 In this video, I want to talk about the derivatives of the inverse trigonometric functions. So we're talking about arc sine, arc cosine, arc tangent, and the other three. We do know how to calculate the derivatives of the trigonometric functions, sine, cosine, tangent, et cetera. And it turns out if you combine that knowledge, the derivatives of the trig functions that is, with implicit differentiation, we actually are in a situation where we can compute the derivative of the inverse trigonometric functions. So let me demonstrate how one does that with arc tangent. I claim that the derivative of arc tangent of x is going to be one over the square root of one minus x squared. That might seem like obscure or random at this moment, but at the end of this argument, I'm going to show you that it's actually quite natural to think of this way. And so the idea is going to be the following. Consider arc tangent, for example. Consider for a moment. We're going to call it y. That is to say y equals sine inverse of x. Well, if y equals sine inverse of x, that actually implies that sine of y is equal to x. Or better yet, it's equal to x over 1. Why is this perspective useful? Well, as we've often played around with trigonometry, is that we can set up a right triangle associated to this angle y. I mean, because y, it's going inside a sine, it's a function. It's, excuse me, it's an angle here. And so what we see is that we get a y over here. Let's construct a right triangle for y. Well, with this triangle in mind, the sine ratio is x over 1, where sine, as we know, is opposite over hypotenuse. And so if you want to do that, we get that x is the opposite side. The hypotenuse is 1. How do we fill in the other side? Well, we can use the Pythagorean equation. So remember that the opposite side squared plus the adjacent side squared is going to equal the hypotenuse squared. For which, if we plug in the stuff we know, we have an x squared. We have the adjacent side squared. That we don't know yet. And then we have the hypotenuse squared, which is 1. So if you solve for the adjacent side, you're going to get the adjacent side squared as 1 minus x squared taking the square root. We see that this remaining side is going to be the square root of 1 minus x squared. So we can construct this triangle, this right triangle, associated to this number y, which is acting like the angle in play right here. Why does this matter? Why does this matter of all? We're going to come back to that in a second. So where's the calculus? This was trigonometry. Where's the calculus coming to play here? Let's take this equation from a different perspective. So we have the equation sine of y is equal to x. We want to find the derivative of y, because remember y is equal to sine inverse of x. We don't know how to take the derivative of sine inverse, but we do know how to take the derivative of sine. What if we took the derivative implicitly? What if we took the derivative of sine of y with respect to x, and we do that to the other side as well? So we're taking the derivative with respect to x. Well, by the chain rule, the derivative of sine of y with respect to x, this would be cosine of y times y prime. And the derivative of x with respect to x would just be a 1. So we get cosine of y times y prime is equal to 1. We can solve for y prime by dividing both sides by cosine. We get y prime is equal to 1 over cosine of y. But cosine of y, remember, y, well, let's actually stop here for a moment. We could write this as 1 over cosine of sine inverse of x. But better yet, we already know what to do with cosine of y, right? Cosine of y, I'm actually going to think of this as a secant of y. We'll just take a look at the triangle that we have up here. The cosine relationship is adjacent over hypotenuse. Therefore, its reciprocal secant will be hypotenuse over the adjacent side. And so we end up with 1 over the square root of 1 minus x squared for the derivative of our sine inverse right here. That's exactly what we said it was going to be. So that's kind of cool, right? We are able to use implicit differentiation to find the derivative of sine inverse because we knew the derivative of sine. Let's try this with a different trigonometric function. Let's try this time to compute the derivative of our tangent of x. So our tangent of x, we claim its derivative is going to be 1 over 1 plus x squared. So let's try the same strategy we did before. What if y is equal to tangent inverse of x? Well, if you apply the inverse function to both sides of y equals tangent inverse of x, that means tangent of y is equal to x. Now, take the derivative implicitly on both sides. So take the derivative d dx on both sides of the equation. We'll start off the right-hand side. It's easier. If you take the derivative of x with respect to x, you're going to get a 1. If you take the derivative of tangent, you're going to get a secant squared of y. And then you times that by y prime, for which then if you divide both sides by secant squared of y, you're going to end up with y prime equals 1 over secant squared of y. Which, as we saw previously, right? Secant squared of y, you divide by secant. That's the same thing as you can times by cosine. You could do that. We're going to take a slightly different perspective here. Using our favorite trigonometric identity, the Pythagorean identity, cosine squared y plus sine squared y equals 1, if you divide both sides by cosine squared, let me point out to you that the left-hand side, you're going to get cosine squared divided by cosine squared, which is the 1. Sine squared divided by cosine squared, which is a tangent squared. And then 1 divided by cosine squared is a secant squared. So you're going to end up with a 1 plus tangent squared, which is the same thing as secant squared. Notice we have a secant squared right here. This tells us that the denominator could be written as 1 plus a tangent squared of y. The reason this is advantageous is, remember, tangent of y is equal to x. So this gives us the formula we were promised right here. And so I want to mention the general strategy. If you have a function where you have some functions like, oh, I know f prime, right? If y equals f of x like this, and you know it's derivative, it's differentiable, then what this tells us is that y equals f inverse of x, its inverse function is also differentiable. Because if you want to find the derivative of f inverse of x, like, hmm, let's just do it implicitly. You get f of y, which equals x. Take the derivative of both sides, the derivative on the left-hand side. So when you take the derivative for your d dx, when you take the derivative of the left-hand side by the chain rule, you're going to get f prime of y times y prime. This equals 1. Divide both sides by f prime of y. You're going to get that y prime equals 1 over f prime of y. But what is y? y is f inverse of x. So you end up with the derivative of f inverse of x. So just be 1 over the derivative of f where you then compose it with f inverse of x. The trigonometric functions we saw here do have the luxury that we were able to remove the inverse description whatsoever. That's why we get these algebra expressions like 1 plus x squared. We could have also done a triangle argument in this situation too, had we wanted to. Let's just see what that would have looked like real quick. Just for the sake of completion, angle y. So if we're using the fact that tangent of y equals x, x over 1, that means that opposite over adjacent side, x over 1, the Pythagorean equation then tells you're going to get the square root of 1 plus x squared right here. And so then as an alternative to what we did before, 1 over secant squared is the same thing as a cosine squared of this thing. What is cosine? Cosine is going to be adjacent over hypotenuse. So you end up with a 1 over the square root of 1 plus x squared. And you're squaring all that. That gives us a 1 over x squared again. So there was some trigonometry that helps us simplify it, but the basic strategy is the same. If we have the derivative of a function, we can also compute the derivative of its inverse using implicit differentiation. So now let me show to you the six derivatives of the six inverse trigonometric functions. Two of them we already saw. So we saw that the derivative of sine inverse of x is equal to 1 over the square root of 1 minus x squared. We also saw that the derivative of tangent inverse is 1 over x squared. By similar calculations, you can show that the derivative of cosine inverse is negative 1 over the square root of 1 minus x squared. The derivative of secant inverse is 1 over x times the square root of x squared minus 1. The derivative of cosecant inverse is negative 1 over x times the square root of x squared minus 1. And then the derivative of cotangent inverse will be negative 1 over 1 plus x squared. So notice the pattern that we saw before took the derivatives of the trigonometric functions. When you look at the formulas here between sine inverse and cosine inverse, their derivatives are the same except for this additional negative sign right here. Otherwise, the formulas are the same. The same is true for secant and cosecant. Cosecant inverse, its derivative has a negative sign. And then the derivative of cotangent inverse with cotangent inverse, cotangent inverse has a negative sign in it. Otherwise, they're the same formulas. And that's because of the similarities between the trig functions, their derivatives with their complementary functions. They're basically off by negative signs and such. So all of the co-functions, their inverses will have these negative signs. That pattern's consistent. So it's not a horrible list of numbers. But with these new derivatives, we can calculate new functions. And so this is something we've been doing a lot in our lecture series, right? There are some derivative rules that we've learned like the power rule, the chain rule, the product rule, et cetera. But there's also certain families of functions that we've learned how to take their derivatives of. So we can take derivatives of power functions by the power rule. We can take the derivative of exponential functions like e to the x or two to the x. We can take the derivative of trig functions. We now can take the derivative of inverse trigonometric functions. And so this new family of functions right here, it's like a brand new generation of Pokemon just came out, right? So the generation of cosine, cosine, right? Pokemon, cosine, sine, whatever. I don't know. I guess you could call that. I haven't done it yet, but they did x and y. And I was pretty close to some algebra right there, wasn't it? We have a whole new generation of functions that's now come out. So whether we're playing the video game or the card game, it's like, hey, this whole new expansion came out of inverse trigonometric functions. You can add that to your deck with all the other derivatives, all the other rules that are already in play. So we can take the derivative now of the function y equals one over arc sine of x. Because when you look at that, this function involves an arc sine, which is the new expansion that just came out. But it also is a quotient. Or if you prefer, you could write this as a reciprocal function. So like arc sine to the negative one, excuse me. Now you have to be a little bit careful here because when you see something like sine inverse of x, it erases the negative one. This might be misleading. Some of us might think this is a sine, which that's not true whatsoever. These two negative ones mean different things. And mathematics and any type of language, this is actually what we call a homograph. We're using the same notation, the same writing to represent two different ideas. How do we tell the difference? Context. It's the same things like if I write this word on the screen, W-I-N-D. What's this word? And it doesn't matter what you pick, you're gonna be wrong. Because if you say when, I'm gonna say wind. If you say wind, I'm gonna say wind. Because out of context, you don't know what this word is. That's the point here as well. So when it comes to trigonometric functions, if you see a negative one between the sine and the x, that means it's an inverse function. We're talking about arc sine. On the other hand, if you see the negative one to the side right here, that actually means the reciprocal. And normally, I would never use both notations simultaneously. The reason is for the power rule with derivatives, it'll be to our advantage to have the negative one power there. So when we calculate y prime here, by the chain rule, we're gonna get negative one times arc sine to the negative two power now. And then we have to multiply that by the derivative of sine inverse, which we see above right here. That's gonna be one over the square root of one minus x squared. And so rewriting this, we're gonna get negative one over, we get two sine squared. So we're gonna sine inverse square times the square root of one minus x squared. And so despite the temptation, do not write sine to the negative two of x. Cause those are, the negative one, there's not an exponent, it's a superscript referring to inverse operation. That would be bad notation right there. This is the derivative of our function. Let's look at another function here. Let's combine our new expansion of Pokemon cards with the cards we already have in our deck. Y equals x times arc tangent and square root of x. So we can see here, there's an arc tangent in play, like we remember from above, the derivative arc tangent is one over one plus x squared. But we also have a product of two things, the product rules into play. We have a chain going on here. So let's put all of this into play here. Y prime, we first are gonna use the product rule. So we're gonna take x prime times arc tangent of the square root of x. And then we add to that x times arc tangent, the square root of x, its derivative. Well, the derivative of x with respect to x is gonna be a one. So we have one times arc tangent of the square root of x. And then you take the derivative arc tangent with the square root of x, the chain rule comes into play. You have this inner function, which is the square root of x. And then you have this outer function, which is arc tangent. So taking the derivative there, we're gonna take the outer derivative, the derivative arc tangent, which we saw from above, the derivative arc tangent is one over one plus x squared, the input squared. So we're actually gonna square root of x squared in this situation. This is our outer derivative. But then we have our inner derivative, the derivative of the square root of x, which by the power rule is gonna be one over two times the square root of x right here. So this is our inner derivative, id. And so if you wanna simplify these things a little bit, you do have an x on top right here, and you have a square root of x on the bottom. They do cancel out a little bit, in which case this would leave us with the square root of x on the top. That is to say x divided by the square root of x is equal to the square root of x. You can rationalize the denominator to see such a thing. So simplifying y prime is gonna be tangent inverse, tangent inverse of the square root of x. And then for the next part, we end up with a square root of x all over two times one plus x in the denominator right there. And so we can calculate the derivatives of these inverse trigonometric functions, combining them with the other derivatives rules we know, the other derivative formulas we know, and it's really great, it's pretty fun. And so that's gonna end our lecture 25. In lecture 26, we are gonna continue to use implicit differentiation, but with a slightly different perspective, focusing on what's called logarithmic differentiation. That is, we're gonna tackle using the same strategy we saw in this video, how to compute the derivatives of logarithmic functions, recognizing they're the inverses of exponential functions for which we already know their derivatives. So check us out in the next video. If you learned something here, please hit the like button. If you wanna see more videos like this, subscribe. And as always, if you have any questions, feel free to post them in the questions below. Bye everyone, see you next time.