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Well hello everyone and welcome to our first class on partial differentiation.

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We'll have two classes and the first class will include the following topics.

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First of all we'll start off with a brief review, brief reminder of differentiation of a function

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of one variable. We'll then look at defining a function of two variables and look at partial

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derivatives of a function of two variables. We'll look at the geometrical interpretation

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and the notation that's used. We'll then work on some examples to calculate the partial derivatives

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of standard functions and after that we'll introduce the chain rule for a function of two variables

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and finally we'll move on and look at the extended chain rule which looks at slightly more complicated

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functions of two variables. So first of all we'll have a look at some motivation for partial

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differentiation. Now we've all seen functions of a single variable before, that's functions of one

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variable for example y equals f of x and they are useful in representing certain physical phenomena.

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For example the area of a circle of radius r we know the formula for the area is given by a equals pi

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r squared and here's a diagram showing the circle of radius r and we calculate its area

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 by calculating pi r squared. So the area therefore depends on one quantity and that quantity is r

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and if r was to vary then the area a would vary just as in this diagram here we've got a blue

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circle of radius r we calculated its area and we obtain a value. Now if we take this other circle

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and we calculate its area because we varied r, we made r smaller then the area of the circle

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will be smaller. So the variable a is called the dependent variable because it depends on r which

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we call the independent variable and the mathematical notation for this area is a is equal to f of r so

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area is a function of r and as I said if r was to vary then the area would vary. In real world

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situations though quantities generally depend on more than one variable and that gives rise to

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functions of several variables and I'm going to give you now a couple of examples of functions

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of two variables. In this course we won't look at any more than two variables we'll look at function

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of two variables but the whole idea generalizes to functions of many variables three four five

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as many variables as you want but we'll stick with looking at functions of two variables.

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So here's an example the volume of a cylinder v because the cylinder has radius r and height h

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is given by the formula v is equal to pi r squared h and that makes sense if you think about it.

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The cylinder has a circular base and a circular top and the area of that is pi r squared and if we

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multiply by the height h that will give us the volume so the formula is v is equal to pi r squared

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h so the volume therefore depends on the two quantities r and h and if either one of these

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or both of them vary then clearly the volume will vary so in this case the volume v is the dependent

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variable and it's a function of the two independent variables r and h and the mathematical notation

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for this is v is equal to f of r comma h so in this example here I've kept the radius fixed but I've

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increased the height of the cylinder and clearly you can see as h increases then so too does the

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volume. Another example comes from a combined gas law and the combined gas law gives the relationship

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between pressure, volume and the absolute temperature of a fixed amount of gas. This combined gas law

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comes from a combination of Boyle's law, Charles's law and Gay-Lusick's law for those of you who've

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done physics so it says the combined last gas law states that pv is equal to k times t where p is the

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pressure v is the volume and t is the absolute temperature of the gas and k is some constant so

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we can rearrange this equation in terms of pressure to write p is equal to kt over v so that pressure

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in this case is a function of the two variables v and t. The v and t can vary k of course is a constant

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so we write it as p is equal to f of v comma t and the pressure therefore depends on the two

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quantities volume and temperature v and t and if either one of both of these varies then p will vary.

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Okay so let's have a look we're going to eventually look at the geometrical interpretation of partial

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derivatives but let's remind ourselves of the partial of a functional one variable and the

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geometrical interpretation of a derivative. So a functional one variable can be written as y is

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equal to f of x and we can plot it as a curve in the xy coordinate system so it's a curve in 2d in

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2 dimensions so we've got the x axis going along the horizontal and the y axis going along the vertical

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so and motion is only possible in the horizontal direction or the vertical direction you can go

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horizontally left or right and vertically up and down and the height y above or even below the

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horizontal axis varies with the position of x on the horizontal axis so for example an x here would

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produce a y value up here and if we move along to here say the x value produces a y value here so the

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x the y value will generally vary with your x value and in this diagram I've just plotted it for

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positive y just for simplicity reason but this graph could easily dip down below the horizontal axis

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and you would get a negative value for y so in this diagram the guy walking along the top of the

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curve is free to move horizontally and vertically as he follows the curve up and down so the curve

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 the slope is constantly changing it's here we've got a positive slope here we'll have a zero slope

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where we reach the turning point and then the slope will become negative it will become zero again

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and it starts going positive again and so on now the geometric interpretation of a derivative so

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you can essentially think of a derivative as a function that allows us to measure the gradient

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or slope of a function and when we substitute a value of x into the derivative we obtain the slope

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of the tangent line at that point and we also obtain the slope of the curve at the point so for

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example if y is equal to f of x of some function of x then the derivative you can either write it as

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df by dx or dy by dx gives a rate of change of f of x with respect to x you know essentially it's

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telling us how y is changing with respect to x so evaluating as I've just said evaluating the

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derivative at a point at some point we'll call x zero gives the gradient of the tangent line

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and hence the slope of the curve at that point so in this diagram we've got a red curve y equals

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f of x if we were to differentiate the formula for this function we would get an expression that

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would allow us to calculate the gradient at any point and if we plug in the point x zero that'll

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give us the slope of the curve at that point and it's a positive slope in this case so let's do an

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example so here is a curve y is equal to f of x is equal to x squared plus 3x so I want to calculate

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the slope of the curve at the point x equals four so remember to do that I've got to differentiate my

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function and the derivative that results will give me a formula for calculating the slope of the curve

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at any value of x and I've been given a specific value of x that I want to plug in to the expression

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for the derivative so let's differentiate the function first of all so it's just the power

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rule that applies x squared when you differentiate down comes the power subtract one of ways you get

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2x and 3x when you differentiate gives us 3 so we now want to find the slope at the point x equals

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four so this expression will allow us to do that so we plug in x equals four and we find

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that the slope at the point x equals four is 11 it's positive so the curve is increasing at x equals

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four so that is an example of differentiation of a function of a single variable and evaluating

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the slope of the curve at a given point you'll be familiar with these these are just this just a

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table of standard derivatives of a function of a single variable as you can see sine differentiates

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to give cos cos becomes minus sine when you differentiate it tan differentiates becomes

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sec squared x remember sec x is one over cos x e to the x is unique in that zone derivative it

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doesn't change when you differentiate it and log x when you differentiate it becomes one over x so

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this is a table of standard derivatives when we come to look at partial differentiation you'll find

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that there are no new rules for differentiation everything you've come across before will hold

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the only thing that gets a little bit tricky is you've got more variables to work with but we'll

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see that in due course so first of all let's have a look at a function of two variables so in this

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diagram here I've plotted the function z is equal to f of x y and that's cos x times cos y so it's a

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function of the variable x and it's a function of the variable y so a function of two variables

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two independent variables x and y so z is a function of x and y that represents a 3d surface in the

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x y coordinate system so if I just go back for a minute let's have a look at a function of a single

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variable so for a function of a single variable y equals f of x look at one independent variable and

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for every value of x you plug into the formula you know it could be something x squared plus three

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like in the example that we had yeah x squared plus three x for every value of x we plug into the

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formula we get a value of y and the value of y if it's positive it would be above the x axis and if

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it's negative it'll be below the x axis so every value of x will generate a value of y and hence

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the resulting curve well in in three dimensions where you get a function of two variables the idea

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similar really except that you know now you've got a z axis we've got our x y plane here so what

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happens is that for every x y coordinate pair you plug them into the formula and out comes a value

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of z if z is positive then you're above the x y plane and if x is negative you're down below the

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x y plane you can think the x y plane has been the floor a positive value will go up towards the

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ceiling and a negative value will be down below the floor so the height as I've just said above

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or below the horizontal axis varies with the x y position in the plane so in general for every x y

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coordinate pair when you plug it in you'll get a different z value and in this case that's what our

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function looks like it's a three-dimensional surface function of the two variables x and y

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okay so that leads us to the idea of partial derivatives we saw in a function of one variable

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the curve where the guy was walking along the slope was changing as he moved from left to right or

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equally well from from right to left and the slope of the curve was changing things are slightly

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different slightly more complicated when we go up to three dimensions because in two dimensions as we

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saw this fellow could only walk to the left or right and he would go up and down along the curve

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depending on what the slope of the function was okay so he could only go left or right and he would

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move up and down along the curve now in three dimensions in 3d of course he can walk off in any

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direction he likes you know he doesn't have to walk parallel with any axis he can walk off in

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n direction and explain that a bit more in a minute so a partial derivative it's a function

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it's a derivative of a function of more than one variable and to calculate a partial derivative

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all variables apart from the variable we are differentiating with respect to our held constant

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i'll explain that shortly and we differentiate as i've said before using the standard rules

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for differentiating a function of a single variable so in this case here in this module

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here as i've said we're only going to look at functions of two variables we don't look at

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but we could look at functions of three variables four variables five variables and so on the ideas

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all carry across but we restrict ourselves to functions of two variables

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now as i mentioned earlier the two dimensional diagram we saw with the guy walking along the

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curve he could only move along in the positive x and positive y direction he would move up and down

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along the curve depending on the slope when we up things to three dimensions so

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a guy standing here he could actually walk off in any direction he wants because he could spin

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around through 360 degrees he could walk off parallel to the y-axis he could walk parallel to the x

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axis but he could also walk out at an angle he could come directly out like this or he could walk

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out like so he could so he's got freedom to walk off in any direction 360 degrees

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so that's an awful lot of different derivatives but we're going to restrict ourselves to only

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consider moving off parallel to the y-axis and parallel to the x-axis so in other words

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we're only going to look at the slope of the surface in the direction of the x-axis and the

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slope in the direction of the y-axis so let's just have a look at a brief geometrical interpretation

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of the partial derivatives with respect to x and with respect to y let's have a look at the partial

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derivative with respect to y first the partial derivative with respect to y requires us to

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keep x fixed and to move in the y direction parallel to the y-axis so that's what i'm going to do here

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so so this surface here in blue we take a slice through it at a fixed value of x at x equals x0

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and if you do that you'll obtain this orange plane here on this orange plane x0 x is constant

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it's got the value x0 all the way along you can think the plane could actually stretch all the way

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along to the left and all the way along to the right but i've only just taken a segment of it just

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to show you what's going on so when we slice through the surface at a fixed value of x when x is equal

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to x0 we obtain this plane here and the plane would have equation x equals x0 and it'll intersect

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our surface here in this yellow line so what you what we need to think about if we're looking at

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the slope in the y direction we're going to keep x fixed and we can walk along this yellow line just

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in a similar way to the guy in 2d was walking along the curve this guy can walk along the yellow line

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and of course depending on the surface the contours of the surface he'll go up and down and up and

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down just following the contours of the surface along the yellow line keeping x fixed at the value

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x0 so x is fixed the only thing that's varying is y and that will lead us to the definition

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of a partial derivative with respect to y so just at the foot of the slide here i say if x is kept

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constant at x equals x0 and y is increased by moving along the yellow curve parallel to the y axis

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the slope of the surface changes and that leads us to the idea of the partial derivative of the

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function with respect to y now the notation the partial derivative of function z equals f of xy

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partial derivative with respect to y is written like this you use a curly d so it's partial dz by dy

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and you use curly d's because it's a function of more than one variable we use curly d's in the work

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you'll have done up to now where we only had functions of a single variable we use a straight

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back d like that we would have a dy by dx so that relates to functions of a single variable

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whereas the curly d notation corresponds to functions of more than one variable and partial

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derivatives okay so to calculate the partial with respect to y that's dz by dy with differentiator

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function z of xy with respect to y while we treat x as a constant and the reason for that is fairly

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obvious because as we said we take the slice through the surface at a constant value of x

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the value of x doesn't change it's it's an x is equal to x0 all the way along the line and all the

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way along that curve so all the way along that curve x is constant the thing that's changing is y

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so that gives us a partial derivative with respect to y and we can do a similar thing for

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calculate the partial derivative with respect to x and here what it looks like geometrically

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so in this case we take a cross section of our surface z equals f of xy at y equals y0

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it's parallel with the x axis so we keep y fixed at y equals y0 and the red curve on the top of

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the surface will correspond to our cross section that's defined by the plane y equals y0 this

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orange plane is the slice through the surface in this direction where we keep y fixed and we intersect

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the surface in this red line and again we can along this line here y is fixed y is kept constant

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at y equals y0 and so along the red line y is constant at y equals y0 the thing that's changing

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is x as the guy would walk out along the red line he would go up and down and up and down depending

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on the contusion of the surface and the slope would be changing as he went along the red line

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so that leads us to the partial derivative with respect to x so if we keep y constant at y equals

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y0 and we increase x or equally well decrease it if we increase it though by moving along the red

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curve in this direction parallel to the x axis the slope of the surface changes and that would give us

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the partial derivative with respect to x and similarly to an earlier definition for partial

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derivative z by dy we can do the same thing with respect to x so the partial derivative of a function

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z with respect to x is written like this with the curly d's curly dz by dx and so rate of change

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the rate at which the function changes as x changes and y is kept constant so y is kept

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constant in this case at the value y0 so this this curve has the value y0 all the way along with the

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thing that's changing though is x and to calculate the partial derivative with respect to x that's

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partial dz by dx we differentiate z equals f of x y with respect to x and keep while keeping y constant

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so just to summarize that i've just finished saying this to differentiate to obtain partial

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dz by dx we differentiate z with respect to x treating y as a constant and the notation we have

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we've seen the curly dz by dx we've seen that alternative notation short more short hand notation

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the z with x as a subscript that means exactly the same as this notation here and because z is

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equal to f you can just replace the z by f if you prefer so any one of these they all mean exactly

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the same thing the partial derivative with respect to x similarly we can do the same thing for y

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where we've got the partial derivative with respect to y which we've seen before in this

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notation here this notation is called Leibniz after the mathematician Leibniz this Leibniz

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notation we can also write it in short hand notation and we're doing it with respect to y with y as a

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subscript and again we can replace z with f and all of these mean exactly the same that the partial

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derivative with respect to y so now we've got the background of the motivation of the

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geometrical interpretation out of the way let's have a look at some examples so suppose we've got

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a function z is equal to x squared minus 2x squared y cubed plus 4 y plus 5 well that function

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represents a surface in three dimensions it's a function of x and a function of y so z is equal

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you could write it like this if you want it you know you could write z is equal to f of x y if

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you like so this is also equal to f of x y so there we are so we want to calculate the partial

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derivative with respect to x and the partial derivative with respect to y so the question

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has to do that first and then we have to calculate the slope in both the x direction and the y

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direction at this point p okay so let's have a look at doing that so

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okay so let's do the partial derivative with respect to x first of all so we've got partial dz

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by dx so all I do here is I replace z by the function given in the question there you go

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that's what we've done now when we used to differentiate functions of a single variable

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as you know you just do term by term if you've got x squared plus 3x you would differentiate the x

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squared and you would differentiate the x and you would add them together so you can just do

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term by term differentiation well you can do exactly the same when you've got a function

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of two or more variables you just differentiate your function term by term so let's have a look

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at doing that so I've got four terms here one two three four so I'm going to just take the derivative

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with respect to x of each of the four terms and that's all I'm doing here remember of course these

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are partial derivatives I use the currently denotation because I've got a function of two variables

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so the partial with respect to x of x squared minus the partial respect to x of 2x squared y

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cubed plus the partial with respect to x of 4y plus the partial with respect to x of the constant

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five so this is actually very easy this one because this is just a function of a single

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variable so there's nothing new here this is just the same differentiation you've seen before

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you differentiate an x squared with respect to x so you would get 2x what about this one here now this

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involves an x and a y and they're at a constant they're all it's a product 2x squared y cubed the

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derivative is with respect to x so in this example just starting out I'm going to take

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everything that does involve x outside the derivative operator and the reason for that is

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remember that when we differentiate with respect to x we treat y and any function of y as a constant

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so if we treat y as a constant then y cubed is a constant and so 2 is 2y cubed so we can take the

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2y cubed outside and that just leaves me x squared to differentiate with respect to x

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now this term here doesn't involve x so I'm just going to leave it as it is it's again it's a function

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of a single variable and the variable being y I'm just going to leave that as it is I'll deal

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with that shortly just leave it as it is and again leave the constant as it is so I've got d by dx

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of the constant five so going through these I've already done this one but let's just say it again

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differentiate x squared with respect to x I get 2x I've got to do the same thing here differentiate

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x squared with respect to x so I get 2x don't forget the 2y cubed you're multiplying by and

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there's a minus there plus now I've got to differentiate 4y with respect to x it's an x

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derivative I'm calculating so as we talked about earlier we treat y as a constant and we also treat

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any function of y as a constant so 4y will be a constant will be thought of as a constant

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and if you differentiate a constant with respect to x you get zero so that's why the zero comes in

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here and again this is purely just a constant five and when you differentiate a constant with respect

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to x you get zero so that's where that zero comes from and I can tidy this a little up a little bit

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multiply them together to get 4xy cubed and here's my answer 2x minus 4xy cubed so that's the

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partial derivative with respect to x and that expression will allow us to calculate the slope

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of the curve at any point in the x direction we're now doing the same thing for y again I'll

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replace my z with a function that's given in the question I'm going to differentiate this term by

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term so it's the derivative with respect to y of each of these four terms this is just a function

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of a single variable so I'll leave it as it is that's the way I'm doing these is whenever I've got a

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function of a single variable I'm just leaving it as it is so that one stays as it is four y's

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significant that'll stay as it is and the five is a constant so that'll just stay as it is this

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one I've got a product of x is in y's so what I'm going to do is like I did in the previous one

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with respect to x I'm going to I'm differentiating here with respect to y so x and any function of x

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is treated as a constant so x squared will be treated as a constant and 2x squared will be

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treated as a constant you can lump all of this together and to all intents and purposes it's

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just a constant term because we'll differentiate with respect to y so the 2x squared I can take

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outside the operator so out it comes and that just leaves me y cubed to differentiate with respect

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to y so I can go through each of these term by term so differentiate x squared with respect to y

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so x and any function of x and therefore x squared is treated as a constant so its derivative is zero

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the derivative of three y of y cubed with respect to y is three y squared so that comes there

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the derivative of four y with respect to y is just four and the derivative of the constant five

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with respect to y is just zero and again I can combine all of this so I'll have four minus six

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x squared y squared so that expression there is for the partial derivative with respect to y

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and it'll allow us to calculate the slope of the surface at any point in the y direction

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okay so let's have a look at calculating the slope in the x and y direction so as we saw in part one

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we found the partial derivative with respect to x and we also found the partial derivative with respect

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to y and we'll ask to calculate the slope of the x direction at a point p three minus 128

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so to do that I use my expression for the partial derivative with respect to x

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and this notation with this notation here means take your partial derivative with respect to x

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the expression that you found and substitute in these values of course it's just the x and y value

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you sub in we just included the z value just to show where you would be standing on the surface

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that's the coordinates of the point on top of the surface so we sub in the x value and the y value

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into the partial d z by dx we found earlier and we find that the value is 18 so the partial

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derivative with respect to x is positive and if we keep y fixed or we keep y constant and very

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x the function would be increasing at the point p so if you look along the surface in the direction

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of x you're you're varying x moving parallel with the x axis and in that case the surface is

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increasing you'd be climbing your pole so we do the same thing now for the slope in the y direction

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so we use these partial dz by dy and we plug in the x and y coordinates into this expression here

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and we find we get a negative value so the partial derivative with respect to y is negative

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so the function is decreasing at the point p as we vary as we vary

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as we vary y and keep x constant okay so we're so we're keeping x constant in this case

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okay so let's have a look at an example of doing that so here's our diagram I've taken

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from this website you can see this was called a saddle point because it does look like a saddle

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so we can see suppose this is our point p here well if we move off parallel with the x axis

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we can see the slope is increasing so at the point p if we move off parallel with the x axis

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we can see that the slope is increasing on the other hand if we move off parallel with the y axis

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you can see the slope is decreasing so it is perfectly possible to stand at a point p on a surface

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and for you to be increasing in one direction and decreasing in another direction at standing at

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that point okay so let's have a look now at another example

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so here's a function of two variables again we're surfacing three dimensions and we want to

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find the partial dz by dx so I replace z by the given function there it is inside the bracket

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I can do term by term differentiation so I'll differentiate the first term plus the derivative

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00:35:28.640 --> 00:35:33.040
of the second term and these derivatives are partial derivatives calculate respect to x

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00:35:34.080 --> 00:35:41.760
so in this one I'm going to take out anything that doesn't involve x so the 27 over 5 y can come

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00:35:41.760 --> 00:35:48.160
outside and that just needs me x to differentiate with respect to x and this is just a function of

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00:35:48.160 --> 00:35:56.000
a single variable function of x so I'll just leave it as it is so I differentiate x with respect to

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00:35:56.000 --> 00:36:04.960
x I get 1 and if I differentiate this term 1 3 1 over 4 x with respect to x all that survives

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as a constant 1 3 1 over 4 so putting all of this together I get the answer in the blue box at the

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foot of the slide so that's the partial derivative with respect to x of this function so we treat y

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00:36:19.440 --> 00:36:27.840
as a constant now in this one we've been asked to find the partial respect to y and that requires us

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00:36:27.840 --> 00:36:37.600
to treat x as a constant so again partial dz by dy I replace z with a given function there it is

294
00:36:38.240 --> 00:36:46.240
I'll do term by term differentiation with respect to y so my first term is here plus the derivative

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00:36:46.240 --> 00:36:52.800
of my second term here it's a y derivative so I can take out anything that doesn't involve y I'm

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treating x as a constant so we treat 27 over 5 times x as a constant so out it comes and that just

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00:37:02.720 --> 00:37:08.160
leaves me y to differentiate with respect to y this is a function of a single variable so I'll

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just leave it as it is because that'll be easy to handle so I differentiate y with respect to y

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that'll give me 1 so I'll obtain this expression here which is 27 over 5 x plus the derivative of

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00:37:24.320 --> 00:37:32.560
this expression with respect to y well this only involves x well it's a constant times x a function

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00:37:32.560 --> 00:37:38.080
of x and we know that x is treats are constant how we differentiate with respect to y and any

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00:37:38.080 --> 00:37:44.320
function of x treats are constant so how we differentiate this term here will get 0 and

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00:37:44.320 --> 00:37:52.800
my answer is as given in the blue box at the foot of the slide okay so let's have a look at another

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example so in this case we've been asked to differentiate the function z is equal to sine x

305
00:38:02.000 --> 00:38:07.520
times sine 2y we want to do partial derivative with respect to x and the partial derivative with

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00:38:07.520 --> 00:38:17.200
respect to y now one thing I'll say here just it's sometimes tempting to use the product rule here

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you could use the product rule perfectly fine that would be perfectly okay and you would get

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00:38:24.160 --> 00:38:29.040
the correct answer but you don't actually need the product rule here because if you look at it

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remember the product rule comes into play when you get two functions

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00:38:36.800 --> 00:38:44.720
involving the same variables multiplied together but this one your first function only involves x

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00:38:44.720 --> 00:38:50.720
and your second function only involves y so you don't actually need the product rule the product

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00:38:50.720 --> 00:38:58.480
rule if you had if this first function had been something like sine of x y sine 2y if you wanted

313
00:38:58.480 --> 00:39:03.760
to calculate the partial derivative with respect to y you would need the product rule because this

314
00:39:03.760 --> 00:39:10.000
is a function of x and y and this is a function of y so you could two functions of y multiply

315
00:39:10.000 --> 00:39:16.000
together so you would need the product rule but we'll see examples of the product rule in the next

316
00:39:16.000 --> 00:39:22.160
lecture in partial differentiation so we'll just leave that for now here we don't need the product

317
00:39:22.160 --> 00:39:27.520
rule because we've just got a function of x times a function of y so let's just go straight into it

318
00:39:27.520 --> 00:39:34.000
and calculate the partial derivative so the partial derivative of z with respect to x we replace z

319
00:39:34.000 --> 00:39:39.920
by the given function there it is now it's a partial derivative with respect to x or y and any

320
00:39:39.920 --> 00:39:46.960
function of y is treated as a constant so the sine 2y can come outside the operator and that just

321
00:39:46.960 --> 00:39:55.360
leaves us sine x to differentiate with respect to x and that's easy we know that it's just cos x

322
00:39:55.360 --> 00:40:03.520
so my final answer is that the partial derivative with respect to x is sine y cos sine 2y cos x

323
00:40:04.720 --> 00:40:10.960
we're then asked to find the partial derivative with respect to y and again we replace z with a

324
00:40:10.960 --> 00:40:18.160
given function it's a y derivative so x and any function of x this treat as a constant so sine x

325
00:40:18.160 --> 00:40:26.400
can be taken outside and that just leaves us sine of 2y to differentiate with respect to y

326
00:40:26.400 --> 00:40:34.080
now I'll just say something about this one here we've got sine of 2y now the derivative of that

327
00:40:34.080 --> 00:40:43.520
is not cos of 2y because you've got a function of a function so we need the chain rule some of you

328
00:40:43.520 --> 00:40:48.080
might know this is the composite rule because you've got a function composed of more than one

329
00:40:48.080 --> 00:40:53.600
function it's also called the composite rule we call it the chain rule so we need the chain rule

330
00:40:53.600 --> 00:40:59.600
here and the chain rule if I'm going to just use this notation for differentiation with respect to

331
00:40:59.600 --> 00:41:05.680
y d with a subscript y so I want to differentiate this function with respect to y so I differentiate

332
00:41:05.680 --> 00:41:13.680
the outside function cos I leave the bracket as it is 2y and I've got to multiply by the

333
00:41:13.680 --> 00:41:20.240
derivative of the term inside the brackets so that's the chain rule for a function of one variable

334
00:41:20.240 --> 00:41:25.760
I'll I'll actually be doing an example of that shortly before we look at the chain rule for a

335
00:41:25.760 --> 00:41:32.480
function of two variables but that's just a quick reminder of how you differentiate a function of

336
00:41:32.480 --> 00:41:39.840
a function differentiate the outside function sine becomes cos leave the bracket as it is

337
00:41:39.840 --> 00:41:47.760
and then multiply by the derivative of the term inside the brackets so I would I would get 2 cos

338
00:41:47.760 --> 00:41:53.200
2y and I've got to multiply that by sine x which I took outside because that was just

339
00:41:53.200 --> 00:42:02.800
been treated as a constant so I get 2 sine x cos 2y okay so there we are so we've got another

340
00:42:02.800 --> 00:42:08.880
example here similar cases to the last one we've got a function of y multiplying a function of x

341
00:42:08.880 --> 00:42:16.160
so we don't need the product rule so let's do the partial with respect to x first of all so I

342
00:42:16.160 --> 00:42:22.800
replace said by the given function it's an x derivative so y and any function of y can get

343
00:42:22.800 --> 00:42:31.200
taken outside and treated as a constant so e to the 3 y comes out and that just leaves me log 4x

344
00:42:31.200 --> 00:42:39.040
squared to differentiate now a common mistake here when people differentiate log of 4x squared

345
00:42:39.040 --> 00:42:43.440
is a function of a single variable so it's just the same differentiation as we've been

346
00:42:44.240 --> 00:42:50.320
you know come across up to now the common mistake here is that people think the derivative of log

347
00:42:50.320 --> 00:42:58.000
of 4x squared is 1 over 4x squared it's not that's wrong that's incorrect you need the chain rule

348
00:42:58.000 --> 00:43:05.440
because you've got a function log applied to another function 4x squared so you get a function

349
00:43:05.440 --> 00:43:12.800
of a function so that means you need to use the chain rule and the chain rule first of all

350
00:43:12.800 --> 00:43:18.720
differentiate the outside function when you differentiate log it's going to be 1 over whatever's

351
00:43:18.720 --> 00:43:25.680
inside the bracket so it's going to be 1 over 4x squared so anyone who did this would be partly

352
00:43:25.680 --> 00:43:32.320
correct but because it's a chain rule you must multiply by the derivative of the term inside

353
00:43:32.320 --> 00:43:38.720
the bracket and that would be 8x when you differentiate 4x squared with respect to x you get

354
00:43:39.680 --> 00:43:47.200
now you can simplify this 4 goes into that once goes in there twice x goes to that once x goes

355
00:43:47.200 --> 00:43:54.080
into x squared once it leaves you x so you'd be left with 2 over x and that's what I've shown

356
00:43:54.080 --> 00:44:00.320
in the green box here I've just shown the differentiation been done in in more detail there

357
00:44:00.320 --> 00:44:09.200
there it is so when we put all of this together I'll have e to the 3y times the derivative of

358
00:44:09.200 --> 00:44:18.720
this term which was 2x so I'll just get 2e to the 3y over x okay so that's the partial derivative

359
00:44:18.720 --> 00:44:24.880
of this function with respect to x so let's do the partial derivative with respect to y and once

360
00:44:24.880 --> 00:44:32.000
again we don't need the product rule because we've got a function of y multiply a function of x

361
00:44:32.000 --> 00:44:39.280
it's a y derivative so we treat anything that involves x as a constant so log of 4x squared

362
00:44:39.280 --> 00:44:47.200
will be treated as a constant so let's set up a differentiation we'll replace said by the given

363
00:44:47.200 --> 00:44:55.040
function y derivative so x and any function of x therefore log of 4x squared is treated as a constant

364
00:44:55.040 --> 00:45:04.880
so out it comes so we've got to differentiate e to the 3y with respect to y okay so this again

365
00:45:06.000 --> 00:45:12.000
is a function of a function it's an exponential function applied to the function 3y so we need

366
00:45:12.080 --> 00:45:19.040
the chain rule so e to the 3y it's a chain rule for a function of a single variable of course

367
00:45:19.040 --> 00:45:24.400
e to the 3y is our function we can write it like this some people find it a bit easier to see the

368
00:45:24.400 --> 00:45:30.640
chain rule been applied if we write you know if we've got brackets present and x is a common

369
00:45:30.640 --> 00:45:38.400
notation for for the exponential function like saying computer programs we can write as x 3y so

370
00:45:39.360 --> 00:45:44.080
let's do the chain rule the chain rule says differentiate the outside function well we know

371
00:45:44.800 --> 00:45:50.640
that the exponential function is its own derivative so the exponential just becomes the exponential

372
00:45:50.640 --> 00:45:57.520
when you differentiate leave the bracket as it is so you get 3y times the derivative of the term

373
00:45:57.520 --> 00:46:08.320
inside the bracket which is 3 so you get 3e cubed y when we write it in this notation here 3x3y

374
00:46:08.400 --> 00:46:15.360
and it's given like so so this is what we've got then we've got our log 4 of x squared on the outside

375
00:46:15.360 --> 00:46:20.240
multiplying the derivative term we've just calculated the derivative term as 3e to the 3y

376
00:46:20.240 --> 00:46:26.400
and I can rearrange them if I want I don't have to but I find it neater to write it like this so I

377
00:46:26.400 --> 00:46:37.280
get 3e to the 3y log 4x squared so that is the partial derivative with respect to y okay so

378
00:46:37.280 --> 00:46:42.480
now we've just finished talking about the chain rule and we saw how it is essentially a shortcut

379
00:46:42.480 --> 00:46:51.200
approach you know to the chain rule in this example you know like differentiate the outside function

380
00:46:51.200 --> 00:46:56.400
the exponential leave the bracket as it is and multiply by the term inside the brackets but

381
00:46:56.400 --> 00:47:04.320
sometimes it's useful to see the more formal definition of the chain rule and it'll help us

382
00:47:04.400 --> 00:47:12.960
when we come to calculating partial derivatives using the chain rule so the chain rule is for

383
00:47:12.960 --> 00:47:19.440
calculating derivatives of what are called composite functions that's a function of a function

384
00:47:20.720 --> 00:47:27.360
and that's when we were working with one variable we saw examples of that just a few minutes ago

385
00:47:27.360 --> 00:47:34.080
but I'm going to give you the full definition here so suppose we've got a function y is equal to f of u

386
00:47:34.480 --> 00:47:42.960
where u is equal to some function g of x so for our example we could have you know if I just jump

387
00:47:44.320 --> 00:47:51.120
you know the example I'm going to do let's just look at this we could have y is equal to sign u

388
00:47:52.320 --> 00:48:03.840
where u is equal to what was it x squared plus 3x x squared plus 3x so that's what we've got

389
00:48:04.720 --> 00:48:12.480
so how do we differentiate this well let's look at our diagram here and hopefully this will

390
00:48:12.480 --> 00:48:22.800
will help so we've got at the at the root node we've got y and at the terminal node we've got x

391
00:48:24.080 --> 00:48:31.440
so we want to calculate dy by dx but y is written as a function of u which is in turn

392
00:48:31.440 --> 00:48:36.960
a function of x and that's what we're showing here y is a function of u which in turn is a

393
00:48:36.960 --> 00:48:44.160
function of x now to calculate dy by dx start at the root node go to the terminal node multiply

394
00:48:44.160 --> 00:48:50.640
derivatives on your way so you would calculate y is a function of u so you could calculate dy by du

395
00:48:51.360 --> 00:48:58.320
u is a function of x so you could calculate du by dx and to get dy by dx you just multiply

396
00:48:58.320 --> 00:49:05.440
these two derivatives together so let's have a look at an example so again I've copied my diagram over

397
00:49:05.440 --> 00:49:15.520
here um there it is there's my tree diagram here I've got y this equal to sign of x squared plus

398
00:49:15.520 --> 00:49:24.480
3x now I could do that using the the method the the short method that I used earlier it's fine

399
00:49:24.560 --> 00:49:28.400
when we're working with simple functions but sometimes as functions become more complicated

400
00:49:29.120 --> 00:49:35.120
it's much better to use this approach so we'll for illustration purposes we'll use this full

401
00:49:35.120 --> 00:49:40.240
approach and that'll help us when we come to using the chain rule for a function of two variables

402
00:49:40.240 --> 00:49:46.000
so I'm going to call the thing in brackets u that's a standard approach called set the dummy

403
00:49:46.000 --> 00:49:52.800
variable u equal to the expression in brackets so u is equal to x squared plus 3x so y is equal to

404
00:49:52.800 --> 00:50:01.520
sign u there you are I've got to differentiate uh y with respect to u so dy by du is just

405
00:50:01.520 --> 00:50:07.600
cause u differentiate sign you get cause and use a function of x so I differentiate u with respect

406
00:50:07.600 --> 00:50:18.480
to x to get 2x plus 3 my formula tells me that dy by dx is just the product of dy by du and du by

407
00:50:18.480 --> 00:50:25.840
dx now my three diagrams over here I've got my my variable y at the root node and I've got the

408
00:50:25.840 --> 00:50:35.600
variable x at the terminal node and I want to find dy by dx so dy so y is at the root x at the terminal

409
00:50:35.600 --> 00:50:44.880
node so I go from y all the way down to x and I multiply derivatives together on my way so y is

410
00:50:44.960 --> 00:50:52.000
a function of u so I get dy by du use a function of x I've got du by dx I've already calculated

411
00:50:52.000 --> 00:50:59.360
them I know what they are dy by du is cause u du by dx is 2x plus 3 so I obtained this expression

412
00:50:59.360 --> 00:51:07.440
here now the original question that was asked was to determine dy by dx so really I can't leave

413
00:51:07.440 --> 00:51:13.120
u in here but that's not a problem because I know what u is I'm the one who defined it I set it equal

414
00:51:13.120 --> 00:51:19.840
to the expression inside the bracket which is x squared plus 3x sorry sorry which was yes x squared

415
00:51:19.840 --> 00:51:28.480
plus 3x so I do cause of u becomes cause of x squared plus 3x times 2x plus 3 so I've written it

416
00:51:28.480 --> 00:51:38.720
in this form here so that's an example of the the chain rule applied to a function of one variable

417
00:51:38.720 --> 00:51:46.320
and you will have seen that before so now we're going to extend that definition of the chain rule

418
00:51:46.880 --> 00:51:53.680
to the case where we've got a function of more than one variable now in that last example

419
00:51:54.800 --> 00:52:01.760
we set u equal to the term inside the bracket so u is a function of the single variable x

420
00:52:02.720 --> 00:52:10.960
y was a function of u and u was a function of x only x both functions of single variables

421
00:52:10.960 --> 00:52:19.040
now when we in this example this chain rule we're going to look at the case when u is a function

422
00:52:19.040 --> 00:52:28.640
of two variables x and y so if I just if I can just take my diagram over to the slide

423
00:52:29.040 --> 00:52:35.600
just bear with me while I copy it there we are so this was the case of a function of a single

424
00:52:35.600 --> 00:52:44.400
variable of one variable y was a function of u which in turn was a function of x in this case now

425
00:52:45.440 --> 00:52:52.160
z is a function of u that's fine it's a function of a single variable but u is now a function of x

426
00:52:52.240 --> 00:53:01.040
and y that means we can calculate as before we can calculate the derivative of the variable at the

427
00:53:01.040 --> 00:53:08.080
root node with respect to u in other words dz by du similar to our single variable case there

428
00:53:08.720 --> 00:53:17.440
we just do z by du but now use a function of two variables so we must calculate the partial derivative

429
00:53:17.440 --> 00:53:23.360
with respect to each variable so we've got partial respect to u and the partial partial respect to x

430
00:53:23.360 --> 00:53:31.120
sorry and the partial with respect to y and then if we want the derivative of our function z with

431
00:53:31.120 --> 00:53:38.800
respect to x that's a partial derivative we simply start at the root node and travel down to the

432
00:53:38.800 --> 00:53:47.680
terminal node multiplying the derivatives on our way so the partial with respect to x becomes the

433
00:53:47.680 --> 00:53:57.840
derivative of z with respect to u times the partial derivative of u with respect to x now I think

434
00:53:57.840 --> 00:54:03.280
I've mentioned this before but why have I got a straight back d there and a curly d there well it's

435
00:54:03.280 --> 00:54:12.480
because z is a function of one variable so we just use the standard notation that we've always

436
00:54:12.480 --> 00:54:17.760
used for a derivative because we only dealt with derivatives of functions of single variables up

437
00:54:17.760 --> 00:54:23.520
to now and this is the notation we'd have seen a straight back d representing the derivative

438
00:54:23.520 --> 00:54:28.960
when we have functions of more than one variable functions like in this case a function of two

439
00:54:28.960 --> 00:54:34.560
variables we've got to introduce the curly d to show that it's a partial derivative that's all it is so

440
00:54:35.680 --> 00:54:42.560
we've covered this one the partial respect to x is the product of the derivatives down this branch

441
00:54:42.560 --> 00:54:50.320
now similarly we can get the partial of z with respect to y by starting at the root node z and

442
00:54:50.320 --> 00:54:58.800
traveling down the right hand branch of the tree to the terminal node y and we multiplied derivatives

443
00:54:58.800 --> 00:55:07.120
as we go along that journey so we'll have the derivative of z with respect to u times the

444
00:55:07.120 --> 00:55:17.760
partial derivative of u with respect to y and that's the formula given down here okay so we can now

445
00:55:17.760 --> 00:55:22.240
go ahead and look at an example

446
00:55:25.760 --> 00:55:31.760
so what do we have here so we've got a function z is equal to f of x y if you like

447
00:55:31.760 --> 00:55:39.040
this equal to 2x cubed plus x y plus y to the four all to the power six now

448
00:55:41.600 --> 00:55:46.400
the question here is do we need the chain rule for this and the answer is yes we do

449
00:55:46.480 --> 00:55:53.920
because we've got a function that's the power six function applied to another function the

450
00:55:53.920 --> 00:56:01.600
function inside the brackets it's probably more obvious to see if you had a sign out there rather

451
00:56:01.600 --> 00:56:06.320
than the power six that you know a function of a function but this is a function of a function

452
00:56:06.320 --> 00:56:11.360
you've got the power six function applied to the function inside the bracket so we're going to use

453
00:56:11.360 --> 00:56:18.080
the chain rule to calculate the partial derivatives with respect to x and with respect to y so we'll

454
00:56:18.080 --> 00:56:23.760
do we'll start off by doing what we normally do we've got function of a function or a composite

455
00:56:23.760 --> 00:56:31.760
function you prefer to call it that so z is going to be equal to u to the power six where u

456
00:56:32.560 --> 00:56:39.920
is set equal to the term inside the brackets which is standard practice you set the w variable u

457
00:56:39.920 --> 00:56:46.800
equal to whatever's inside the brackets in general that's the approach that we take we'll set u equal

458
00:56:46.800 --> 00:56:54.480
to 2x cubed plus x y plus y to the four and if that is u then z is equal to u to the six okay

459
00:56:55.760 --> 00:57:02.480
so uh i've copied that over to the slide here i've copied over my diagram my three diagram that i'm

460
00:57:02.480 --> 00:57:09.840
going to use so i need the full derivative that's a derivative with the straight back d of the single

461
00:57:10.560 --> 00:57:14.960
of z which is function of the single v able u i need that derivative and i need the two

462
00:57:14.960 --> 00:57:22.240
partials that's u with respect to x and u with respect to y and my formula first of all i'm

463
00:57:22.240 --> 00:57:27.600
going to calculate the partial derivative with respect to x so there's my formula there which

464
00:57:27.600 --> 00:57:34.320
i obtained by starting at the root node traveling down to the terminal node at x and multiplying

465
00:57:34.320 --> 00:57:41.680
the derivatives together as i go so first of all the derivative of z with respect to u well that's

466
00:57:41.680 --> 00:57:49.120
easy single variable to be the power power low it'll be six u to the five when i differentiate that

467
00:57:49.120 --> 00:57:55.040
there we are and i've got to multiply that with a partial derivative with respect to x so i want

468
00:57:55.200 --> 00:58:01.520
partial derivative of this term with respect to x now i could break it up and you know till

469
00:58:01.520 --> 00:58:07.440
become more confident with these type of things it's probably a good idea to break it up into the

470
00:58:07.440 --> 00:58:14.560
separate three derivatives and differentiate with respect to x so but let's see if we can do this

471
00:58:14.560 --> 00:58:20.800
without doing that just uh we've got to differentiate two x cubed with respect to x well that's easy

472
00:58:20.800 --> 00:58:26.400
because you know it's just a function of a single variable it's going to be six x squared

473
00:58:28.240 --> 00:58:34.240
leave the middle term for now differentiate y to the four with respect to x now it's an x derivative

474
00:58:34.240 --> 00:58:41.440
i'm calculating so y and any function of y will be treated as a constant and that includes y to the

475
00:58:41.440 --> 00:58:47.600
power four so y to the power four has been treats a constant differentiate that with respect to x

476
00:58:47.600 --> 00:58:54.400
that'll be zero so that just leaves me the middle term to differentiate with respect to x

477
00:58:55.120 --> 00:59:01.440
so here it is i've just isolated it here there you go it's an x derivative so y

478
00:59:01.440 --> 00:59:07.200
treats the constant so out it comes i've then just going to differentiate x with respect to x so

479
00:59:07.200 --> 00:59:15.680
that's just one and y times one is y so that will just be y when we differentiate so this

480
00:59:16.240 --> 00:59:24.000
term when i differentiate this term with respect to x i get six x plus six x squared sorry plus y

481
00:59:24.000 --> 00:59:32.160
there it is there and i've got to multiply that by six u to the five u was defined up here so i

482
00:59:32.160 --> 00:59:39.120
just plug that in there so i get six times this expression all to the power five don't forget it's

483
00:59:39.120 --> 00:59:44.720
raised to the power five when you differentiate so that's the answer there so i've now got to do

484
00:59:45.440 --> 00:59:56.560
the derivative with respect to y same approach really i look at my diagram and i've got to travel

485
00:59:56.560 --> 01:00:02.800
down from the root node to the terminal node that includes y multiplying derivatives on my way so

486
01:00:02.800 --> 01:00:10.160
i'll go down here z is a functional single variable u so i've got the full derivative dz by du and i've

487
01:00:10.160 --> 01:00:17.680
got to multiply that by the partial derivative du by dy so let's calculate these so i've got um

488
01:00:18.960 --> 01:00:26.880
dz partially z by dy is the full derivative z by u times the partial du by dy okay full derivative

489
01:00:26.880 --> 01:00:35.680
of dz by du z is equal to u to the six so that's we six u to the five there we are now this one i need

490
01:00:35.680 --> 01:00:41.680
a partial derivative with respect to y of this this term inside the bracket which involves

491
01:00:41.680 --> 01:00:50.080
three separate terms so i'll differentiate them one by one so the first term is a function of x only

492
01:00:51.520 --> 01:00:56.800
and it's a y derivative i want so we treat x and any function of x is a constant so you differentiate

493
01:00:56.880 --> 01:01:02.560
two x cubed with respect to y you'll get zero so that won't contribute let's do the last one first

494
01:01:02.560 --> 01:01:09.440
of all differentiate y to the four with respect to y you get four y cubed okay so that's fine

495
01:01:10.080 --> 01:01:15.680
four y cubed what about the middle term well let's have a look at it over here this is very

496
01:01:15.680 --> 01:01:21.680
similar to the last one except now we differentiate with respect to y so it's a y derivative so x gets

497
01:01:21.680 --> 01:01:28.000
three to the constant so we can take the x outside the operator there it goes differentiate y with

498
01:01:28.000 --> 01:01:36.880
respect to y i'll get one x times one is x so that's where that x comes from so i'll have six u to the

499
01:01:36.880 --> 01:01:46.880
five times x plus four y cubed i need my answer in terms of x so that requires me to substitute for

500
01:01:46.880 --> 01:01:52.880
u in here but u is just as given up here to the term that's inside the brackets there you go so i'll

501
01:01:52.880 --> 01:02:01.200
get six times this expression to the power five times x plus four y cubed so that's uh

502
01:02:04.000 --> 01:02:10.560
that's the derivative with respect to y of my function my original

503
01:02:11.120 --> 01:02:22.080
composite function my function of a function okay so let's have a look at another of these

504
01:02:22.080 --> 01:02:29.280
so here's a function z equals f of x y is sine of three x to the five y squared minus two x to the

505
01:02:29.280 --> 01:02:36.720
four y and i want to find the partial derivative with respect to x of that function so we'll do

506
01:02:36.800 --> 01:02:44.080
the same thing as we did before we'll use our diagram here we've got z will be a function of

507
01:02:44.080 --> 01:02:53.200
u which in turn will be a function of x and y so we set u equal to the expression inside the

508
01:02:53.200 --> 01:03:02.800
bracket so u will be as given here so then z will be equal to sine u okay i've set u equal to this

509
01:03:03.440 --> 01:03:12.320
bracket term so clearly z will just be sine u i want to get the partial respect to x so i start at

510
01:03:12.320 --> 01:03:18.880
the root node with z and i travel down to the terminal node with x and i multiply derivatives

511
01:03:18.880 --> 01:03:27.360
as i go along so let's do that so first of all i need dz by du well that's pretty easy

512
01:03:27.360 --> 01:03:38.160
um i need the derivative of sine u which is just cos u just a function of a single variable so sine

513
01:03:38.160 --> 01:03:44.160
differentiates to cos so it could cause you i've then got to do the derivative of this term here

514
01:03:44.160 --> 01:03:51.840
with respect to x so um you can do it term by term so let's have a look at that so i'm going to do

515
01:03:52.480 --> 01:04:09.360
d by dx of 3x to the 5 y squared minus partial d by dx of 2x to the 4 y like that

516
01:04:10.640 --> 01:04:19.840
no it's an x derivative so anything involved in y can come outside so i can actually take the

517
01:04:19.840 --> 01:04:23.120
3 y outside and that just leaves me

518
01:04:25.920 --> 01:04:35.680
x to the 5 to differentiate minus it's an x derivative so i can take out anything that

519
01:04:35.680 --> 01:04:41.920
does involve x so the 2 y can come outside and it leaves me x to the power 4 to differentiate

520
01:04:42.000 --> 01:04:50.000
so if i do this i've got 3 y squared times differentiate x to 5 i get 5x to the 4 minus

521
01:04:50.000 --> 01:05:02.240
2y times 4x cubed and if i work these out i get 15x to the 15x to the 4 y squared you know when i

522
01:05:02.240 --> 01:05:12.640
multiply that out 15x to the 4 y squared minus it'll become 8x cubed y and that's exactly what

523
01:05:12.640 --> 01:05:24.320
i've got on the at that line there okay so uh i'm almost done i'm almost done all i have to do now

524
01:05:24.320 --> 01:05:31.840
is replace u with the bracketed term the thing that i set equal to u at the very start so if i do

525
01:05:31.840 --> 01:05:41.440
that that will give me my partial derivative of this function with respect to x okay so that's

526
01:05:41.440 --> 01:05:49.520
fine so now we can do the same thing with respect to y so i set up my problem in a similar way i

527
01:05:49.520 --> 01:05:55.040
look at my diagram i've got the partial derivative with respect to y the root node is z the terminal

528
01:05:55.040 --> 01:06:00.960
node is y so i travel down the right hand branch and i multiply derivatives as i go along

529
01:06:01.680 --> 01:06:07.360
so i'll need the full derivative d set by d u because u is a function of a single variable

530
01:06:07.360 --> 01:06:14.240
and i'll need the partial derivative of u with respect to y partial derivative because u is a

531
01:06:14.240 --> 01:06:21.600
function of two variables x and y so let's set all of that up so i'm i'm i'm multiplying the

532
01:06:21.600 --> 01:06:28.560
derivatives on my way so i get this expression here this formula here for the partial derivative

533
01:06:28.560 --> 01:06:40.160
with respect to y so i need d by d u of sine u which is just cos u and i need partial d by dy

534
01:06:40.160 --> 01:06:47.120
of the bracketed expression there we are so i can do that just as on the previous slide with

535
01:06:47.120 --> 01:06:52.320
respect to x i can do this term by term and differentiate with respect to y so i've got d

536
01:06:52.960 --> 01:07:03.200
by dy of the first term which is 3x to the 5y squared from that i do minus the partial with

537
01:07:03.200 --> 01:07:13.280
respect to y of 2x to the 4y like that so that is just the next line after this one

538
01:07:14.240 --> 01:07:23.440
so i can do that you know i can it's a y derivative so x and any function of x can get pulled outside

539
01:07:23.440 --> 01:07:30.320
the operator so that just leaves me i can take the 3x to the 5 out so it'll just leave me partial

540
01:07:30.880 --> 01:07:39.840
d by dy of y squared minus well i want to it's a y derivative so x and any function of x can get

541
01:07:39.840 --> 01:07:47.920
taken outside so the 2x the 4 gets taken outside and that just leaves me y to differentiate so if i

542
01:07:47.920 --> 01:07:58.000
do this i will get 3x to the 5 times y squared minus 2x to the 4 the derivative of y with respect to

543
01:07:58.000 --> 01:08:06.480
y is just 1 so i can simplify this to give me 3 well it's it's it's it's fine as it is it's it's

544
01:08:06.480 --> 01:08:19.360
done really so that will give me oh sorry of course i didn't actually do the y derivative

545
01:08:20.160 --> 01:08:30.240
dear dear so i had to of course one thing i didn't do was differentiate the y squared to give me 2y

546
01:08:30.240 --> 01:08:39.840
so i can put all of this together to get 6x to the 5 y minus 2x to the 4 sorry about that so

547
01:08:40.800 --> 01:08:46.160
yeah don't forget of course to differentiate the term inside the bracket so i get obtained this

548
01:08:46.160 --> 01:08:51.840
expression here that just leaves me to replace the u by the bracket term and that would give me

549
01:08:51.840 --> 01:09:04.240
my answer for the partial derivative of z with respect to y okay so that's one example of the

550
01:09:04.240 --> 01:09:14.480
chain rule being applied to a function of two variables here's another slightly more complicated

551
01:09:14.480 --> 01:09:26.720
example in that last example we said u equal to the term in the bracket and that was a function

552
01:09:26.720 --> 01:09:39.520
of x and y in this case what do we have we've got a function that said that is a function of x and y

553
01:09:39.760 --> 01:09:50.160
but x and y are themselves the term a function of t so you know that could happen if you just think

554
01:09:50.160 --> 01:09:59.760
about you know you think about say a particle moving about you could have its x y you'd have

555
01:09:59.760 --> 01:10:08.320
its x y z coordinates but you could have that x and y dependent time for example so if you

556
01:10:08.320 --> 01:10:14.480
calculated the x y coordinates at a given time you would get certain values if you calculated let's

557
01:10:14.480 --> 01:10:21.600
say a second later you would get a different value so the x and y coordinates can vary with time for

558
01:10:21.600 --> 01:10:31.440
example so let's have a look at what goes on there then so here's my diagram at my root node i've got

559
01:10:31.520 --> 01:10:40.000
my z function which i've just said is a function of x and y it's a function of the two variables

560
01:10:40.000 --> 01:10:48.240
x and y set to c equal to f of x y x and y in turn are functions of the single variable t

561
01:10:49.600 --> 01:10:56.000
so that's what we've got now what derivatives do we need well set is a function of x and y so we

562
01:10:56.000 --> 01:11:00.720
need partial derivatives so we need the partial respect to x coming down the left branch and we

563
01:11:00.720 --> 01:11:09.120
need the partial respect to y coming down the right branch then x is a function of t only so we'll

564
01:11:09.120 --> 01:11:18.080
need the full derivative that's a straight back d dx by dt also y is a function of t only so we'll

565
01:11:18.080 --> 01:11:25.520
need the full derivative dy by dt so that's our diagram and that's going to allow us to calculate

566
01:11:25.520 --> 01:11:35.280
the derivative of our function z with respect to t and the way we do that is very similar to what

567
01:11:35.280 --> 01:11:45.040
we're doing earlier we we start at the root node and we travel to the nodes the terminal nodes that

568
01:11:45.040 --> 01:11:53.520
involve t and we add the left branch expressions to the right branch expressions in other words

569
01:11:53.520 --> 01:11:59.760
you go down the left branch first let's go down the left branch first you get partial dz by dx

570
01:12:00.640 --> 01:12:12.080
times dx by dt that's it there plus you must do this plus similar journey down the right branch

571
01:12:12.080 --> 01:12:22.080
partial dz by dy times dy by dt and that will give you the derivative of the original function

572
01:12:22.080 --> 01:12:30.240
z with respect to t so that's what we're going to do and it's always easier to illustrate this with

573
01:12:30.240 --> 01:12:38.800
 an example so suppose we've got a function z is equal to f of x y and that's given by x squared

574
01:12:38.800 --> 01:12:47.520
plus y squared i've copied my diagram over z at the root node function of two variables x and y

575
01:12:47.520 --> 01:12:53.680
so i can differentiate with respect to x down the left branch and i can differentiate z with

576
01:12:53.680 --> 01:13:02.320
respect to y down the right branch and these are of course partial derivatives now x is a function

577
01:13:02.320 --> 01:13:09.120
of t only and it's given by t to the power four so i can differentiate x with respect to t

578
01:13:10.000 --> 01:13:16.560
similarly y is a function of t only and it's given by y is equal to sine t so i can differentiate y

579
01:13:17.280 --> 01:13:24.000
with respect to t and that would be a full derivative just like dx by dt with a full derivative

580
01:13:24.000 --> 01:13:29.040
because it's a function of a single variable there they are x is t to the four y sine t

581
01:13:29.040 --> 01:13:35.120
only one variable they're t but z on the other hand is a function of x and y so you need partial

582
01:13:35.120 --> 01:13:42.000
derivatives for that one so let's calculate the four derivatives that we need so we need the partial

583
01:13:42.000 --> 01:13:49.680
dz by dx well that's easy we treat y and any function of y is a constant so this is really

584
01:13:49.680 --> 01:13:56.400
just x squared plus a constant differentiate that with respect to x you get 2x plus 0 when you

585
01:13:56.400 --> 01:14:03.440
differentiate the constant so you get 2x similarly when you differentiate z with respect to y we treat

586
01:14:03.440 --> 01:14:09.600
x and therefore x squared is a constant so when you differentiate a constant you get 0 differentiate

587
01:14:09.600 --> 01:14:15.120
y squared with respect to y you get 2y so there's the partial with respect to y so we've got the

588
01:14:15.120 --> 01:14:21.440
partial respect to x which is 2x and we've got the partial respect to y which is 2y all that's left

589
01:14:21.440 --> 01:14:27.520
for us now to do is to calculate the partial respect to the partial sorry the full derivative of

590
01:14:28.320 --> 01:14:35.280
x with respect to t and the full derivative of y with respect to t x is for t to the fourth so

591
01:14:35.280 --> 01:14:42.960
differentiate that you get 4t cubed y is sine t differentiate that you get cos t so let's put

592
01:14:42.960 --> 01:14:53.120
all of this together we want the full derivative of z with respect to t okay so there it is dz by dt

593
01:14:53.120 --> 01:15:01.440
and the formula tells us it's the derivative stone the left branch plus the product of the

594
01:15:01.440 --> 01:15:06.640
derivative stone the right branch so that's what we've got there's my formula and I've got all the

595
01:15:06.640 --> 01:15:16.400
expressions I need up above so it's going to be 2x times 4t cubed plus 2y times cos t okay I know

596
01:15:16.400 --> 01:15:23.840
what x and y are so they're given to me in the question so I can substitute for x and y so what

597
01:15:23.840 --> 01:15:31.920
I'm going to get I'm going to get x was t to the fourth so sub that in so I'll get 8t to the four

598
01:15:32.560 --> 01:15:41.040
times t cubed and here I've got 2y cos t but I know that y is sine t so I'll get substitute

599
01:15:41.600 --> 01:15:50.800
y sine t in for y I'll get 2 sine t cos t multiply this out or simplify it if you prefer I get

600
01:15:50.800 --> 01:15:56.640
8t to the seven just the law of indices there multiplying t to four times t cubed you just add

601
01:15:56.640 --> 01:16:06.880
the powers to give you t to the seven and you've got plus two sine t cos t so that's how you do that

602
01:16:08.080 --> 01:16:14.720
this one actually since the functions we've been given are relatively simple we could actually

603
01:16:15.600 --> 01:16:21.680
have just substituted for x and y in here and we would just get said to be a function of

604
01:16:21.680 --> 01:16:27.760
the single variable t let's have a look at that so I'm saying we know what x is it's t to the fourth

605
01:16:27.760 --> 01:16:34.160
so sub it in there y is sine t so sub it in there so we get a function z that only involves t and

606
01:16:34.160 --> 01:16:40.720
then it's just differentiation of a function of a single variable in this case we're lucky because

607
01:16:40.720 --> 01:16:48.000
the functions are relatively simple and they allow us to do that without too much trouble but quite

608
01:16:48.000 --> 01:16:53.760
often the functions can be fairly complicated and doing this substitution approach will just

609
01:16:53.760 --> 01:16:59.120
make things quite difficult and you should really use the chain rule but let's just see it for this

610
01:16:59.120 --> 01:17:05.680
particular example and it'll check your answer at the same time so that's what I said so there's the

611
01:17:05.760 --> 01:17:13.120
function z we know what x is it's t to the fourth so sub it in and we know what y is it's sine t so

612
01:17:13.120 --> 01:17:20.800
sub that in so that's our line here simplify that we get t to the eight plus sine of t all squared

613
01:17:22.720 --> 01:17:29.360
differentiate that you get eight t to the seven we need the chain rule here so it's a function of

614
01:17:29.440 --> 01:17:36.880
a function so we're applying the square function to sine t so first of all let's see how do we

615
01:17:36.880 --> 01:17:42.000
differentiate the power we use the power law bring the two down take one away from so you get two

616
01:17:42.000 --> 01:17:49.360
sine t times the derivative of the thing in brackets so the derivative of sine t is cos t so that is

617
01:17:49.360 --> 01:17:54.800
my answer and if you look back to the previous slide you'll see that that is exactly what I had

618
01:17:54.800 --> 01:18:05.280
so we have verified our answer okay so um that's fine so that was an example

619
01:18:07.760 --> 01:18:15.280
if I just take over my diagram if I can take this over yeah and I'm just going to paste it down the

620
01:18:15.280 --> 01:18:22.880
corner here because I'm going to compare it with the next case going back to that example we've

621
01:18:22.880 --> 01:18:30.480
just completed we saw that z was a function of x and y where x and y themselves were in turn a function

622
01:18:31.280 --> 01:18:38.880
of a single variable t now there's no need that x and y have to be function of a single variable

623
01:18:38.880 --> 01:18:43.680
they could be a function of two variables three variables whatever we're going to look at the

624
01:18:43.680 --> 01:18:52.560
case when x and y are a function of two variables called u and v okay the idea is exactly the same

625
01:18:52.640 --> 01:19:03.120
as before so this is just slightly more complicated than the example we just did we've now got that

626
01:19:03.120 --> 01:19:10.240
x and y are a function of two variables rather than just the single variable t so let's see how we do

627
01:19:10.240 --> 01:19:19.360
that so I'm just saying up here let z equals f of x y b a differentiable function of x and y where

628
01:19:19.360 --> 01:19:28.960
x is a function of u and v and y is a function of u and v and u and x and y are both differentiable

629
01:19:28.960 --> 01:19:37.760
as well so if we have that we want the partial respect to u so let's say we're going to look at

630
01:19:37.760 --> 01:19:44.960
our root node and we're going to travel down the left branch until we reach u the terminal

631
01:19:44.960 --> 01:19:54.320
node with u and we're going to add to that the product derivatives as we travel in the right

632
01:19:54.320 --> 01:20:01.920
branch from the root node z to the terminal node with u same ideas in the previous example

633
01:20:04.160 --> 01:20:09.840
okay and we add the left branch to the right branch so what do we got so we want to go from

634
01:20:09.840 --> 01:20:16.480
root node to terminal node terminal node has to be u so and multiply derivatives on our way so we get

635
01:20:16.480 --> 01:20:23.600
the partial of z with respect to x multiplied by the partial of x with respect to u so there we are

636
01:20:23.600 --> 01:20:31.120
so that's that but there it is there plus similar journey here the partial of z with respect to y

637
01:20:31.760 --> 01:20:38.240
times the partial of y with respect to u so you're going from root node with z terminal node with u

638
01:20:38.240 --> 01:20:44.240
multiplying derivatives on your journey so that gives us the partial of z with respect to u

639
01:20:44.880 --> 01:20:52.480
we can do a similar thing for the partial of z with respect to v we're going to do the left branch

640
01:20:52.480 --> 01:20:59.520
we'll do the left branch first and we'll add to that the result from the right branch so let's do it so

641
01:20:59.520 --> 01:21:07.440
the partial of z with respect to x brings us to here times the partial of x with respect to v

642
01:21:08.400 --> 01:21:16.160
so that gives us that expression there plus similar journey in the right branch partial of z with respect

643
01:21:16.160 --> 01:21:22.480
to y times the partial of y with respect to v and that gives us that tip so these are the

644
01:21:22.480 --> 01:21:28.480
expressions that we use to calculate the partial of z with respect to u and the partial of z with

645
01:21:28.480 --> 01:21:36.400
respect to v when z is a function of the two variables x and y which themselves are both functions

646
01:21:36.480 --> 01:21:46.720
the two variables u and v right so let's have a look at an example so z is equal to f of x y

647
01:21:46.720 --> 01:21:55.120
and it's given by this function here x cubed times y squared where x is a function of uv given by

648
01:21:55.120 --> 01:22:01.360
two uv squared and y is also a function of the two variables u and v given by three u squared v

649
01:22:02.080 --> 01:22:08.400
and we're going to ask to find partial dz by du so we just take our formula that's our given formula

650
01:22:09.040 --> 01:22:16.400
and we'll take we'll start at the root node with z and we'll travel to the terminal node

651
01:22:17.520 --> 01:22:23.520
 with u multiplying derivatives as we go we'll do that in the left branch and we'll do that in the

652
01:22:23.520 --> 01:22:30.080
right branch and we'll add the left branch result to the right branch result and that gives us our

653
01:22:30.080 --> 01:22:36.720
formula okay so we just follow the green path from root to terminal node in both the left branch

654
01:22:37.520 --> 01:22:50.800
and in the right branch so in this example I need to find the partial with respect of z with respect

655
01:22:50.880 --> 01:23:03.040
to x so I've got d by dx of x cubed y squared so I can take the y squared outside because it's an x

656
01:23:03.040 --> 01:23:09.680
derivative I'm calculating like it x cubed so if I differentiate that get y squared times

657
01:23:09.680 --> 01:23:18.960
three x squared so it becomes three x squared y squared and that's the result there now I can also

658
01:23:20.080 --> 01:23:32.640
calculate the partial derivative of z with respect to y so I've got d by dy of x cubed y squared so

659
01:23:32.720 --> 01:23:40.720
that means I treat x and any function of x as a constant so I get x cubed d by dy of y squared

660
01:23:40.720 --> 01:23:51.600
and that is just x cubed times two y so I get two x cubed y okay so now I also have to

661
01:23:52.400 --> 01:24:02.560
calculate the partials of x and y with respect to u so I do d by du of x which is

662
01:24:03.600 --> 01:24:10.960
two u v squared so anything that does involve u can come outside so about two v squared comes out

663
01:24:10.960 --> 01:24:20.240
du by d d by du partial derivative of u because I've taken the two v two v squared outside so it's

664
01:24:20.240 --> 01:24:28.160
just two v squared times one so that's just two v squared so that's that expression obtained okay so

665
01:24:28.240 --> 01:24:38.960
the next one that I need to calculate is the partial of y with respect to u so I get d by du of

666
01:24:39.680 --> 01:24:48.400
y which is three u squared v cubed and it's a u derivative so I can take anything that involves

667
01:24:48.400 --> 01:24:59.680
v outside so we get three v cubed d by du of u squared that's a u in there so it becomes three v cubed

668
01:25:01.120 --> 01:25:09.120
times two u when you differentiate that and that becomes six u v cubed and there we go that's

669
01:25:09.120 --> 01:25:16.000
that expression so we obtained the four expressions that we need and now we can use them to calculate

670
01:25:18.560 --> 01:25:27.840
the derivative partial d z by du okay so that's the formula remember we travel down from the root node

671
01:25:28.480 --> 01:25:33.600
to the terminal node in the left branch and we do the same thing in the right branch and we

672
01:25:33.600 --> 01:25:37.040
multiply the derivatives as we go and we add them together so that's what we're going to do now we've

673
01:25:37.040 --> 01:25:44.240
got all the information we need up above partial d z by dx is three x squared y squared times partial

674
01:25:44.240 --> 01:25:54.880
d x by du is two v squared plus partial d z by dy is two x cubed times partial dy by du which is six

675
01:25:54.880 --> 01:26:01.360
u v cubed I can multiply all this out and I get this expression here now question asked us for the

676
01:26:01.360 --> 01:26:08.000
partial d z by du and I've got x's and y's still mixed up here but I know what x is and I know what

677
01:26:08.000 --> 01:26:15.920
y is so I can substitute in for x and y so I'll have an x squared there so there's my x term up here

678
01:26:15.920 --> 01:26:22.720
so that goes in here and it's squared because x squared y squared so I take this term here

679
01:26:22.720 --> 01:26:30.640
and I square it there you go that's it there times v squared plus 12 x cubed sub in for x and cube it

680
01:26:31.440 --> 01:26:42.480
times y there's y there times u times v cubed there and if you simplify that you're to simplify

681
01:26:42.480 --> 01:26:47.840
that you can check this yourselves you would get this answer at the foot of the slide here 504

682
01:26:47.840 --> 01:26:56.640
du to the 6 v to the power 12 okay so we can do a similar calculation to calculate the partial

683
01:26:57.520 --> 01:27:05.760
derivative of z with respect to v so let's let's have a go at that we want the partial derivative

684
01:27:05.760 --> 01:27:14.000
of z with respect to v I'll actually leave that one as an exercise for yourselves um okay but one

685
01:27:14.000 --> 01:27:22.080
thing I want to show you before we finish is that suppose now in this example we would look at this

686
01:27:22.080 --> 01:27:27.920
and we say well our functions are reasonably straightforward we could actually just substitute

687
01:27:27.920 --> 01:27:36.720
x in here and y in here and calculate the partial derivatives directly with respect to u and v

688
01:27:38.080 --> 01:27:47.520
so if I do that said it's equal to x cubed y squared x is equal to this two u v squared so

689
01:27:47.600 --> 01:27:54.240
substitute in there and cube it times y squared so I'll have to square that term and if I do all

690
01:27:54.240 --> 01:28:00.720
that I obtain this expression here and I can easily differentiate this with respect to you

691
01:28:00.720 --> 01:28:08.320
because I treat v as a constant and any function of these so v to the power 12 gets treated as a

692
01:28:08.320 --> 01:28:17.120
constant and all I have to differentiate is the u to the 7 so I get 72 times 7 u to the 6 times v

693
01:28:17.120 --> 01:28:23.120
to the 12 and I simplified out I get this expression here which is exactly the same answer that I got

694
01:28:23.120 --> 01:28:28.960
using the full chain rule um in this case the functions were relatively straightforward that I

695
01:28:28.960 --> 01:28:36.160
could directly substitute them if I wanted to but in general they won't be and you should know how to

696
01:28:36.160 --> 01:28:44.320
use the full chain rule so as I said earlier there's an exercise for you find the partial derivative

697
01:28:44.320 --> 01:28:50.480
with respect of z with respect to v the answer is given you can do it both ways using the full

698
01:28:50.480 --> 01:28:58.400
chain rule and you can do it by substituting to check your answer so that's um the end of this

699
01:28:58.400 --> 01:29:06.320
lecture now you should be able to attempt the Mobius questions one and two and from the lecture

700
01:29:06.320 --> 01:29:12.480
notes you should be able to attempt tutorial questions one and two so these are the questions

701
01:29:12.480 --> 01:29:18.960
you should be able to attempt in the second class in the second lecture on partial differentiation

702
01:29:18.960 --> 01:29:23.360
we're going to look at the product rule for a function two variables we'll also look at the

703
01:29:23.360 --> 01:29:29.280
quotient rule for a function of two variables and we'll also look at higher order partial derivatives

704
01:29:29.280 --> 01:29:35.280
you know you'll remember that if you had a function for talking about functions of single variables if

705
01:29:35.280 --> 01:29:43.840
you had uh I don't know 3x to the 4 you could calculate the first derivative as 12x cubed

706
01:29:43.840 --> 01:29:50.560
then you could calculate a second derivative just by simply differentiating again 36x squared

707
01:29:50.560 --> 01:29:55.680
and so on if you wanted you can do the same thing for partial derivatives slightly more

708
01:29:55.680 --> 01:30:02.400
complicated of course but we're not going to look at derivatives higher than second order so

709
01:30:02.400 --> 01:30:08.720
that's what we'll be doing in the next class so I'll just stop the recording at that