Derivative of arcsec(x)

Derivative of arcsec(x)

[johnnolen]

It isn't necessary to redefine the adjacent side as 1/x and the hypotenuse as 1. In order for an inverse function to exist the original function must be one - to - one. Cosine is one - to - one only if its domain is restricted to quadrants I & II of the unit circle. Since secant is the reciprocal of cosine its domain is likewise restricted to quadrants I & II but with the exception that sec(π/2) is undefined. d sec(y)/dy = sec(y)·tan(y), which can be rewritten as sec^2(y)·sin(y). In quadrants I & II sine is strictly positive; therefore, we can rewrite sec^2(y)·sin(y) as sec^2(y)·|sin(y)| = x^2·sqrt{x^2 - 1}/|x|. Therefore d sec(y)/dy = |x|·sqrt{x^2 - 1}. By the inverse function theorem then d[arcsec(x)]/dx = 1/|x|·sqrt{x^2 - 1} ◼