Differential Calculus part 4

Ajouté : 2026-05-17

[00:00:01]All right. Hello everyone.

[00:00:05]If you can hear my voice, please. I would like to see I would like to see response in the chat box. If you can hear me, I would like to see response in chat box.

[00:00:22]I'm waiting for responses.

[00:00:32]All right. So, I still want to see at least five other persons.

[00:00:42]All right. So, I'm going to be sharing my screen right now.

[00:00:48]Okay.

[00:00:50]Awesome.

[00:01:03]So I believe we can all see my screen right?

[00:01:10]If you can see my screen type yes type yes type yes type yes.

[00:01:22]All right. Right. So let's let's fire up. Let's fire up.

[00:01:30]So we've talked about a couple of things in um calculus, differential calculus precisely. We've talked about first principle general formula.

[00:01:39]All right. So let me just go straight.

[00:01:41]We want to talk about um parametric equation differentiation of parametric equation. That is the next thing I want to do right now.

[00:01:54]Okay, now we've been doing so when you talk about parametric differentiation, what does it mean?

[00:02:01]Sometimes variable x and y are given as a form of a third variable t. Now look at the example we have here. Now we have x given to us to be equal to 1 + 4 sin one year y = 2 cos t. So we say that variable X is given as a function of another variable T and at the same time variable Y also is given as a function of that same variable T. So whenever you have um equations like this you call it parametric equations. So the variable t is called the parameal and the two.

[00:02:40]So whenever you have a parametric um equation how do you go?

[00:02:45]So when a curve is given in parametric form in terms of parametric t we can use shin rule to find the derivative of y with respect to x. So it's easy and straightforward just make use of the idea of shin rule. Now let's start with example now.

[00:03:05]So let me increase my screen so we can see it well. All right. So let's start with examples.

[00:03:12]one example I'll move on.

[00:03:18]So let's say the parametric equation of a curve r x = t ² and y = 5 - 2 t. Then the question is to find dy by dx in terms of the parametric t parameter t sorry parameter t.

[00:04:03]All right. Like I said, whenever you have equations like this such that two variables x and y is given as a function of the third variable t, then you make use of parametric equation and the idea you use is shin rule.

[00:04:16]So you can go to previous video solution. So first of all, let's write the first equation x = t².

[00:04:28]I believe you're following.

[00:04:31]All right.

[00:04:35]So then the other equation let's say y = 5 - 2t. So I will advise that you differentiate t you differentiate x with respect to t. dx dt that will give us 2 t and at the same time you differentiate y with respect to t dy dt that should give us minus 2.

[00:05:02]All right like I said we are trying to find the y by the x. So what's the formula x according to general dy by dx is equal to okay so we can have it as dy by dt time dt by dx I believe you know that this is the formula for now since the other third variable we are dealing with is t that's why we have dy by dt multiply by dt by dx all right so what's d y by dt is - 2 * what is dt by dx? See what we have here? We have dx over dt as 2t, right?

[00:05:49]So that means we need to if we're going to have dt by dx. So d t by dx reciprocate it will now become what? 1 / 2t. So the t by dx will be what? 1 / 2 t.

[00:06:05]All right. So, let me rub off what we have here.

[00:06:13]So, you can divide 2 and 2 that will give us one. So, we have -1 / t. So, that is the answer. Parametric equation is very very simple and straightforward.

[00:06:27]So, let me just say it again. You have you given two equations x = t² and y = 5. So you're told to find the y by dx.

[00:06:35]So different x with respect to t that will give you 2t and at the same time do the same thing differentiate y with respect to t. Now to find your dy by dx you use the formula for shin rule which is dy by dtip.

[00:06:54]Now we have that is why we have it as 1 / 2t. then okay so do we get that now all right so I'm going to give us a question to practice we can practice this particular question so let me just say practice this question all right so let x and y be Two parametric functions defined as x to be = 4 t - 6 and y = t ² + 7.

[00:07:53]So find dy by dx. All right, so let's go ahead and practice that question. So that's it for parametric equation. There are lots of questions you can get in parametric equation, but at least with this knowledge, you can be able to solve every kinds of parametric equation. All right. So let's move let's move down to differentation of inverse trigonometry.

[00:08:19]So differentation of inverse trigonometry function.

[00:08:25]If you're with me, can you just um on the chat box?

[00:08:30]Drop something on the chat box if you're with me. Probably your favorite emoji.

[00:08:38]All right. Now, so now let's move to differentation of inverse trigonometric function.

[00:08:43]Video we dealt with differentation of trigonometric function, right?

[00:08:48]and those videos if you do not understand.

[00:08:52]So now let's let me just go straight examples now.

[00:09:01]So find the derivative of respect to x for the following function. So the first one I'm going to deal with here, let me say when your y is equal to sin inverse of x and secondly when your y is equal to tan inverse of x.

[00:09:31]All right, we already know that normal trigonometric function is like maybe y= sin x, y = cos x and um tan x and normal se x and the likes. So when you talk about inverse is either you call it sin inverse of x or you call it a sin x a sin x. So whichever one you want to call it either you call it a sign or you call it sin inverse you're still talking about the same thing. All right. So how do we differentiate inverse trigonometric function? Let's look at differentiate inverse trigonometric function. It's not hard. Let's let's move. So for the first one we have that y is equal to sin inverse of x. So this can also be written as x = sin y. Yeah, you can reverse the process. If you take the sign to this other side, it gives us what? Sin y. So sin y is equal to x.

[00:10:37]Do we get that? And the next thing for us to do now is to differentiate x with respect to y.

[00:10:47]So when you find the derivative of x with respect to y, that's the x by the y. This is going to give us cos y.

[00:10:56]But if you observe according to the question, what we are looking for is dy by dx.

[00:11:03]Right?

[00:11:07]Okay. So we can reverse the process and make it dy by dx to be equal to 1 / cos y.

[00:11:16]All right. So that is not the answer yet. Then you now make use of the trigonometry identity. So let me write the trigonometry identity here for those of us that have forgotten.

[00:11:29]We have three force trigonometry identity. We have three trigonometry identity.

[00:11:35]We have it as sin square. You can use theta or you use any variable sin square theta equal to 1.

[00:11:44]Then we also have um C square theta to be equal to cos² theta. This is the second trigonometric identity. Let me call this one the first.

[00:12:00]And then the last one is if you have um tan square theta + 1 to be equal to se square theta. This is the third trigonometric identity. So these are the three trigonometric identity we have.

[00:12:13]um in some other videos I might take us trigonometry as a whole topic. So we'll see how to prove this but just know this. All right. Now let's let's come here. Let's come back here. So we can recall from the first trigonometric identity. Recall that from the first trigonometric identity that sin square okay we're using y this time around. So s square y cos square y is equal to 1.

[00:12:42]So I will advise that you make cos y the subject of the formula here. Make cos y of formula here. So if you make cos square y formula. So you bring this one down here. That will become what? 1 - sin square y. Then you can take the square root of both sides. So our cos y will be equal to the square root of 1 - sin square y.

[00:13:12]All right. So you go back to this place, this place here, you replace the value of cos y with this square root with this roo<unk> 1 - sin square x. So our dy will now so let me let me call this one star.

[00:13:28]So we can substitute star into star.

[00:13:35]So just take this equation and put it um dy by dx to be equal to 1 / instead of cos y now instead of cos y now we're going to replace it with this that will be the square t of 1 - sin square y.

[00:13:54]All right so the final step is this. Go back to this particular So go back to this particular point where we have x = to sin y. So wherever you have sin y you replace it with value of x there. So we are going to have dy by = 1 /<unk> 1 - x². So that is our final answer.

[00:14:32]Okay.

[00:14:34]So whenever you have y to be equal to either sin inverse of x y = a sin x is the same thing. And I believe we know that if you differentiate it, your dy by dx is always going to be 1 / - x². Please take note of that. I'm still going to write the standard um differential for inverse trick.

[00:15:05]All right. So let me solve the second question. y = tan inverse of x. Let's solve the second question now.

[00:15:14]y = tin inverse of x.

[00:15:24]Okay. So we do the same here.

[00:15:32]So first of all this can be written as x equal to tan y if you take the tan inverse to the other side.

[00:15:42]All right. And then the next thing is to find the derivative of x with respect to y. Differentiate x with respect to y. If you differentiate tan y, you're going to have um se square y.

[00:15:54]I think I I showed us in the previous class when I was dealing with um trigonometry differentiation of trigonometric function. Yeah, let me take it down. Let's let me quickly.

[00:16:05]Okay, differentiation of trigonometry function.

[00:16:10]Put differentiate it. This is going to give us x² x. So that is in the previous video. We can go if we need it and also give us some standard differential coefficients if you remember this.

[00:16:25]So this is it where tan x = 6 x if you differentiate it. All right, let's go back to what we're dealing with.

[00:16:33]All right, so when you differentiate x with respect to y, you're going to have sex² y.

[00:16:41]So like I said we need dy by dx. So reverse the process or reciprocate it.

[00:16:45]We have dy by dx to be equal to 1 / x².

[00:16:52]All right. Now go back to your trig identity. I wrote three trigonometric identity.

[00:17:00]Let copy it again and paste it here so that we can see what we Okay.

[00:17:13]All right.

[00:17:15]So this time around what can guess or what can box what trigonometry identity are we going to use now to simplify this? Are we using the first one or the second one or the third one? I want to see our comments please. Am I using the first trigonometry or the second one or the third one?

[00:17:34]I'm waiting for someone to say something.

[00:17:40]beautiful.

[00:17:42]So I will be using the so you can write it that recall that tan square theta + 1 = 6² theta let's let's use y sorry let's use y here * square y + 1 = square y thank you for the comment right so let me call this one star and let me call this on star.

[00:18:13]So you substitute star into So wherever you see x square y you're going to slot tan square y + 1.

[00:18:27]So we have dy by dx to be = 1 / instead of se square y we're going to have what?

[00:18:35]tan square y + 1.

[00:18:40]Awesome.

[00:18:42]Now go back to the go back to this place where we have our x to be equal to tan y.

[00:18:51]We're going to slot in the value for x as tan y.

[00:18:57]So we have dy by dx to be equal to 1 / instead of having tan y now we're going to have what? x² + 1. So that is our answer. Or you can still have it as 1 + x² whichever one you want to do.

[00:19:15]Okay. So do we get that now?

[00:19:27]All right.

[00:19:29]So I think next thing we can talk about is um application of differentiation.

[00:19:34]So, I don't want this video to be too long. And so, we're going to stop here now. We'll stop here now.

[00:19:43]So, let's let's have a lovely day.