Oxford Calculus: Partial Differentiation Explained with Examples

hello maths fans this is dr tom crawford from the university of oxford and today we're talking about partial differentiation partial differentiation is what happens when you differentiate a function of more than one variable so if we start by considering the standard derivative so if i want to know df by dx for a function f of x then this is equal to the limit as h tends to zero of f at x plus h minus f at x divided by h if we were to look at the graph of this function which maybe would be something like the following so f of x and then x perhaps it would look something like this then what we're doing here is we are taking a point here let's call this x another point at x plus h and what we're doing is we're approximating this gradient and we're taking the limit as we allow this point to move closer and closer to x so we're working out how does our function f change in the x direction because f is just a line what this means is that f can only change in the x direction so as we move along the x-axis we're interested in how does the value of f or the value on the y-axis change so the only real derivative we can do here is the f by dx because that is the only direction in which f is changing now if we have a function of several variables and let's just start with two for example we have f of x and y so now f depends on both x and y and so what this is actually going to give you is a surface so if you were to plot z is equal to some function of x and y you get a three dimensional surface which will change in both the x direction and the y direction so if we consider the example z equals x y squared plus y x cubed so when we plug this into our maple calculator app we are given the three surface plot of this function and we can see that it is indeed a surface it's changing in both the x and the y directions simultaneously and so therefore it now makes sense that you might want to know how the function is changing in more than one direction as an example from this particular viewpoint we can see that as we move along the x-axis the function is decreasing so in this direction in the x-direction our function is decreasing but if we rotate the image we can now see that along the y-axis so in the y-direction our function is increasing so our function is behaving differently it is changing in a different way depending on the direction in which we are investigating and so what partial derivatives really do is allow us to calculate the rate of change for a given direction so the x partial derivative tells us how does our function f of x y how does that change specifically moving in the x direction and then the y partial derivative will tell us how our function changes if we only move in the y direction so mathematically our definitions are going to be as follows if i want to know the partial x derivative of f where f is a function of x and y first of all we use this curly d notation and this just says that this is a partial derivative compared to the standard or full derivative in this situation and we define this to be the limit as h tends to zero so unchanged so far of f of x plus h comma y minus f of x y all divided by h so you'll notice it's very similar in fact almost identical to our full or standard derivative df by dx but in the partial case what we're doing is we're adding our h our small increment this is added only to the x coordinate of our function y remains unchanged so this is telling us that the y-coordinate doesn't change the value of y remains constant and we're just looking at what happens if we change the x value only the partial y derivative is of course very similar so here we're interested in how f changes in the y direction so we take the same limit as h goes to zero but now we leave x unchanged so it's f of x and now we add h to the y value then we subtract f x y and then we divide by h so very similar to the partial x except now we do the plus h on the y coordinate because we're interested in how the function changes when we change y and only change y so the x coordinate here remains constant and we are purely moving a small amount in the y direction and then taking the limit to calculate the y partial derivative now that we know the formal mathematical definitions of our partial x and y derivatives let's actually work through an example of the function we looked at earlier f of x y is x y squared plus y x cubed so let's compute the partial derivatives to see how this works in practice and the key thing here is to remember that when you're doing your x partial derivative y doesn't change so that means you can basically treat y as a constant and similarly when you do your y partial derivative x isn't changing so therefore you treat x as a constant so if f of x y is x y squared plus y x cubed if we calculate first df by dx so our usual rules of differentiation apply we still differentiate any function of x as though it is a full or standard derivative however remember y is constant in our x partial so these y terms just treat them like any other constant they are just a number that we just ignore pretty much so what i mean by that is if i do df by dx this first term x times y squared when y is constant this is just constant times x so when i differentiate a constant function times x i just get the constant so the derivative on x behaves as normal the derivative of x is just one and then i'm still multiplying by my constant which here just happens to be y squared the next term the y is again constant so i just keep that and then what's going to happen is i differentiate x cubed exactly as i would normally so i bring the 3 down bring the power down reduce the power by 1 x squared so the x partial derivative is just y squared because this is linear in x y is constant plus the cube's term becomes a squared bring the three down and again just leave the y there because it's a constant now the y derivative is very similar but this time we just treat x as a constant so the x is constant i've got a y squared so if i differentiate y squared i get 2y so overall the x remains where it is i get 2xy and then the next term so x again is constant this is just constant times y so a linear function in y just gives me the constant which here would just be x cubed and that's pretty much it this would be our partial x derivative and this is our partial y derivative now we can go a little bit further we can calculate the second order derivatives so d2f by dx squared and so that just means take this derivative here df by dx and differentiate again with respect to x where we remember y doesn't change so y is a constant so this one the first term is a constant differentiates to give zero and then if i do my x derivative on x squared the 2 comes down so i get 6 y x so that is my second x partial derivative of the function f we can also calculate our second y derivative in the same way so d to f by d y squared so that is going to be the y derivative of d f by d y remembering that x is a constant here so this x cubed term goes to zero because it's just the derivative of a constant and here again x is the constant we differentiate the linear function of y so we're just going to get 2 x and we could carry on we can calculate higher and higher derivatives we can even mix the derivatives so we could do d2f by dxdy or even d2f by dy dx where you're changing the order so one of the means do x derivative first then the y derivative and the other one would mean y derivative first then x derivative and i do recommend trying this out for yourself to get practice with doing partial derivatives of higher orders maybe with mixed derivatives so just pick any function f of x y you can plug it into the maple calculator app and it will automatically tell you the x and y partial derivatives of the function once you enter it into the app so you should do it yourself first and then you can use the app to check your answer and sort of mark your own work and see if you're beginning to grasp how partial differentiation works all of the other rules about differentiation that you may have learned for example the product rule the quotient rule the chain rule these all still apply when doing partial derivatives so i've got another example here which is a little bit more complicated but we'll go through the first partial derivatives to give you a feel for how the product quotient and chain rules work in this context so f of x y is going to be x squared plus 2x all multiplied by sine of x squared plus y plus e to the y minus 2x so first of all let's calculate our x partial derivative df by dx we remember that y doesn't change y is now a constant so this is just now a constant as is the y in the exponential now i still need to do the x derivative using product rule and also a chain rule as well so the first term let's do a product rule so let's differentiate the first bracket with respect to x so that's 2x plus 2 then i have the second term which is unchanged now i want to differentiate the second term and leave the first term unchanged as i would for any normal product so that will give me plus x squared plus 2x multiplied by the x derivative of sine of x squared plus y so we're going to have to do a chain rule as well so we differentiate sine which gives us cos of x squared plus y now having differentiated the sign we get cos but then the chain rule says i have to multiply by the x derivative of the argument of sine so i've got x squared plus y y is a constant in this situation because i'm doing an x partial so x squared plus constant if i differentiate that with respect to x i just get 2 x and that will be the product rule done on that first term now we also have to differentiate the exponential so what i'm going to get here is the exponential is unchanged so e to the y minus 2x and then from the chain rule i have to differentiate the argument so i need to do the x partial derivative of y minus 2 x y is a constant so that disappears and we just get minus 2 from the chain rule finally if we now do the partial y derivative in much the same way this one is actually a lot easier because we remember that when we're doing a y partial derivative we've got over here that x isn't changing x is constant so this whole term x squared plus 2x is just a big old constant so that just stays where it is and now what we actually differentiate is the sine term because that has a y dependency so the derivative of sine gives me cos of its argument and then the chain rule tells me to differentiate the argument but when we're doing this with respect to y x is a constant so the y partial of x squared plus y it's just one so you don't actually get an additional factor and you're going to get the same thing with the exponential in fact because the derivative of the exponential gives you back the same exponential multiplied by the y partial of the argument but again x is a constant so that just disappears and we just get back the derivative of y which is 1. so this x squared plus 2x times cos of x squared plus y plus e to the y minus 2x that is our much simpler y partial compared to the slightly more complicated x partial derivative if you want some more practice with partial differentiation i've actually set up a worksheet in maple learn which you can access completely for free by clicking the link on screen so there's plenty of questions for you to have a go at plenty of partial derivatives for you to calculate and remember you can always check your answers by entering the function in the maple calculator app as this will automatically calculate those partial derivatives for you thank you everyone for watching please do remember to subscribe to my channel if you've enjoyed this video and i'll be back soon with some more maths fun you

Channel: Tom Rocks Maths

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Keywords: partial differentiation explained, partial derivatives explained, partial differentiation visualised, partial differentiation, partial derivatives, oxford calculus, university of oxford math, oxford maths, oxford mathematics, oxford mathematician, tom crawford, tomrocksmaths, tom rocks maths, tom crawford math, tom rocks oxford, university of oxford professor, oxford university lecture, oxford university tutorial, partial differentiation example, partial derivative example, TRM

Id: RVwcBGzQcT8

Length: 18min 21sec (1101 seconds)

Published: Wed Nov 25 2020