The derivative of arcsec(x)

The derivative of arcsec(x).

[messedupworld]

Thanks, in class most of professors skip this, very well explained. 

[sarebear]

as a high schooler this was very helpful

[oguzhantopaloglu]

Thank you sir, this video was very useful.

[zullyzulema]

Dude you are a Crack ! ...well I am studying for my next Real Analysis ' partial exam, well.. Hard work pays off!

[jan-willemreens]

... y = arcsec(x) <---> x = sec(y) ---> dx/dy [sec(y)] = sec(y)tan(y) <----> dy/dx = 1/(sec(y)tan(y)), draw a right triangle, choose one of the acute angles as angle (y), sec(y) = hypotenuse/adjacent side = x/1---> opposite side = SQRT(x^2 - 1) ----> tan(y) = opp/adj = SQRT(x^2 - 1)/1 = SQRT(x^2 - 1), thus resulting for the derivative of arcsec(x): dy/dx = 1/(sec(y)tan(y)) = 1/((x)(SQRT(x^2 - 1))) or dy/dx = 1/( IxI SQRT(x^2 - 1)) ...

[faraharaku]

Pardon Sir, why in the end of the answer is use absolute x or |x|? Why that isn't just use x? Thank you

[fredericoyj]

Do the 'x' in the triangle represent the hypotenuse at all times?

[parthsarathidixit]

Hey can you tell me which grade mathematics is this in the US