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 one divided by the derivative of the original function with the f inverse of x being the input so let's let f(x) equal sine x. We know that first the 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 one divided by cosine of sine inverse x. We also know that by the Pythagorean identity that cosine theta is equal to square root one minus sine squared theta. Thus one divided by the square root of one minus sine squared and instead of theta we are gonna put in what it equaled which was sine inverse of x and the sine and the sine inverse, because of composition functions are gonna cancel so we're gonna get f, we're gonna get the derivative of the inverse equaling one divided by square root one minus x squared and that's true when the original function was given as sine x another other 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 equaling cosine y, y prime get y prime by itself one divided by cosine y. Again remember sine squared y plus cosine squared y equal one so this cosine y we could think of a square root one minus sine squared y but up here we knew that sine y is really just x. So we get one divided by the square root of one minus x squared is y prime, so given the original function sine inverse of x its derivative is one divided by the square root of one minus x squared if we look at tangent, we have f(x) equaling tangent x so f prime of x is secant squared x. The inverse of the original function is gonna be tangent inverse of x so now using the definition, we have that the derivative of the inverse is one divided by the derivative of the original with the inverse being the input. So one divided by secant squared tangent inverse x, and remember we have it 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 plus one. The tangent and the tangent inverse because of composition functions are gonna cancel leaving us x squared plus one 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 one equal secant squared y, y prime. Divide each side by that secant squared y. Remember your Pythagorean identity, so we get one divided by one plus tangent squared y equaling y prime Well, tangent y right here is really just x, so one divided by one plus x squared 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 over one 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 over 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 that definition we'd have one divided by secant, secant inverse x times tangent secant inverse x secant is secant inverse, composition functions gonna leave us x. Tangent secant inverse of x, remember Pythagorean identity again, one plus tangent squared theta equal secant squared theta So tangent theta is positive negative square root secant squared theta minus one now if we actually look at the graph of our secant inverse, paying attention to it needing to be one the one, we can see that the slope is always gonna be positive so we're only gonna care about the positive case here we always have increasing slopes for this. So we're gonna get secant secant inverse being x square root secant squared of whatever the angle was in this case secant inverse x minus one The secant and secant inverse because they're inverse functions are gonna cancel so one divided by the absolute value of x, 'cause remember slopes always positive we need it to be a positive square root of x squared minus one. 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 one divided by the absolute value of x, square root of x squared minus one Another way to do it, oh didn't do it this time. Okay, 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 over one Thus sine inverse of x would have to equal some beta, because that would say sine of beta would equal opposite over hypotenuse So cosine inverse, adjacent over hypotenuse, sine opposite, and those would both equal one so we know that alpha plus beta is pi halves So now we can come up with some relationships, that if we know alpha is cosine inverse of x and beta is sine inverse of x that's got to equal pi halves So to find the derivative of the cosine inverse, we can just think of that as finding the derivative of pi halves minus sine inverse of x the derivative of a constant is just zero, the derivative of sine inverse x, we found a moment ago to be one over the square root one minus x squared. So zero minus one over square root of one minus x squared or just the derivative of cosine is negative one divided by square root one minus x squared We're gonna do the same thing for the other two cotangent inverse of x is just gonna be pi halves minus tangent inverse of x, so the derivative of that constant pi halves is zero the derivative of tangent inverse x we developed a minute ago, so we really have the derivative of cotangent inverse x being negative one divided by one plus x squared Cosecant inverse of x, the same thing we're gonna take the derivative of pi halves minus secant inverse x. So the derivative of pi halves is zero. The derivative of secant inverse x was one divided by the absolute value of x square root of x squared minus one so the derivative of cosecant inverse of x is really just negative one going over that idea again if we have alpha plus beta equaling pi halves, we can see that cosecant alpha is one over x and we can see that secant beta is one over x. So alpha is cosecant inverse of y beta is secant inverse of y Alpha plus beta is pi halves so we can also think of letting one over x just be some value of y the integrals by the way, this works for, all of these formulas work for, um composition of functions so if instead of x it was u, we'd have the derivative cosine inverse of u equaling negative one divided by the absolute value of u square root of u squared minus one du dx so the integrals of du over the square root of a squared minus u squared is just gonna be sine inverse of u divided by a plus c where u squared is less than a squared du is the integral of du over a squared plus u squared as one over a tangent inverse of the quantity u divided by a plus c, for all u and then the last one integral of du over u square root of u squared minus a squared is one over a secant inverse absolute value u over a plus c. Where the absolute value of u is greater than a, is greater than zero. We typically don't use the cosine cosecant, and cotangent because they're really the same function with just a negative. And 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 c's thank you and have a wonderful day