Oxford Calculus: Partial Differentiation Explained with Examples

Hello math fans, I'm 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, if I want. To find df times dx for a function f of x, then this is equal to the limit as h approaches 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 would perhaps be something like the following then f of x and then x would maybe look like this so what we're doing here is we're taking a point here, let's call this We're taking the limit while allowing this point to get closer and closer to x, so we're figuring out how our function f changes in the x direction because f is just a line.

What does this mean? is that f can only change in the x direction, so as we move along the x axis we are interested in how the value of f or the value on the y axis changes, so the only real derivative What we can do here is f times dx because that's the only direction that f is changing now if we have a function of several variables and let's start with two, for example, we have f of x and y, so now f depends on both x and of and, and so what? this will actually give you a surface, so if you were to plot z equals some function of x and y you will get a three dimensional surface that 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 get the three surface graph of this function and we can see that it is, in fact, a surface that is changing in the x and y directions simultaneously, so So it now makes sense that you would want to know how the function changes in more than one direction.

As an example, from this particular point of view, we can see that as we move along the x axis, the function decreases, so in this direction in the x direction our function is decreasing, but if we rotate the image now we can see that along the y axis, so in the y direction our function is increasing, so our function behaves differently, it changes in a different way depending on the direction. in which we are investigating and then what

partial

derivatives really do is allow us to calculate the rate of change for a given direction, so the partial derivative of x tells us how our function f of x works and how that changes by moving specifically in the x direction and then the partial derivative of y will tell us how our function changes if we just move in the y direction, so mathematically our definitions will be as follows if I want to know the partial derivative of x from 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 when h tends to zero and has not changed so far from f of x. plus h comma and minus f of x and all divided by h, so you'll notice that it's very similar, in fact, almost identical to our full or standard derivative df times dx, but in the partial case what we're doing is adding our h our small increment, this is added only to the x coordinate of our function and remains unchanged, so this tells us that the y coordinate does not change, the value of y remains constant and we are just seeing what happens if we change the value of x only the partial derivative of y is of course very similar, so here we are interested in how f changes in the y direction, so we take the same limit when h goes to zero, but now we leave x unchanged, so it is f of x and now we add h to the value of and then subtract f x y and then divide by h, much like partial x except now we add h at the y coordinate because we are interested in how it changes the function when we change y and just change y so that the x coordinate here remains constant and we are simply moving a small amount in the y direction and then taking the limit to calculate the partial derivative of y now that we know the formal mathematical definitions of our partial derivatives of x and y, let's actually work with an example of the function that we saw earlier f of x and is x y squared plus y x cubed so let's calculate the partial derivatives to see how this works in practice and the key here is to remember that when you're doing your partial x the derivative y doesn't change which means you can basically treat y as a constant and similarly when you do your partial derivative y doesn't change then you treat squared plus y x cubed If we first calculate df times dx so that our usual differentiation rules apply, we still differentiate any function of x as if it were a complete or standard derivative;

However, remember that y is constant in our x partial, so we simply treat these y terms like any other constant. are just a number that we just ignore, so what I mean by that is if I do df times dx this first term x multiplied by y squared when y is constant, this is simply constant multiplied by x, so when I differentiate a function constant multiplied by x i just get the constant so that the derivative of x behaves normally, the derivative of so I keep that and then what's going to happen is differentiate x cubed exactly like I normally would, so I go under the 3, under the power, I reduce the power by 1 x squared so that the partial derivative of x is just y squared because this is linear in x and it is constant plus the cube term becomes a square, lower the three and again leave the y there because it is a constant.

Now the derivative of y is very similar, but this time we simply treat x as a constant, so x is constant. I have a y squared. so if I differentiate and square I get 2y, then in general x stays where it is, I get 2xy and then the next term, so x again is constant, this is just a constant multiplied by y, so a linear function in and it just gives me the constant here just be x cubed and that's it, this would be our partial derivative of x and this is our partial derivative of y now we can go a little further, we can calculate the second order derivatives, so d2f by dx squared and that just means taking this derivative here df by dx and differentiating again with respect to x where we remember that y doesn't change, so y is a constant, so this, the first term is a constant, differentiate to give zero and then if I do my derivative of x at x squared, 2 goes down, so I get 6 and x, so that's my second partial derivative x of the function f.

We can also calculate our second derivative y in the same way, so d a f times d y squared, so this will be the y derivative of d f. remembering that x is a constant here so this term x cubed goes to zero because it is just the derivative of a constant and here again x is the constant we differentiate the linear function of y so we will only get 2 we could go 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 are changing the order, so one of the means does the derivative of first x and then the derivative of y. and the other would mean derivative y first, then derivative x and I recommend trying this yourself to practice with higher order partial derivatives, maybe with mixed derivatives, so just pick any function f of x and y you can plug into maple. calculator app and it will automatically tell you the x and y partial derivatives of the function once you enter it in the app, so you need to do it yourself first and then you can use the app to check your answer and mark your own work and see if you are Starting to understand how partial differentiation works, all the other rules about differentiation that you may have learned, for example, the product rule, the quotient rule, the chain rule, all of them still apply when doing partial derivatives, so I have another example. here, which is a little more complicated, but we'll go over first partial derivatives to give you an idea of ​​how the quotient product and chain rules work in this context, so f of x y will be x squared plus 2x, all multiplied by sine of x squared plus y plus e raised to y minus 2x so first of all let's calculate our partial derivative of and in the exponential now I still need to do the derivative of x using the product rule and also a chain rule, so the first term let's do a product rule, so let's differentiate the first parenthesis with respect to x, so that's 2x plus 2 so I have the second term that hasn't changed now I want to differentiate the second term and leave the first term unchanged like I would with any normal product so that gives me plus x squared plus 2x multiplied by the derivative x of sin of x squared plus y, so we'll also have to do a chain rule to derive the sine, which gives us cos of x squared plus y.

Now that we have differentiated the sign we get cos, but then the chain rule says that I have to multiply by the derivative of x of the argument. of sine so I have x squared plus y and it is a constant in this situation because I am doing a partial of x so x squared plus constant if I differentiate that with respect to x I only get 2 x and that will be the product rule made in that first term now we also have to differentiate the exponential, so what I'm going to get here is that the exponential doesn't change, so e to the power of y minus 2x and then from the chain rule, I have to differentiate the argument so I need to do the partial derivative x of y minus 2 actually much easier because we remember that when we're doing a partial derivative of y, we have here that x doesn't change x is constant, so this whole term x squared plus 2x is just a big old constant, so it stays where is and now what we really differentiate is the term sine because it has a dependence on y, so the derivative of the sine gives me the cos of its argument and then the chain rule tells me to differentiate the argument, but when we do this with respect a and x is a constant so the partial y of x squared plus y is just one so you don't actually get an extra factor and in fact you'll get the same thing with the exponential because the derivative of the exponential gives you back the same exponential multiplied by the partial y of the argument, but again x is a constant, so it just disappears and we get the derivative of y which is 1. so this x squared plus 2x times the cos of x squared plus and plus e to the power of y minus 2x, that's our much simpler method. y partial compared to the slightly more complicated partial derivative There are many questions for you to try many partial derivatives to calculate and remember that you can always check your answers by entering the function in the maple calculator app as this will automatically calculate those partial derivatives.

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