so we're now ready

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to use our formula. So the partial of z with respect to x is f sub x times g plus fg sub

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x. And I've got all of these things up here. So f sub x is three x squared, and I can multiply

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that by g. So it's a diagonally opposite term that I multiply together. There you go there. And I

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also multiply, it's plus the product of those two diagonally opposite elements, f times g sub x.

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So that's what I've got here. Then, if I want to simplify the answer, I spot that I've got the same

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exponential term appearing in both of these expressions. So I could take this e to the two

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xy out as a common factor. That'll leave me three x squared here. And leave me this term x cubed plus

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four y squared all times two y. I've taken the exponential out of here, out of this term, and

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I've taken it out of that term. So that just leaves me this term, and that term. So I've got three x

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squared plus x cubed plus four y squared all times two y. I can then multiply out this term here. So

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I'll get two x cubed y plus eight y cubed, and that's all going to multiply the exponential e to the

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two xy. And that's it. That is the product rule applied to our function z. And we've calculated

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partial d z by dx. We've now got to get the partial d z by dy. So it's a similar process, right down

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my f and my g and differentiate both of them with respect to y. So if it's a y differentiation, I treat

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x and any function of x as a constant. So you differentiate a constant, you get zero. So that

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won't play any part. Differentiate four y squared with respect to y, I get eight y. Differentiate

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this exponential, it's a function of a function. So I need the chain rule. So this is what I'm going

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to do in this green bubble over here. I've got g equals e to the two xy. I'm going to write it as x

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to xy. So the rule, of course, for the chain rule is differentiate the outer function. Well, the

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exponential is its own derivative. So it stays the same. Leave the bracket as it is. So I'll have two

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xy in brackets, multiplied by the derivative with respect to y of the function inside the brackets.

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That's two xy. And again, I can write I've got d by dy. That's the partial d by dy of two xy. It's a y

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derivative. So x and any function of x can be taken outside. So that's what I'm going to do. And that

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just leaves me y in there. So I'm going to have two x the derivative of y with respect to y is just one.

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So it's equal to two x. And don't forget, of course, that's multiplying the exponential e to the two xy.

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So that's where that expression for the derivative of g with respect to y comes from. Then it's just a matter

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of applying the formula. So laid out in this form with f and g in the top line and f sub y and g sub y below

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them. I just multiply the diagonally opposite terms. So I'll have eight y times e to the two xy plus this term

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xq plus four y squared times two x e to the two xy. So that's what we've got. Again, I'll go and tidy this up a

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little. I've got my exponential function e to the two xy appears in both terms. There's one term there plus the

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second term, the exponential appears in both. So I can take that out as a common factor. And I'll have eight y plus this

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term xq plus four y squared times two x. I can then multiply out this bracket here to get two x to the power four

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plus eight xy squared. And all of that then multiplies the exponential. And that is my partial derivative with

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respect to y obtained using the product rule of my original function z. So there we go. So that is an example of the

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product rule being applied to differentiate the product of functions of two variables. Okay, let's do another example.

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So here we've got z to the z to log x cos xy. Do we need the product rule for this? Well, just let's just pause for a

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minute and look at this function a little closer. We've been asked to calculate the partial with respect to x to start

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with. Well, yes, we do need the product rule because our first function is log x, and that's multiplying another

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function of x cos xy. So one function of x log x, multiplying another function of x cos xy. So we definitely need the

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product rule for that. Okay, so let's do that. Let's call one of them f and the other one g. So I'm going to call the log

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x, I've called that f and g I'll call cos xy. So there's my f and there's my g. I've got to differentiate them both with

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respect to x. So differentiate log x with respect to x. Well, that's a standard derivative of a function of a single

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variable. It's just one over x. I've got to differentiate g now with respect to x. And if I look at this, I've actually got a

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function of a function. I've got the cosine function applied to the function xy. So I need to use the chain rule. So the chain

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rule, calculating the derivative with respect to x of cos xy, what does the chain rule say? Well, it says differentiate the

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outer function. So cos when you differentiate becomes minus sign, leave the bracket as it is. So you get minus sign xy and

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differentiate what's inside the bracket with respect to x. So if I do that, you know, I've got d by dx, if I just write it down

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here, I've got the partial d by dx of xy. So I can take the y outside if I like, and it's just d by dx of x, which is y times

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one, which is just y. So I'm going to have minus sign y times y. Or in other words, I can rearrange it if I want to as minus

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y, sign y. And that's where that comes from. And then it's just a straightforward matter of applying the formula. So here's your formula. The

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partial of z with respect to x is f sub x times g plus f times g sub x. And if you laid out like this, as I've said before, the f and the g

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on top and the partial derivatives down below, it's just a matter of multiplying the diagonally opposite terms. So I'll have one over

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x times cos xy to get that cos xy over x. And I'm going to have plus the product of those two. So it's going to be plus log x times

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minus y, sign xy. And the plus and the minus together gives a minus. So you're going to get minus y log x, sign xy. And that's it. That is the

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partial derivative with respect to x of this function z, which was a product of two functions of x. Okay, so now we'll have to find the

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partial derivative with respect to y. And you can use the product rule here if you like, you'll get the correct answer. But you might just

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notice something that yes, we do have two functions multiplied together. But the first function only involves x, and the second function

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involves x and y. So if we want the partial with respect to y, we don't need to use the product rule because this first function,

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log x can just be treated as a constant when we calculate the partial derivative with respect to y. So that's what I say here.

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I say note that we do not need the product rule to calculate partial descent by dy since we can treat log x as a constant. So if we do that,

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we've got partial descent by dy, we replace z with the given function. There it is. We've just finished saying that it's a y derivative. So x and any

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function of x like log x is treated as a constant. So that can be taken outside of the derivative. So there it is. And then all we have to do is

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differentiate cos xy with respect to y. And that's fairly straightforward. It's the chain rule again. Cos becomes minus sign when you

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differentiate it, leave the bracket as it is xy and differentiate xy with respect to y. You know, we can take the x outside and then get the derivative

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of y with respect to y which is 1. So the x times 1. So you'll end up with minus x sign xy. And you've got this product here. And we can just simplify that.

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Just rearrange it, move the bracket, just have minus log x, minus x log x, sorry, times sign xy. So that's yet another example of using the product rule with functions of two variables.

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Okay, moving on, that's the product rule. So we've seen the chain rule. We saw that last week and we've seen it quite a few times in this class already. And we've just looked at the product rule.

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That leaves us a quotient rule to look at. And we'll remind ourselves of the quotient rule for a function of one variable to start with and then look at how it's applied when you get a function, when you get

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functions of two variables. So let's just look at that. Okay, so the quotient rule for a function of one variable, it's f dash g minus f g times g dashed all over g squared. So, you know, if you

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remember a product rule, the way I'd written it, the way I'd written it, let's see, let's go back, there it is, product rule, the way I'd written it, it was, I had f dash g plus f g dash now written in this order to get the

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quotient rule, you just change the plus to a minus and you divide by g squared. That's all you have to do.

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Okay, so let's see if this can be extended to function of two variables and what the answer is yes it can. Here we've got a function z, which is formed by dividing a function f, which is a function of x and y by another function g, which is also a function of x and y.

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And the product rule tells us that what we have to do is, well, it's essentially the same as the single variable case, except here, you know, we can't just write f dashed g minus f g dashed because we won't really know which, you know, what f dashed refers to.

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We've referred to a differentiation with respect to x or we've referred to differentiation with respect to y. So again, what we do is we use the shorthand notation for the derivative, partial derivative with respect to x, it's f with a subscript x and we do the same thing for g, the partial

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derivative with respect to x is g sub x and that's our formula there. It's very similar to the single variable case and the same thing happens for y, except now that is a derivative with respect to y, you want to calculate of f and for g, you want to differentiate g with respect to y and we plug these into the formula and that's what we've got.

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So we've got a formula for the partial of z with respect to x and a partial formula for the partial of z with respect to y when we're using when we've got a quotient and we need to use the quotient rule.

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So let's just do an example and see how this works. So here's my function said is equal to x squared y divided by x squared plus y squared. Well, if you look at this, this is clearly

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on top, we've got a function which we'll call f. So this is clearly, I'm just get the pen and the tablet. So this is clearly z is equal to f of x, y. So x squared y is clearly a function of x and y as is the function below and that'll be g of x, y.

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So our function said is clearly a quotient involving a function on the top. That's a function of x, y, that's f and down below we've got g, which is a function of x, y. So we need the quotient rule.

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So that's all I've said here. I've decided that f is going to be equal to x squared y and g will be x squared plus y squared. So I've got my formula that I know that I discussed in the previous slide for

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the partial derivative with respect to x and the partial derivative with respect to y of a quotient f over g, this function here.

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So let's do partial respect to x first of all. So the partial with respect to x, there we are there, f and g. So I want to find and write f here and g here and down below I'll calculate the derivatives.

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So the partial with respect to x of x squared y. Well, there we are in the green bubble here. I can, it's an x derivative I'm calculating so I can take the y outside.

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All I've got to do then is differentiate x squared with respect to x and that gives me 2x and the derivative will be 2xy. So that's there.

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And similarly, I've got to calculate the partial derivative with respect to x of g. So x squared with respect to x, 2x.

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It's an x derivative. So we treat y and any function of y as a constant, differentiate a constant, you get zero. So all that survives here is

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the 2x. So the partial derivative of g with respect to x is just 2x.

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Okay, so let's put all of this together in our formula. So the formula says the top line. Well, again, if we've righted in this form with f and g on the top and the derivatives down below, it's going to be

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this diagonal, the opposite pair multiplied together. So we'll get 2xy times x squared plus y and it's minus the product of those two diagonally opposite ones.

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f times g sub x, so it's going to be 2x times x squared y. And what do we got in the denominator? We've got g all squared. So we've got to square this thing and that's what we have here.

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So we can tidy up the numerator a little. So let's multiply out these brackets. We'll get 2x cubed y plus 2xy cubed, okay, minus 2x cubed y. So that's fine. And now you might spot the first term, 2x cubed y minus 2x cubed y will give me zero.

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So the answer would just be 2xy cubed over x squared plus y squared all squared. So that's a partial derivative of our function said by using the quotient rule. That's a partial derivative with respect to x.

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I've now got to do the same thing to get the partial derivative with respect to y. So I'll define f as the top function and g as the bottom function. Down below, I'll calculate the derivatives. Now, f sub y, that means I've got to take the partial derivative with respect to y of this function, x squared y.

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It's a y derivative, so the x squared can be taken outside and that just leaves me x squared times the derivative of y with respect to y, which is just one. So it's just going to be x squared. So that's where that comes from.

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I've then got to differentiate the g function with respect to y, but that's easy because that x squared is just a constant. So that would be zero when you differentiate with respect to y. And this will become 2y. And again, I use my formula and basically it would be the product of those two, my f sub y times g.

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It would be the product of those two, x squared times x squared plus y squared minus the product of those two, minus x squared y times 2y.

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I can simplify the top, that become x to the power of 4 plus x squared y squared minus 2x squared y squared.

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Okay, there we are, that's it there. Can we simplify this in any way? Well, we can't do much with the x to the 4 term, so that stays, but look at this one.

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We've got one lot of x squared y squared minus two lots of x squared y squared, so it'll be minus one lot of x squared y squared. And if we want to, we can simplify the numerator a little by taking out a common factor.

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You've got an x squared here and x to the 4 is just x squared times x squared, so we take an x squared out as a common factor.

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So we'll need an x squared here to make our x to the 4 term and we must have minus y squared so that when we multiply it out, we'll get a minus x squared y squared and we'll put the second term in the numerator.

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So that is the partial derivative with respect to x of our original function z, which was the quotient of two functions of x and y and we applied the quotient rule to calculate partial dz by dy.

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So that's it calculated.

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Okay, so we've now seen the product rule for function of two variables. We've seen the quotient rule for function of two variables and we know from the examples we've done that it's a very similar process to when we're working with functions of a single variable when we're using the product of quotient rules for functions of single variables.

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We're now going to move on and look at something we've also seen before when we've got a function of one variable, higher order derivatives, particular higher order partial derivatives.

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But first of all, let's rewind a reminder cells of the higher order derivatives of functions of a single variable.

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So suppose you've got a function like y is equal to x cubed. If we differentiate y with respect to x, well, it's just the power rule, it'll become 3x squared. That would be the first derivative.

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We can differentiate this again to get the second derivative. So that will give us, so this is really, if I wrote this down, I've got my first derivative dy by dx.

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And I'm now going to differentiate this thing again with respect to x, so I get d by dx of that, and that gives us d squared y dx squared.

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So that is just the second derivative of our function y. So it's another word, it's a higher order derivative, a second order derivative.

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We are now going to extend this idea to functions of two variables and calculate the partial derivative, second order partial derivatives.

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So you might think that we would calculate the first order partial derivative with respect to x, and then the second order partial with respect to x, and do the same for y, the first order partial with respect to y, and get the second order partial with respect to y, and that would be it.

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Well, that's not quite true, not for second order partial derivatives. There's two other derivatives we've got to calculate, these are called the mixed second order partial derivatives, and I'll explain that through this diagram.

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So there are four types of second order partial derivatives. So let's start with our function.

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z is equal to f of x, y, z is some function of x and y.

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And we know that we can calculate, we can differentiate with respect to x to get the first order partial with respect to x, and we can differentiate with respect to y to get the first order partial with respect to y.

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We can then, with the first order partial with respect to x, we'll differentiate again with respect to x to get the second order partial with respect to x, and we can do the same thing with the first order partial with respect to y, differentiate that again with respect to y to get the second order partial with respect to y.

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Now as I said there are two other parts we must take into account. So over here in the left branch of our tree we differentiated z with respect to x and we differentiated again with respect to x.

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Now there is no reason why we can't take the first order partial respect to y with respect to x and differentiate z with respect to y. And that gives us one of the mixed second order partial derivatives.

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So coming down this branch we have done first derivative with respect to x, second derivative with respect to x. We have done first derivative with respect to x and instead of differential with respect to x we now differentiate that function there with respect to y to give us this mixed second order partial.

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Similarly in the right branch of the tree we can calculate the first order partial with respect to y, go down, differentiate again with respect to y to get the second order partial with respect to y.

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Again coming down we would differentiate with respect to y but this time we are going to differentiate this result with respect to x and that will give us the other second order mixed partial derivative.

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Now there is a result and it's based on something called Clairot's theorem or Schwartz's theorem that these two partial derivatives for the types of function we look at will always be equal. They will always be the same.

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It doesn't matter if you differentiate with respect to x first and then y or you differentiate with respect to y first and then x the result you get for the mixed second order partial will always be the same.

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Okay so let's just look at some alternative notation and have a little bit more explanation on this notation here.

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So the first order partial with respect to x we saw last time in the last lecture that we can write this in short hand notation as z sub x and the first order partial with respect to y we can write it as z sub y.

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So the x and the y in each case denote the variable with different with respect to.

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Well we can easily write that the second order partial is z sub xx and the second order partial with respect to y is z yy.

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What about these mixed partial derivatives? Well this one here what does this say? This was coming down the left branch get the partial derivative with respect to x first and then differentiate that with respect to y.

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So that was coming down here get the partial derivative with respect to x and differentiate that result with respect to y.

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So that was the notation we had here in short hand notation we write this as z sub xy.

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Now how do we know the order here? Well we work from the inside out with differentiate with respect to x first so we write the x first followed by a differentiation with respect to y so we have the y second.

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So that's why z sub xy. Similarly for this second order mixed partial derivative in this case we came down the right branch of our tree we differentiate the function z with respect to y first and then this result with differentiate with respect to x.

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So back to your tree remember we differentiate z with respect to y and then we differentiate this result with respect to x to give us this mixed second order partial.

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And in short hand notation because we do this the y derivative first we write y first and then x because it's a y derivative first followed by an x followed by differentiation with respect to x.

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And as we know these mixed second order partial derivatives must be equal if they're not you've made a mistake.

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Let's have a look at an example. So what do we got here? So we've got z is equal to 3x squared y cubed plus 10y to the 4 minus 8x to the 6.

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So that is certainly a function of x and y so z is equal to f of xy that's it given there that's a function z is a function of x and y.

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So I want to calculate determine all the first and second order partial derivatives just write pd's for partial derivatives.

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The first order partial derivative to start with we did these in the first lecture and it's fairly straightforward.

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We're just adding and subtracting terms so we can go in and differentiate them one at a time just to term by term differentiation.

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It's an x derivative so we treat y as a constant and any function of y.

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So in this case the y cubed can be taken outside and actually we'll take the 3y cubed outside.

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I was just going to differentiate x squared and of course multiplied by 3y cubed in this one.

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Well, it's a single variable so we just leave it as it is and we'll do the same for that just leave it as it is because it's a single variable.

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I differentiate x squared with respect to x I get 2x so I simplify this I get 6xy cubed so that's that expression.

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Now I differentiate this thing here which is purely a function of y it's an x derivative so we treat this as a constant so you differentiate a constant you get 0.

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I differentiate 8x to the 6 with respect to x just the power law down it comes gives me 48x take one away 48x to the 5 and that's minus of course so I can simplify my expression to give me my answer at the foot of the page.

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So that's my first order partial derivative with respect to x.

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And now I could go and calculate the first order with respect to y but I think since I've got this one I'll just differentiate it again with respect to x to get the second order partial derivative with respect to x.

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So the second order partial with respect to x simply says take the first order partial with respect to x and differentiate it again with respect to x.

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So that's what this rotation here means.

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So the second order partial with respect to x is the derivative with respect to x of the first order partial which we just calculated on the previous slide there it is.

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And again I can do it term by term it's an x derivative so I take y in any function or y outside so the 6y cubed will come out and I've just got to differentiate x with respect to x so that's just one 6y cubed times one 6y cubed.

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My second function is just a function of x so that's just the derivative or function of a single variable using the power law down it comes and take one away from it and if I do that I get 240x to the power 4.

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So that's that one done as well and I put them together and that gives my final answer for the second order partial derivative with respect to x.

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So let's look at the first order with respect to x and the second order partial derivative with respect to x.

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I now need to do the same thing for y so let's do the first order partial derivative with respect to y and again I replace it with a given function there it is.

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It's just adding and subtracting so I'll just do them term by term that's all I'm doing here and wherever I've got a product involving x and y I'm going to look at which variable might differentiate with respect to.

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Well it's y so I treat x and any function of x as a constant so I'm taking the x squared out I could have actually taken the 3x squared out it doesn't matter I'll get the same answer anyway.

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And these are just functions of a single variable so I just leave them as they are.

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So let's differentiate 3y cubed with respect to y gives me 9y squared that's times x squared so that gives me 9x squared y.

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Differentiate 10y to the 4 with respect to y gives me 40y cubed so I've got plus 40y cubed.

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I've got a function of x 8x to the 6th with differentiate with respect to y so that's just 0 because we treat x and any function of x as a constant so you differentiate with respect to y it becomes 0 so there's my final answer at the foot of the slide here.

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So that's my first order partial derivative with respect to y and I now want to differentiate this once again with respect to y to get my second order partial derivative with respect to y.

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So here we are there's my first order with respect to y but I calculated on the previous slide and I'm going to differentiate that one more time with respect to y.

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So the partial derivative with respect to y from the previous slide there it is I'm adding them together so I can just do this term by term.

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So I'm going to do this term first I got a product of x and y so it's a y derivative so I'm going to take x and any function of x outside the operator.

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So there it goes 9x squared comes outside the derivative operator and that's just a requirement to differentiate y squared with respect to y so it's just 2y and I'll get 9x squared times 2y will give me 18x squared y and then I'm going to differentiate my second term with respect to y.

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It's just a function of a single variable and since the power law it'll give me 120y squared so there at the foot of the slide is my second order partial derivative with respect to y.

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So I've now got my two second order partial derivatives one of them with respect to x so I've come down the left branch on the outside of the left branch of my tree diagram to get the second order partial with respect to x.

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I've come down the extreme right branch of my tree to get the second order partial with respect to y that just leaves me the mixed second order partial to calculate so I must come down here get the first order partial with respect to x which I've already done and differentiate that with respect to y and here I've got to get the first order partial with respect to y again I've done that already and I must differentiate that with respect to x.

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These two functions in here these two expressions must be the same so let's go back and do that.

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So there we are so first of all I'm going to take I'm going to go in the left branch of the tree and I'm going to take my first order partial derivative with respect to x and I'm going to differentiate that with respect to y.

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So my first order partial derivative that was the first calculation I did actually it was 6x y cubed minus 48x to the 5 and I've got to differentiate this with respect to y again additional subtraction so we can do it term by term.

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My first term is going to product of x and y so I know that it's a y derivative so I can take x and any function of x outside so the 6x comes out and it just leaves me y cubed to differentiate with respect to y so it will be 3y squared I can simplify this to give me 18xy squared.

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My second term well it only involves x it's a function of a single variable only involving x and it's a y derivative I want so x and any function of x is treated as a constant you differentiate a constant you get 0 so my final answer is 18xy squared so that's my first mixed second order partial derivative.

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That was taken the partial respect to x and differentiate that with respect to y now I have to do the other one now I've got to take the first order partial respect to y and differentiate that with respect to x and I should get the same answer I should get 18xy squared if I do everything correctly.

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Now my first order partial respect to y I calculated earlier on it was 9x squared y squared was 40y cubed so we're just adding two terms so we can just differentiate term by term it's an x derivative so in here any function of y can be taken outside so if I want to I can take out the 9y squared and that's exactly what I've done here so I'm just differentiating term by term.

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And I'm taking out any function of y because it's an x derivative so outcomes of 9y squared that leaves me x squared to differentiate that's just 2x of course when I differentiate respect to x and that becomes 18xy squared.

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My second term was 40y cubed and I differentiate that with respect to x so y and any function of y is just a constant so when I differentiate it with respect to x I get 0 so my answer is 18xy squared and my look back at the previous slide that's exactly the same and that is what I expected that the mixed second order partial derivative is equal they have to be otherwise.

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Something's gone wrong so I've just verified that they are the same so that is the answer the full answer to that question I've calculated all second order partial derivatives all four of them.

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The second order partial respect to x the second order partial respect to y and the two mixed second order partial derivatives which are equal to each other.

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Okay so that's so we can represent this in a tree diagram if we want just so we can see we've already seen the tree diagram.

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So there's my original function z I differentiate respect to x to get this function here I differentiate that again with respect to x to get this function these are the calculations I've done in the previous slides.

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I then do the same thing for y differentiated once with respect to y and differentiate by first order derivative with respect to y differentiated again to get the second order partial with respect to y.

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I now do the mixed second order partial so I differentiate by function z with respect to x to get the first order partial derivative with respect to x.

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I differentiate that with respect to y to get the second mixed order partial derivative the second mixed order partial derivative.

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And I do the same thing in this branch of the tree I take my first order partial derivative with respect to y and I differentiate that with respect to x to get the second order the the other second order mixed partial derivative.

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partial derivative and we saw from the previous slide that these two were in fact equal. So this

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just summarizes everything that we just did in a tree diagram. So we've got another example here,

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another very similar example really. So we can go through it. It's a function of two variables,

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such x and y appearing in the function. So again, we want the first order partial respect to x.

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So z is given by this function to replace z by the given function. First order partial respect

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to x to term by term. It's respect to x or y in any function or y can be taken outside. So the

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7y to the 5 comes out and that just leaves me x cubed to differentiate respect to x and that's

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3x squared. So I can simplify this first term. I'll get 21x squared y to the 5. That's fine. My

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second term, well it's only a function of a single variable and it's a power loss. It'll be 24x cubed.

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So minus 24x cubed, there it is. My last term involves y only. We've got a function of y,

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8y squared. So we treat it as a constant differential respect to x, we get 0. So that's the first order

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partial derivative with respect to x. I now take that result and I differentiate it again with

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respect to x to get my second order partial derivative with respect to x. Once again,

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we've got just two terms subtracted. So I can do term by term differentiation.

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Again, it's an x derivative. So with this first term, I can take any function of y out

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and I'll take the 21y to the 5 outside and that'll just leave me x squared to differentiate

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respect to x and we know that that's 2x. So I can simplify this expression here to get 42x y to the

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5. My second term is 24x cubed. Well it's a function of a single variable. It's a power loss. So it'll be 72x

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squared. So my answer will be 42x y to the 5 minus 72x squared. So now I've got the first

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order partial derivative with respect to x. I've also got the second order partial derivative with

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respect to x. So I'll repeat the calculation and get the second order partial with respect to y.

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But first of all, of course, I've got to get the first order with respect to y. Again, I can replace

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said by the given function. We're just adding or subtracting terms. So it's going to be term by

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term and the derivatives are all with respect to y. So it's a y derivative. So x and any function

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of x can be taken out. So the 7x cubed gets taken out, leaving me y to the power 5 to differentiate

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respect to y. So it'll be 5y to the 4. I can simplify this to give me 35x cubed y to the 4. So that's

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that first term. My next expression, my next term, this one here, only involves x. So I've got a function

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of x 6x to the 4 differentiate now with respect to y with 3 to x constant. So when you differentiate it,

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you get 0. My third term is 8y squared. It's a function of one variable. 8y squared differentiate

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with respect to y is just 16y. Don't forget the minus. So I'll get 35x cubed y to the 4 minus 16y.

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So that's the first order partial derivative with respect to y. And I now want to take this

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expression and differentiate it again with respect to y to get the second order partial

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derivative with respect to y. And that's what this notation here is saying. My first order partial

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derivative with respect to y, differentiate that one more time with respect to y. So we got this

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first order partial with respect to y. We've got that on the previous slide. There it is. I can

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differentiate this term by term. That's what I've got here. Both the derivatives are of course with

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respect to y. It's a y derivative. So x and any function of x, like 35x cubed, can be taken outside.

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There it is. And that just leaves me y to the power 4 to differentiate with respect to y. And that is

309
00:51:44.960 --> 00:51:55.280
just 4y cubed. I can simplify this expression to give me 140x cubed y cubed. That's fine. Back

310
00:51:55.280 --> 00:52:02.320
now to the second term. 16y with differentiate with respect to y will be just 16. Don't forget

311
00:52:02.320 --> 00:52:11.600
there's a minus there. And so that comes through to my answer. 140x cubed y cubed minus 16. So that

312
00:52:12.160 --> 00:52:18.720
is now the first order partial derivative with respect to x and y calculate. The second order

313
00:52:18.720 --> 00:52:23.520
partial derivative with respect to x and y calculate it. I now need to calculate the two

314
00:52:24.160 --> 00:52:31.520
mixed second order partial derivatives. So the first one, for no particular reason,

315
00:52:31.520 --> 00:52:36.720
I've decided to take the first order partial with respect to y and differentiate that with

316
00:52:36.720 --> 00:52:42.640
respect to x. So the first order partial with respect to y we found earlier. There it is.

317
00:52:42.640 --> 00:52:48.880
Again, I can do term by term differentiation. And the differentiation is with respect to x. So both

318
00:52:48.880 --> 00:52:55.200
terms get differentiated with respect to x. So I look at my first expression.

319
00:52:57.360 --> 00:53:04.320
It's an x derivative. So y and any function of y can get taken outside. So the 35y to the 4

320
00:53:04.320 --> 00:53:10.240
gets taken outside. And that just leaves me x cubed to differentiate with respect to x. And

321
00:53:10.240 --> 00:53:18.080
that's just 3x squared. Now I can simplify this to give me 105x squared y to the 4. My second term

322
00:53:19.360 --> 00:53:27.120
16y. It's a functional y and I'm differentiated with respect to x. So we treat any function of y

323
00:53:27.120 --> 00:53:35.120
as a constant. And when differentiated with respect to x we get 0. So my final answer for the mixed

324
00:53:35.120 --> 00:53:45.200
order partial derivative is 105x squared y to the 4. So that was when I calculated the

325
00:53:46.080 --> 00:53:51.360
first order with respect to y first and differentiated with respect to x. Let's do it the other way

326
00:53:51.440 --> 00:53:57.040
around. Let's get the first order partial with respect to x and differentiate that with respect

327
00:53:57.040 --> 00:54:04.880
to y. So there you are. Partial dz by dx is now going to be differentiated with respect to y. I

328
00:54:04.880 --> 00:54:11.840
know what this is. It's the first thing I calculated in this question. There it is there. I'm differentiated

329
00:54:11.840 --> 00:54:19.840
with respect to y. There's two terms. So I can do them term by term. It's a y derivative. So any

330
00:54:19.840 --> 00:54:26.640
function of x gets taken outside. So the 21x squared gets taken outside and that just leaves me

331
00:54:26.640 --> 00:54:35.120
y to the 5 to differentiate with respect to y and it becomes 5y to the 4. Multiply all this together.

332
00:54:35.120 --> 00:54:42.240
I get 105x squared y to the 4. But I've also got my second term to differentiate here.

333
00:54:42.240 --> 00:54:49.520
But it's a function of x. It's 24x cubed and when we differentiate the function of x with respect

334
00:54:49.520 --> 00:54:55.840
to y we treat it as a constant and when you differentiate a constant you get 0. So this

335
00:54:55.840 --> 00:55:03.120
will not contribute. And my final answer would be 105x squared y to the 4. So that's our other

336
00:55:03.120 --> 00:55:08.240
mixed second order partial derivative. And if I compare it with the one on the previous slide

337
00:55:08.240 --> 00:55:14.880
I find it's exactly the same and that's really what we would have expected. That the mixed second

338
00:55:14.880 --> 00:55:25.520
order partial derivatives are equal to each other and they certainly are. So that's that one. So

339
00:55:25.520 --> 00:55:34.880
let's do another example. So we've got set is equal to x squared plus e to the xy. Well it's

340
00:55:34.880 --> 00:55:42.560
definitely a function of x and y. It's certainly a function of two variables and we've got to

341
00:55:42.560 --> 00:55:51.760
determine all first and second order partial derivatives. So I can calculate each of these

342
00:55:51.760 --> 00:55:56.960
down here. So that's the first order and the second order but I haven't done the mixed order

343
00:55:56.960 --> 00:56:04.320
once yet. But I'll explain these first of all. So in red I'm doing the differentiation with respect

344
00:56:04.320 --> 00:56:09.280
to x. The first order derivative with respect to x and the second order derivative with respect to

345
00:56:09.280 --> 00:56:15.920
x. And similarly in the blue I do the first and second order partial derivative with respect to y.

346
00:56:16.640 --> 00:56:24.800
So first of all let's do the first order partial with respect to x. So differentiate x squared I get

347
00:56:24.800 --> 00:56:32.080
2x. Now I'm going to differentiate the second function e to the xy. I'm going to differentiate

348
00:56:32.080 --> 00:56:37.360
that with respect to x. So it's the chain rule because we've got a function of a function the

349
00:56:37.360 --> 00:56:44.960
exponential function being applied to the function xy. So the rule says in the chain rule says leave

350
00:56:46.000 --> 00:56:51.840
differentiate the outside function. So the exponential just differentiates to give us the

351
00:56:51.840 --> 00:56:57.680
exponential it doesn't change. So the exponential stays the same how we differentiate. Leave the

352
00:56:57.680 --> 00:57:04.800
bracket as it is and multiply by the derivative of the thing in brackets. So this if you like is

353
00:57:04.800 --> 00:57:12.400
just like a constant we're doing the derivative of a constant times x because it's a derivative

354
00:57:12.400 --> 00:57:18.480
with respect to x so we treat y as a constant. So if we differentiate a constant times x that'll

355
00:57:18.480 --> 00:57:23.520
just leave a constant. So y is playing the role of the constant here so we get a y there. So we

356
00:57:23.520 --> 00:57:33.760
tidy this up a little bit and we get 2x plus y e to the xy. Okay so we now want to differentiate

357
00:57:34.400 --> 00:57:40.400
this guy one more time with respect to x to get the second order partial derivative with respect

358
00:57:40.400 --> 00:57:46.400
to x and that's what this notation here represents. So I know what that first order partial derivative

359
00:57:46.400 --> 00:57:54.000
does I just calculated on the previous line and I've written it like this 2x plus y e to the xy

360
00:57:54.000 --> 00:58:02.240
I've just written xxy. Okay so let's do this term by the two terms so I can just do a term by term

361
00:58:02.240 --> 00:58:10.240
differentiate 2x with respect to x well that gives me two. Now I've got to differentiate y e to the

362
00:58:10.240 --> 00:58:22.160
xy with respect to x so y and any function of y can be taken as 3 is a constant so the y can get

363
00:58:22.160 --> 00:58:29.200
taken outside we can't take this out because we've got x appearing in here it's a it's a product

364
00:58:29.200 --> 00:58:35.440
involved in x so we can't take that outside if it was just e to the y it could come out but there's

365
00:58:35.440 --> 00:58:45.920
an x involvement there so we've got y times the partial derivative of xxy now we don't actually

366
00:58:45.920 --> 00:58:53.520
need to do this again because if you remember we've already done this we're already differentiated

367
00:58:53.520 --> 00:59:04.240
e to the xy here partial d by dx of xxy we've found to be y e to the xy so it's exactly the same

368
00:59:04.320 --> 00:59:12.480
calculation here partial d by dx of xxy so it must be equal to this thing so there it is y

369
00:59:12.480 --> 00:59:18.080
e to the xy don't forget we've got a y on the outside here we'll multiply by so we can tidy this

370
00:59:18.080 --> 00:59:26.240
up and you get 2 plus y squared e to the xy so the thing here just to be aware that you've already

371
00:59:26.240 --> 00:59:35.440
done this calculation you've already done the derivative of xxy with respect to x we did it up here

372
00:59:37.120 --> 00:59:42.800
so hopefully we can we can get something like that happening when we're doing the partial

373
00:59:42.800 --> 00:59:48.640
respect to y as well so let's see if we can so we've got to do the partial with respect to y

374
00:59:48.640 --> 00:59:55.120
of our function here so let's do this so first of all we're going to do the partial

375
00:59:55.840 --> 01:00:02.480
of x squared with respect to y well it's a y derivative so we treat x and any function of x

376
01:00:02.480 --> 01:00:07.600
like x squared is a constant differentiate a constant you get zero so that's where that zero comes

377
01:00:07.600 --> 01:00:16.320
from then we're going to differentiate e to the xy with respect to y so I'll write it as x xy

378
01:00:17.040 --> 01:00:24.240
so the rule says chain rule says differentiate the outside function so the exponential just stays

379
01:00:24.240 --> 01:00:30.240
the same that's the rule for differentiating exponential doesn't change its own derivative so

380
01:00:30.240 --> 01:00:40.000
x leave the bracket as it is so you get x xy and I want to differentiate the thing inside the brackets

381
01:00:40.720 --> 01:00:46.880
with respect to y now x has been treated as a constant so that's just like something like

382
01:00:47.600 --> 01:00:53.840
differentiating 3y with respect to y and the answer there all that would survive would be the

383
01:00:53.840 --> 01:01:00.720
constant so x is playing the role of the constant here so when you differentiate xy with respect to

384
01:01:00.720 --> 01:01:08.080
y with x being a constant the only thing that survives is the x so you know I mean I can just

385
01:01:08.080 --> 01:01:14.240
maybe quickly write something down just to explain that a little bit more so let's just maybe do it

386
01:01:14.800 --> 01:01:26.080
read so I've got d by dy of xy differentiate nothing in brackets it's a y derivative so the x

387
01:01:26.080 --> 01:01:35.200
comes out so you get d by dy of y so that just becomes one so it's x times one so it's just x so

388
01:01:35.200 --> 01:01:46.480
that's where that x comes from so if I just simplify this I can write this x e to the xy

389
01:01:47.360 --> 01:01:54.160
okay so I now want to find the second order partial derivative with respect to y and to do that I

390
01:01:54.160 --> 01:02:00.160
differentiate the first order partial derivative with respect to y once more with respect to y

391
01:02:00.160 --> 01:02:06.720
I know this thing I just found that the calculation above so what I've got to do is I've got to

392
01:02:06.720 --> 01:02:14.160
calculate the partial derivative with respect to y of x e to the xy and I've written it like this

393
01:02:14.160 --> 01:02:20.400
now it's a y derivative so the x can get taken outside so there we are and all we have to do

394
01:02:20.400 --> 01:02:27.280
here is calculate the partial with respect to y of e to the xy and I look back up and say well I've

395
01:02:27.280 --> 01:02:36.560
already done that there it is and the answer was x e to the xy so the this becomes x e to the xy

396
01:02:36.560 --> 01:02:42.560
when you do the derivative you're multiplying by x don't forget that so the answer is x squared

397
01:02:42.560 --> 01:02:50.960
e to the xy so let me just get rid of that just to tidy it up a little okay so now we've got the

398
01:02:50.960 --> 01:02:58.000
first and second order partial derivatives with respect to x here in red and with respect to y

399
01:02:58.000 --> 01:03:07.840
in blue I've now got to get the mixed second order partial derivatives so to do that which one

400
01:03:07.840 --> 01:03:13.920
I do first well I'll take the first order partial with respect to x and I'll differentiate that

401
01:03:13.920 --> 01:03:20.400
with respect to y there from the previous slide the first order partial partial dz by dx

402
01:03:21.120 --> 01:03:30.720
was here 2x plus y e to the xy there it is 2x plus y e to the xy there it is there so I've

403
01:03:30.720 --> 01:03:41.360
got to differentiate this with respect to y so I can do a term by term the first term 2x is just

404
01:03:41.440 --> 01:03:47.040
a function of x so we treat it as a constant when we differentiate with respect to x so it

405
01:03:47.040 --> 01:03:54.080
becomes zero we differentiate so that would be zero now the second part is slightly more complicated

406
01:03:54.800 --> 01:04:03.680
because it's a y derivative and we've actually got two functions both involved in y so we need

407
01:04:03.680 --> 01:04:13.200
the product rule here so we need to use the product rule and we'll actually need the chain rule as

408
01:04:13.200 --> 01:04:18.640
well to differentiate e to the xy so let's have a look at this now I'm going to differentiate

409
01:04:19.440 --> 01:04:27.040
y e and as you can see I've just called that I've just given a name and called it w w is equal to y

410
01:04:27.520 --> 01:04:36.000
e to the xy I'm going to you I need the product rule I'm going to let f be equal to y so that one

411
01:04:36.640 --> 01:04:49.840
I'll let f so I'll just say let f equal y and g equal e to the x y so it's a y derivative I'm

412
01:04:49.840 --> 01:05:00.960
calculating so f sub y of y with respect to y you know that's just your d by dy of y so differentiate

413
01:05:00.960 --> 01:05:10.480
 y with respect to y it's just one and g sub y is equal to what well it's a derivative of e to the xy

414
01:05:10.480 --> 01:05:20.400
with respect to y have we done that before well yes we have haven't we yes we have we've calculated

415
01:05:20.400 --> 01:05:35.600
the derivative of e to the xy with respect to y we did it here in blue the derivative of e to the

416
01:05:35.600 --> 01:05:49.760
xy with respect to y is x e to the xy so there we are so what have we got we know that it's x e to

417
01:05:49.760 --> 01:05:55.520
the xy we just calculated we've done that calculation before and then I want to use the

418
01:05:57.200 --> 01:06:03.280
product rule so it's just the diagonally opposite once multiplied together so it's one times e to the

419
01:06:03.280 --> 01:06:15.200
xy plus y times x e to the xy like that and that's where this result comes from this thing here was

420
01:06:15.200 --> 01:06:21.120
a bit more complicated needed the product rule because it's a product of two functions that each

421
01:06:21.120 --> 01:06:29.520
involve y and this bit the exponential needed us to use the chain rule but fortunately we remembered

422
01:06:29.600 --> 01:06:38.640
we'd actually done that before the derivative of g equals e to the xy there with respect to y we found

423
01:06:38.640 --> 01:06:45.520
was x e to the xy so we've done that before so that saves us a little bit of work and then if we really

424
01:06:45.520 --> 01:06:53.040
wanted to we could tidy this up a little bit and we spot that the exponential function appears in here

425
01:06:54.000 --> 01:06:59.760
in this term and in this term so we can take it out as a common factor so we take it out as a

426
01:06:59.760 --> 01:07:08.720
common factor so we need two terms in the bracket now to get e to the xy we must multiply e to the xy

427
01:07:08.720 --> 01:07:13.840
by one so that's why that's a one there sometimes people get a little bit confused about that you

428
01:07:13.840 --> 01:07:23.360
need a one in here and then to get this term obviously you will need xy to multiply the e to the xy to

429
01:07:23.360 --> 01:07:31.280
get this term so that's the first of our mixed second order partial derivatives and we can now go on

430
01:07:31.280 --> 01:07:36.720
and calculate the second one and we'll hope that it's equal to this if it isn't we made a mistake

431
01:07:36.720 --> 01:07:47.200
so let's do that then so this one involves taking the first order partial derivative with respect to

432
01:07:47.200 --> 01:07:55.680
y and differentiating that with respect to x so if we look back let's go and find the first order

433
01:07:55.680 --> 01:08:03.360
partial with respect to y we'll have it here there it is it's x e to the xy so this is the

434
01:08:03.360 --> 01:08:10.160
thing that I now want to differentiate with respect to x and I've sported something I've

435
01:08:10.160 --> 01:08:17.440
sported that this will need the product rule because I've got a function of x multiplying another

436
01:08:17.440 --> 01:08:24.800
function that involves x so this is just similar to the calculation I did over leaf so I've got

437
01:08:24.800 --> 01:08:38.160
w is equal to x e to the xy so I'm going to let f equal x and g equal e to the xy now I'm

438
01:08:38.160 --> 01:08:45.200
count it's an x derivative that I want here so f sub x well that's an easy one it's just one

439
01:08:46.320 --> 01:08:54.320
and g sub x well I wanted to differentiate e to the xy with respect to x have I done that before

440
01:08:54.800 --> 01:09:02.000
well yes I have there it is that's the derivative of e to the xy with respect to x

441
01:09:02.560 --> 01:09:11.680
is given by this and tied it up it's just y e to the xy so thankfully I've done that before

442
01:09:11.680 --> 01:09:18.720
so that saves me a bit of work it's y e to the xy there we are I can now use my

443
01:09:19.040 --> 01:09:28.640
um product rule I use my product rule so it's these diagonally opposite ones f sub x times g

444
01:09:28.640 --> 01:09:38.640
so it's one times e to the xy there you are plus x times y e to the xy so that's it that is the

445
01:09:38.640 --> 01:09:44.400
product rule applied to this function here and of course we had to use the chain rule on the

446
01:09:44.480 --> 01:09:49.760
exponential part of it so we can write this like so

447
01:09:52.320 --> 01:09:58.240
one times e to the xy is just e to the xy again I've spotted that I've got the same exponential

448
01:09:58.240 --> 01:10:06.640
term appearing in both the terms so I can take e to the xy outside as a common factor

449
01:10:07.600 --> 01:10:12.800
to get this here I need to have a one in here so that will give me e to the xy times one will

450
01:10:12.800 --> 01:10:23.840
give me that and here I'll have xy so that e to the xy times xy will give me my second term and

451
01:10:23.840 --> 01:10:29.440
if I look back at the previous slide where I got the other mixed second order partial derivative

452
01:10:29.440 --> 01:10:40.240
I can see that they are in fact equal as we expected so that's how we do that so that's um

453
01:10:41.200 --> 01:10:47.280
as we've now looked at the product rule and the quotient rule and we've also looked at

454
01:10:48.000 --> 01:10:56.000
higher order derivatives from working with a function of two variables so just to summarize

455
01:10:56.000 --> 01:11:02.000
all of this the unit it's extended the idea of differentiation of a function of one variable

456
01:11:02.560 --> 01:11:08.080
and we've introduced the concept of a partial derivative of a function of two variables so

457
01:11:08.080 --> 01:11:15.280
we're now able to understand what we mean by a partial derivative remember in the last class we

458
01:11:15.280 --> 01:11:22.560
looked at some a geometrical interpretation of calculating partial derivatives of a function

459
01:11:22.560 --> 01:11:30.960
of two variables and we're also after the examples were just done we're also able to determine partial

460
01:11:30.960 --> 01:11:37.280
derivatives up to second order of a function of two variables and we're also able to apply the

461
01:11:37.280 --> 01:11:43.200
standard rules of differentiation like the chain product and quotient rules to calculate

462
01:11:44.880 --> 01:11:53.120
partial derivatives of a function of two variables so you should now be able to in addition to the

463
01:11:53.120 --> 01:11:58.480
questions we saw that you'd be able to attempt in the last lecture I've just gathered everything

464
01:11:58.480 --> 01:12:03.280
together here so you should be able to attempt all the mobius questions those questions one to four

465
01:12:03.280 --> 01:12:10.000
associated with partial differentiation and working on the tutorial questions from the notes

466
01:12:10.000 --> 01:12:18.800
you should be able to attempt these ones that I've listed here so that is the end of our partial

467
01:12:18.800 --> 01:12:24.560
differentiation lectures we had two lectures on partial differentiation the next topic that we'll

468
01:12:24.560 --> 01:12:32.480
move on to is looking at solving ordinary differential equations so we'll leave it at that for now

469
01:12:32.480 --> 01:12:36.880
okay bye