Differential Equations Review of Partial Derivatives

Hi, welcome back to the differential equations lecture here on educator.com.0000

My name is Will Murray, and we are starting a chapter on partial differential equations.0004

So, the next few lectures are all going to be covering partial differential equations we will also get into Fourier series as well.0009

Before we really start partial differential equations I thought it would be worthwhile to review some partial derivatives just make sure that everybody is comfortable with the whole idea of partial derivatives so, that we can study partial differential equations.0016

Now partial derivatives are something that you really covering in detail in multivariable calculus so, if you are very rusty on partial derivatives what you might want to do is go back and look at the course on multivariable calculus on educator.com0029

If you go back there it really cover partial derivatives in lots of detail, what I am doing this lecture is just going to give you a real quick crash course on 0103 Partial derivatives hopefully you have seen them before and hopefully will kind of remind you how partial derivatives we are.0046

So, let us go ahead and see what the definition of a partial derivative is the idea is that you have a function of 2 variables and sometimes this is given as F of XY for the sometimes these different variables are F XT I will use U of XT so, my function is going to be U my 2 variables are going the X and T.0067

If you are using different variables and your course of differential equations , then you just have to translate between X and T whatever variables you are using we are always getting used U of XT0084

We will talk about the partial derivative with respect to X and the formal definition is in terms of a limit use these curly D here that is that is a curly D being on a regular old straight D like a regular derivative.0096

We also, use U X to represent the partial derivative in fact that is the most common notation U of X because it is the simplest and the idea is that what we are going to do is walk over a little distance in the X direction.0111

So, that is why we change X to X + H and we are to see how much that changes the U values so, this is sort of like δ U / δ X so, be seen how much U change is will we change the X value by a little bit or just keeping the T values constant we are not changing the T values at all.0125

That is why it is the partial derivative with respect to X and the T so, we take that limit and that is the formal definition of partial derivatives.0149

In practice we will not really use that limit definition to actually calculate partial derivatives so, bit more in just a moment on how to calculate them .0157

Let me mention what they represent geometrically if you have a function U of XT and you graph it out in 3-dimensional space if you have X and T here then it represents a surface in 3-dimensional space .0166

So, I graph a surface in 3-dimensional space so, there is some surface and they that is the graph of U of XT and what the partial derivative represents geometrically is if you are standing on that surface and you walk in the X direction and represents the slope of the surface.0183

Walking strictly in the X direction so, holding the T variable constant and we are walking in the X direction the partial derivative the number that you calculate there represents that slope.0205

So, that is what it represents geometrically course that still does not really tell you very much about how to calculate it so, thought ahead and talk about how you calculate partial derivatives.0217

The way you do it is you treat the other variables of your finding U have actually treat the other variable T is a constant and then take the derivative with respect to X so, partial derivative with respect to X you just treat T is a constant and take the derivative just like you learning calculus 1.0227

What if you take the partial derivative with respect to T then you treat the X as a constant and take the derivative just like you did in calculus 1.0246

You can also, take second partial derivatives you can take U X X it means you take into partial derivatives with respect to X U XT would mean you take the partial with respect to X and then the partial with respect to T U TX is the other way around take the partial with respect to T and T the partial with respect to X .0255

UTC means you take 2 derivatives with respect to X with respect to T sorry .0275

Very nice property that says these 2 metal 1s the UXT and the UT X are the same there always equal to each other that is really nice what does it matter which order you take the partial derivative with respect to first.0281

If you are taking the mix partial derivative UXT does not really matter if you take the X derivative first and then the T derivative or if you take the T derivative first and then the X derivative .0300

So, you can do that in either order so, been doing this lesson is practice calculating a lot of partial derivatives and will also, check the theorem on a fairly complicated functions and make sure that it works out right.0313

You can also, use this as a check to make sure that your taking the derivatives right so, let us go ahead and see some examples.0326

In the first example it is a fairly easy 1 harder from here U of XT = X ² x T so, want to find the first partial derivatives U of X and U of T .0334

So, for U of X remember that means you treat T is a constant when we want to take U of X means you take treat T is a constant same thing in this X ² T is this being a constant x X ² .0345

So, the derivative of that is just that same constant x 2 X remember 2 X is the derivative of X ² with respect to X and the T comes along as a constant say to write that as 2 XT .0361

So, that is my U of X my U of T I treat the X as a constant which means X ² is a constant so, that just comes down when taken derivative because that is what happens with concerts and the derivative of T is just 1 so, the derivative with respect to T is just X ² .0376

Think of the X ² is being constant derivative with respect to T of T is just 1 so, that is my UX and my UT.0397

For that 1 just to recap there remember each time you take a partial derivative you are holding the other variable constant so, when you take U of X, that means T is constant think of T is being constant and so, that T comes down derivative of X ² is 2 X and you get 2 XT.0406

When you take the derivative with respect to T that means you think of the X is being constant so, X is constant constants and so, take the derivative with respect to T which means we have a constant x TX ² x the derivative is just that constant X ².0429

Let us go on and look at another example here that U of XT be sin of X x cosine T ² + 3T a little more complicated here .0454

To find the first partial derivatives U of X and UT so, remember we are going to find U of X first so, we can think of T is being constant which means cosine T ² is just 1 big constants.0469

Just and bring that down cosine T ² just 1 big constant now the derivative of sin X is okay derivative with respect to X is just cosine X now here is a mistake that lots of students make in multivariable calculus.0485

We have + 3T now remember if she is a constant that is just a constant so, the derivative of a constant is 0 so, that just drops out not 3T it is just 0 so, I write that is cosine X cosine T ² cosine X cosine T ² and I am done with finding the partial derivative with respect to X .0502

Now for U of T that means that I think of X is being constant so, sin X is just 1 big constant sin X but now I have the derivative of cosine T ² cosine is - side so, - sin of T ² - sin of T ² and now I have to multiply by the derivative of T ² so, that is 2T and I have + 3T and I am taking the derivative with respect to T so, +3 .0530

So, let me simplify this down a bit of the - 2T on the outside - 2T sin X x sin of T ² +3 and that is my partial derivative with respect to T .0566

So, recap how we found those things out when you find the derivative with respect to X your thinking of T is constant think of T is being constant so, that means that cosine T ² is nothing but 1 big constant .0588

Bring that down or take the derivative with respect to X a sin X we get cosine X and the key thing here is a T mistake that a lot of students make but I am training you not to is that the derivative of 3T with respect to X means you think of T is constant.0608

So, 3T is just a big constant and the derivative of a constant is 0 so, that drops out and just get cosine X x cosine T ² or take the derivative with respect to T that means you think of the X as being constant so, sin X is just 1 big constant.0627

That we have to take the derivative of cosine T ² remember cosine is - sin of T ² but by the chain role we have to put the derivative of 2T so, that is the chain role coming in right there and then the derivative of 3T is just 3.0645

So, if we sort things out we get - 2T times sin X + T ² +3 as our partial derivative with respect to T .0665

So, example 3U of XT is X ² + T ² will find all the first and second partial derivatives and we are in a check LaRose theorem in the context of this function .0675

So, when we find U of X start out with that used T is a constant so, I see through the X ² is 2 X derivative of T ² is just 0 because I think of T is a constant so, I just get just get 2X there .0690

U of T is the derivative with respect to T remember the X ² is a constant that goes to 0 and so, I just get 2 T now we take the second derivatives U X X is derivative of second derivative with respect to X so, I do the derivative of to X is just 2.0708

UXT is the derivative of 2 X with respect to T so, that is the derivative of 2X is just a big constant since worth checking the derivative with respect to T so, we just get 0 there.0733

Look over at UT UT X is a derivative of 2T with respect to X so, 2T is just a big constant now so, that 0 and UTT is the derivative of 2T with respect to T that is 2.0754

So, I found all my first and second partial derivatives the theorem says I want to check that UXT is the same as UT X and so, if I look at U XT and UT X they do agree with each other got 0 each time and so, hold at least for this function U .0777

So, let me recap there start out with X ² + T ² take the derivative with respect to X means the T is a constants would drops right out we get 2 X second derivative of that while the first rout of that is the second derivative of the original function is just 2.0801

But the derivative of U X with respect to T means you think of T is a constant so, the derivative of 2X is 0 on the other side UT think of X is a constants of the X ² drops out so, we get through the T ² is 2T.0818

Derivative of that with respect to X is 0 because the T is a constant derivative of 2T with respect to T is 2 check at UT X and UXT and making sure that we got the same thing both ways.0835

We get we got 0 either way so, UT UXT = UT X and example 4 we have a more complicated function U of XT = X / X + T again have to find all the first and second partial derivatives and check for this U so, I remind you here with me using the quotient rule a lot and if you don't remember the quotient rule we got a set of lectures on calculus 1 here on educator.com.0855

Go check them out and you will see lectures on the quotient role but meantime I have a too little mnemonic to remember the quotient rule.0889

If you think of the top as high / in the bottom is ho and then U′ it is the bottom derivative the top so, ho x the derivative of high - the top x through the bottom high x the derivative of ho may right be high x the derivative of ho / the bottom ² so, ho² and so, there is a cute way to say this.0897

So, that can be really useful for taking our partial derivatives Let us go ahead and try it out .0935

U of X is ho hi so, bottom are the top X + T derivative of X with respect to X remember everything here really thinking of X as our variable and T is our constant as long as we are differentiating with respect to X so, derivative X is 1 - hi is the top the derivative of the bottom is the derivative of X + T .0949

Since working derivative with respect to X that is just 1 and in the derivative of T is 0 all / ho of the bottom ² X + T ² and so, on the top of X + T - X that justifies down the T .0978

The bottom is X + T ² so, that is our derivative with respect to X .0996

The second derivative with respect to T see how that works out so, UT now many is the same quotient rule formula ho hi - hi ho / hoho but now I am thinking of T is the variable and X as a constant.1006

So, the bottom ho high X + T is at the bottom the high now the high part is X and the derivative of that with respect to T0 - high is X and the derivative the bottom is 0+1 all / the hoho of the bottom ² X + T ² .1024

So, this is X + T - X so, X + T is made some mistake here and I got a figure out what it is so, the bottom x the derivative of the top know I have not made any mistakes this is correct so, on the top and X + Tx 0.1050

That drops out - X / the bottom ² is X + T quantity ² that is my derivative with respect to T so, is my first 2 .1082

My 2 first partial derivatives and the problem also, says I need to find all the second partial derivatives So, find U X X you XT UT X and U TT .1097

So, let's work those out UX X you XX means I look at U of X and take its derivative with respect to X so, the bottom x the derivative the top ho-high X + T ² now the derivative the top is the top is T but I think of that as a constants that 0 - the top that is T x the derivative the bottom .1111

So, that is X + T ² derivative of that is 2 x X + T x the derivative of X + T using the power rule and the chain rule here.1136

So, derivative of X + T is 1+0 and on that now I have / hoho so, the bottom ² is X + T to the 4th and I see that this this term drops out because multiplied by 0 and now and X + T that will cancel with 1 of these and so, get a cube down there.1148

So, I see it simplifies down to is - 2T / X + TQ X + TQ so, that is my U of X X and now let me calculate U of X T so, ho high is the bottom x derivative of the top I am taking the derivative of U of X now.1174

Derivative the top is derivative of T which is 1 because T is our variable now X is our constant - the top x the derivative of the bottom so, the top x derivative of bottom is T x 2 x X + T using the power rule .1211

X + T ² is 2 x X + T x 1+0 and then / hoho so, X + T to the 4th looks little messy but I see I have X + T factor everywhere so, cancel out X + TX + T and 1 of my X + T is here.1229

So, I see what I have got here is X + T I see I had X + T on ² on the bottom so, this should been X + T ² the bottom x the derivative of the top .1251

So, if I cancel out the change that a little bit if I cancel out 1 X + T the whole thing does not cancel it just cancels out into X + T to the 1 so, let me fix that.1276

So, what I have got is X + T - 2T / X + T Q and so, X + T - 2T is X - T still / X + TQ.1289

So, that is my second mixed partial derivative U of XT we go to the other side and look at U of T and take a couple derivatives of that U of T X that means them to take the X derivative of U of T.1310

So, I am going to go ho-high bottom x derivative the top X + T ² is the bottom x the derivative of the top is the derivative of - X so, that is -1.1333

Since X is my variable right now - the top x the derivative of the bottom so, - - X x the derivative of the bottom is power rule okay 2 x X + T x the derivative of X + T with respect to X is just 1+0 all / the bottom ² all /is that was ho-high - high-ho X + T ².1350

I see you got X + T everywhere again so, we cancel out the X + T everywhere sorry the bottoms in the next should be x + T to the 4th.1381

Because we are squaring a ² and when I cancel out 1 of those X + T .1392

- X + T in the numerator - and - is + 2 X and in the denominator as you got X + T quantity Q and if I simplify that a bit on the top you got 2X - X lots of X - T all / X + T + not T there X + T quantity Q .1398

So, that is UT X something or U TT so, let me workout U TT so, you may go back to UT and take its T derivative so, ho-high x derivative of the top X + T ² x the derivative of the top and the top is - X.1434

But I am taking the derivative with respect to T now something of X is being a constant and is derivative is just 0 - the top x the derivative the bottom - - X.1463

So, now the derivative of the bottom is 2 x X + T x the inside derivative which is 0+1 all / hoho so, all / the bottom ² but the bottom by itself is X+t² when I ² a get to the 4th and so, that term drops out because multiplied by 0.1474

I see you got X + T here cancel of 1 of my X + T and I still got - X at with a - outside so, that is + 2X and I have got X + TQ on the bottom there.1505

So, that is my U of TT so, got all these partial derivatives and second partial derivatives now disposed to confirm Claro theorem says that U XT and UT X are supposed to come out equal.1526

Let us compare those I see in each 1 of those boxes that I have X - T / X + TQ and so, it holds because those 2 boxes are equal U XT = UT X.1549

So, Claro's theorem holds those with a 2 boxes that we are supposed to be equal so, let me go back and show how we did each 1 of these.1567

First I wrote how the chamber of because are the quotient rule because we are using the quotient rule over and over again .1589

My version the quotient rule is ho-high - Heidi Ho / Ho ho that is shorthand to help you remember bottom x the derivative of the top - the top x the derivative the bottom all / the bottom ² .1608

So, we apply that to the initial U when we take derivative with respect to X for that means T is constant T is constant and so, we work out the bottom x derivative of the top X are variables of the derivative is is 1 top derivative the bottom and the bottom ² and that is if I down the T / X+T² .1610

That we take the derivative with respect to accept that so, again T is constant and so, we get the bottom x the derivative of the top but since the top was T and T is constant that is what we get that 0 from .1633

And then to find the top x the derivative of the bottom that is the top right there T remember the bottom we have to use the power rule so, that is as the derivative of that anything ² is 2 x that original thing x the derivative of that things we are using the chain rule as well there.1650

And the bottom ² a Ho Ho is X + T to the 4th but then that X + T and that X + T cancel each other out so, that is why we ended up with X + TQ in the denominator in our final answer.1672

We did UXT we start with UX we started with T / X + T ² but then we held X constants and we took the T derivative X is constant and ran through the quotient rule again so, there is my Ho D high and high and all of this is the Ho so, ho-high - Heidi Ho and again we are holding X constant and taking derivatives with respect to T and there is my Ho Ho right there.1688

But when I look at this I can cancel out and X + T from everything and so, that simplifies down to just a single is X+T² to single X + T - 2T / X + TQ and then that simplifies a little more than numerator simplifies X + T - 2T just reduces down to X - T / X + TQ.1726

That was UXT / on the other side finding U of T, that means X is constant so, we do ho hi the D high gives me just 0 there because the derivative of X is 01 we are assuming X is constant - Hi D Ho so, that is that is the high part and that is the D Ho part.1748

And then that is Ho Ho right there since and find down to - X / X + T ² and we take the derivative of that with respect to X settings we are holding T constant T is constant.1779

So, we take the derivative of that using the quotient rule again so, there is our Ho the high since our X are variable now derivative of X of - X -1 - the high and there is her de-ho again using the chain the chain rule and the power rule.1791

And there is Ho Ho but again we have an X + T canceling everywhere so, we go down to X+T cubed in the denominator numerator simplifies to - X + T just a single power there cancel off 1 power and + 2X in the simple find down to X - T / X + TQ only take the T derivative of UT holding X constant.1814

And so, we are again getting is the chain rule but what we do so, Ho x D high D high .1840

High was X but since X is constant derivative is 0 - there is my high and there is my de-ho again so, Heidi Ho and then there is Ho Ho so, hody high - Heidi Ho / Ho Ho.1856

So, Ho Ho is X + T the 4th but again we have X + T canceling so, just goes down to X + TQ wind up with to X / X + TQ .1871

That is we are all those partial derivatives came from the last step in this problem was to confirm Claro theorem which means your checking to see whether UXT is the same as UT X.1881

Now not all these derivatives we are the same but if you look at U XT here and UT X we did calculate those independently but after we some find them down we got the same thing on both sides.1892

Got X - T / X + TQ and so, we can say for sure that Claro theorem did hold for this particular function.1906

In our example 5 we are going the opposite direction from taking partial derivatives this time we are given a partial derivative or asked to figure out what the original function could be that have used that.1918

So, essentially what we are going to do is working integrate with respect to X so, I am to integrate = the integral to think of it this way as EXT cosine T the X at but now that means we are integrating with respect to X.1929

So, T is constant and just like when we took the derivative with respect to X the help T is constant and so, that means that cosine T is just in the constant which means you can pull it out of the integral.1950

So, cosine T is not equals and pull that out of the integral E XT DX now really integrating that with respect to X me just remind you for example E to the to the 4 X D X would be 1 4th E 4 X + the constant that is not going back to the old calculus 1 technique of substitutions of that looks foreign to you.1963

Check back to calculus 1 lecture here on educator.com it is doing a little substitution U equals 4 X and DU = 4 DX and then doing that little substitution and up with 1 4th in the 4 X.1995

So, here we got cosine T now you want to think about this as E the T X where T is a constants was kind behaving like the 1s the 4 here so, is that 1 4th E 4 X .2032

I get 1 / T x E TX just like the 4 was before now we have T + now be careful here I want to say + C but remember I am integrating with respect to X which means that I am thinking of all T as being constants.2039

So, what I am going to say is + any function of T because if I took the derivative of that if I took the derivative if I took U of X of that it would just go back to 0 so, can really have any function of T here that I like .2049

So, let me collect my terms and simplify that 1 / T x E TX x cosine T + any function of T so, this could be any function of T this is my U of XT any function of T could be your CT here and the reason I can put any function of T in their cosine T in the T natural on the TT ² is because if I took the X derivative it would just cancel away to 0.2066

So, that is my most general form for my U of XT if he took the derivative the X derivative you get back to what we started with .2108

You might ask where is the constant here you can think of the constant is being built into the function of T so, this includes that any possible constant that you would care to add on there.2123

So, let me recap how we figure that out.2140

Basically we are doing the integral but we are doing the integral with respect to X and so, that means that I can think of cosine T as being a constant and I can plug pull it out of the integral and I am integrating E as even the TX remember T is a constant so, the integral E TX is just 1 / T x D TX .2143

Still have that cosine T and normally I would tack on an arbitrary constant here but since I am doing up the opposite of a partial derivative I contact on any function of T because any function of T be treated as a constant.2164

When we take the partial derivative so, this would be considered would be treated as a constants when taking the derivative with respect to X .2181

When finding U of X so, any function of T could be included there and so, just can I include that arbitrary function of T C of T instead of including an arbitrary constant and I can think of an arbitrary constant being built into it.2204

So, my final answer there is what I got from the integral + any arbitrary function of T and the reason that works is because if you took the X derivative that C T would just completely drop out and we get back to that in the E XT cosine T that we started with .2221

So, that wraps up our review of partial derivatives will be all set to go now for learning about partial differential equations starting in the next few lectures.2239

If this was not enough of you partial derivatives, if you are still feeling a little rocky little bit rusty and when you taking partial derivatives then we have a whole set of lectures on multivariable calculus and you get more practice from taking partial derivatives in those lectures.2252

You can go back and watch those lectures again the 1s on multivariable calculus you get lots of practice with partial derivatives.2268

This was just meant to be kind of a quick review practice brushing the rust off so, that when we start doing partial differential equations in the next lecture you will be ready to go.2275

So, that is that is the end of our lecture on reviewing partial derivatives and you are watching the differential equations lecture series here on educator.com.2286

My name is Will Murray, and I very much appreciate your watching, take care.2297