Derivative of inverse cosine function from 1st principles

In this video, I showed how to find the derivative of inverse cosine function using the definition of the dereivative

[kingbeauregard]

Nice!

When you get to the sec and csc, I'll appreciate a good explanation of the absolute values. I've heard explanations that are only half-convincing; I want to believe.

[abhirupkundu]

Or we can say cos^-1x= y, and say cos(y)=x. Since we need dy/dx, we can first find dx/dy through the first principle, and then take its reciprocal, to make it dy/dx, since dy/dx is a fraction(Change in y w.r.t change in x)

[miltonpassanha]

When we have f(x) =arc cos(x) …. then we have ….x= cos y Differentiating both sides we get dx/dy = negative sin y … But sin^2 y + cos^y = 1 which means …. sin y = sq root of 1- cos^2 y which is sq. root of 1- x^2 ( bcoz x=cos y)

Taking reciprocal of dx/dy gives dy/dx which is negative sin y

Hence dy/dx = negative 1/sin y = negative 1/ sq root of 1-x^2

This gets done under 1 minute or so …. then why the fundamental approach via limit 😮

But due respect to you Sir 🙏👍👍

[hervesergegbeto]

Merci beaucoup
Tu fais un exposé limpide

[Michael-sbjf]

I only remember being taught to do this implicitly
funny enough it leads to the same outcome of -1/sin(x) =1 then solving via trig identities

[masoudhabibi]

Thank for your video...sr
Please say difference between arccos and cos^-1(x)

[tomvitale]

Nice. But I need to understand how you transformed cos(A) - cos(B) to -2sin((A+B)/2)sin((A-B)/2). For now, I'll take your word for it, but I need to find out for myself. I assume you reverse engineered the trig identities for the sum of 2 angles.

[Thoughtsprobe]

I'm not seeing well really
The video and calculations are blurred