Calc Topic 3 4 Derivative of Trig Inverses

Añadido: 2026-05-09

[00:00:06]okay this section is the last of the derivative rules that you need to memorize um it is the inverse trig function derivative rules now unfortunately they are one of the hardest to memorize and they are the least to occur on the actual ap exam you never know so we have to go through them okay so you have your inverse sine or your arc sine 1 over the square root so you can read through these that's a negative one over the square root of one minus x squared so notice this is a positive one this is a negative one okay um tangent and cotangent um well of course there's a typo there that is a negative so let's make sure that we get the negative in there okay and then you have arc secant and cosecant again negative so negative negative negative these are positive positive positive okay so maybe that will help you out a little bit okay this is the general rule if you're taking the arc sine of x this is if you're taking the arc trig functions of something other than x so we have the derivative rule times the correction factor the derivative rule times the correction factor okay so put those on note cards try to memorize them and let's learn how to use them okay so if i were to ask you um to find the derivative of this without using an inverse trig function um how would we do that well one of the things that we know is that the inverse sine can be undone by taking the sine of both sides so the sine and remember this notation arc sine or inverse sine of x arc sine or arc sine same thing so these two undo each other if you recall we have the sine of y equals x so now we can take the derivative now and we know the derivative rule for sine is cosine of u times u prime the derivative of x is one so y prime equals one over the cosine of y well what is the cosine of y well let's take a look at our picture over here okay this is our angle y right here we know the sine is x over one that's what was given so if we do pythagorean theorem that gives us our adjacent side a squared plus b squared equals c squared so let's take the cosine of y off of our picture here the cosine is the adjacent over the hypotenuse so the cosine of y is the square root of one minus x squared so y prime the derivative of arc sine which is what you just calculated the derivative of our sign you check up here and it matches so that's the long way of doing it in case you forget if you don't memorize the formulas i wouldn't recommend that but let's try it one more time here the long way of doing the arc sine so if we have um color here we have y equals i just prefer to write arcs i just prefer to write it this way it's just personal preference so i can take the sine of both sides i can take the derivative now with respect to x so that is cosine of u times u prime equals two so y prime equals two divided by the cosine of y well we have to go to our picture here it's not going to be the same picture because the given here was that the sine is equal to x that's what arc sine means so this is saying the sine of our angle we'll call that angle y the sine is 2x over one that's what's given in the original problem so if i do um pythagorean theorem it's going to be c squared minus a squared equals b squared so this side ends up being the square root of 1 minus 2 squared i'm sorry 2x squared which is the quant 2x the quantity squared i'm sorry which is 4x squared so we have just found that the derivative of arc sine is two times the cosine on the picture is one minus four x squared so that's doing it the long way again this means an angle whose sine is x this means an angle whose sine is 2x okay but let's use the derivative rule instead okay let's use the derivative rule so if we're going to take the derivative of arc sine of 2x okay so again here's notation the derivative of the inverse okay so if this is the u right here so if we use the derivative rule it's 1 over the square root of 1 minus our u squared times the derivative of u and that gives me 2 over the square root of 1 minus 4 x squared same answer so i recommend using the derivative rule so let's go ahead and try the derivative rule on this one go ahead stop the video and try it yourself okay the derivative of inverse tan of u okay when you have it memorized inverse tan is one over one plus u squared times the derivative of u so your answer is three over one plus nine x squared okay go ahead and try c i would again stop the video and come back and check your answer okay let's see how you did the derivative the inverse tan derivative we just did that one it's 1 over 1 plus your u squared times the derivative of your u i'm going to put on the side just to make a little bit easier the derivative of our u is the square root rule so your answer is top times top this is going to be two square roots of x and over here we have what one plus x so no need to do the long way if you have the rules memorized all righty challenge problem go ahead and stop the video and see if you can do that all right here we go i see a lot of things i see four functions in here i see a square i see an inverse trig function i see a exponential function and i see a linear function so that sounds like um not double chain but uh triple chain yeah four functions that would be a triple chain problem so go ahead and give it a shot if you haven't and then come back try it on your own all right welcome back fun problem here fun fun okay we're going to start with the outside we're going to use the derivative rule the power rule so it's 2 times the inverse secant of e to the 2x raised to the first power so it's 2 times our u raised to the first power times the derivative of our u which is inverse secant of e to the 2x so all of that is our u so we already did this so let's just carry that down with us and let's go ahead and find the derivative of this well this is again this is got a lot going on here so let's pick the outer function which is our inverse secant or arc secant so the rule for arc secant is 1 over the absolute value of our u this is our u times the square root of u squared minus 1 times the derivative of our u times the derivative of our u you can either write derivative of our u down here and then keep going or can you now do this in your head the derivative of our u this is a chain rule e to a different u times the derivative of our u so your final answer let's see here if you get two times two on the top so that gives you a four um this absolute value is going to be positive so we're gonna divide it by this one so that's those are gonna cancel and you're left with arc secant of e to the 2x all divided by the square root of e that is power of a power so you multiply them together and there's your final answer and if you got that good job triple chain good job and that is practice full of the inverse trig rules and i will give you some more