Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Derivatives of inverse trigonometric functions. The derivative of the inverse of a function is just equal to 1 divided by the derivative of the original function, with the F inverse of X being the input. So let's let F of X equals sine X. We know that the first derivative would be cosine X. We also know that the inverse of the original function would be sine inverse X. So by that definition we have F inverse or the derivative of the inverse function equaling 1 divided by cosine of sine inverse X. We also know by the Pythagorean identity that cosine Theta is equal to square root 1 minus sine squared Theta. Thus 1 / sqrt 1 minus sine squared, and instead of Theta we're going to put in what it equaled, which was sine inverse of X and the sine and the sine inverse because a composition functions are going to cancel. So we're going to get F We're going to get the derivative of the inverse equaling 1 divided by square root 1 -, X ^2. And that's true when the original function was given a sine X. Another way to think about this is sine inverse of X equal Y. So X equals sine Y. So if we take the derivative of each side, we get one equal in cosine YY prime get Y prime by itself 1 divided by cosine Y again. Remember sine squared Y plus cosine squared Y equal 1. So this cosine Y we could think of as square root 1 minus sine squared Y. But up here we knew that sine Y is really just X. So we get 1 / sqrt 1 -, X ^2 is Y prime. So given the original function sine inverse of X, its derivative is 1 / sqrt 1 - X ^2. If we look at tangent, we have F of X equaling tangent X. So F prime of X is secant squared X. The inverse of the original function is going to be tangent inverse of X. So now using the definition we have that the derivative of the inverse is 1 divided by the derivative of the original with the inverse being the input. So 1 divided by secant squared tangent inverse X. And remember we have a Pythagorean identity that says one plus tangent squared X equals secant squared X. So the secant squared we could think of as tangent squared of that angle tangent inverse X + 1. The tangent in the tangent inverse because of composition functions are going to cancel, leaving us X ^2 + 1. Or you could think of this as starting with your Y equaling tangent inverse X, so X would equal tangent Y. Take the derivative of each side 1 equals secant squared YY prime. Divide each side by that secant squared Y. Remember your Pythagorean identity. So we get 1 / 1 plus tangent squared Y equaling Y prime. Well, tangent Y right here is really just X. So 1 / 1 + X ^2 equal Y prime. Another way to look at it is you could draw yourself your triangle and this tangent Y would be opposite over adjacent or X / 1 so that when you come down here, no matter what your trig function is in this, you could then look at this triangle and figure out the sides tangent opposite of our adjacent or X. If we wanted to look at secant X next, we'd have F prime of X equaling secant X tangent X. The inverse would be secant inverse of X. So by the definition we'd have 1 divided by secant secant inverse X times tangent secant inverse X secant a secant inverse composition functions going to leave us X tangent secant inverse of X. Remember our Pythagorean identity again, one plus tangent squared Theta equals secant squared Theta. So tangent Theta is positive negative square root, secant squared Theta -1 Now if we actually look at the graph of our secant inverse, paying attention to it needing to be 1 to one, we can see that the slope is always going to be positive. So we're only going to care about the positive case here. We always have increasing slopes for this. So we're going to get secant secant inverse being X square root, the secant squared of whatever the angle was. In this case secant inverse X -, 1. The secant and secant inverse because their inverse functions are going to cancel. So 1 divided by the absolute value of X because remember slopes always positive. We need it to be a positive sqrt X ^2 - 1. This is understood to be positive so we don't have to worry about the absolute value around that. So the derivative of the inverse function is 1 divided by the absolute value of X sqrt X ^2 -, 1. Another way to do it. Oh, didn't do it this time. OK, let's look at another property. If we let alpha equal cosine inverse of X, we know that cosine X cosine alpha would then equal X. So cosine is adjacent over hypotenuse or X / 1. Thus sine inverse of X would have to equal some beta because that would say sine of beta equal opposite over hypotenuse, so cosine inverse adjacent over hypotenuse sine opposite and those would both equal 1. So we know that a alpha plus beta is π halves. So now we can come up with some relationships that if we know alpha is cosine inverse of X and beta sine inverse of X, that's got to equal π halves. So to find the derivative of the cosine inverse, we can just think of that as finding the derivative of π halves minus sine inverse of X. The derivative of a constant is 0. The derivative of sine inverse X we found a moment ago to be 1 / sqrt 1 - X ^2. So 0 - 1 / sqrt 1 - X ^2 or just the derivative of cosine is -1 divided by square root 1 - X ^2. I'm going to do the same thing for the other two. Cotangent inverse of X is just going to be π halves minus tangent inverse of X. So the derivative of that constant Pi halves is 0. The derivative of tangent inverse X we developed a minute ago. So we really have the derivative of cotangent inverse X being -1 / 1 + X ^2 cosecant inverse of X. The same thing. We're going to take the derivative of π halves minus secant inverse X. So the derivative of π halves is 0. The derivative of secant inverse X was 1 divided by the absolute value of X ^2 of X ^2 -, 1. So the derivative of cosecant inverse of X is really just -1. Going over that idea again, if we have alpha plus beta equaling Pi halves, we can see that cosecant alpha is 1 / X and we can see that secant beta is 1 / X. So alpha is cosecant inverse of Y, beta is secant inverse of Y, alpha plus beta is π halves. So we could also think of letting 1 / X just be some value of Y the integrals. By the way, this works for all of these formulas work for composition of functions. So if instead of X it was U, we'd have the derivative cosine inverse of U equaling -1 divided by the absolute value of U sqrt U ^2 -, 1 du DX. So the integrals of DU over sqrt a ^2 -, U ^2 is just going to be sine inverse of U / a + C where U ^2 is less than a ^2. DU is the integral of DU over a ^2 + U ^2 is 1 / a tangent inverse of the quantity U / a + C for all U, and then the last one integral of DU over U sqrt U ^2 -, a ^2 is 1 / a secant inverse absolute value U / a + C where the absolute value of U is greater than a is greater than 0. We typically don't use the cosine cosecant and cotangent because they're really the same function with just a negative. So if we know our sine inverse, tangent, inverse, secant inverse, the integrals that go with those, we actually can just use those three and not have to use any of the ones that start with CS. Thank you and have a wonderful day.