Question

What is the derivative of `y="arcsec"(x)`?

Answer 1

`dy/dx=1/[x^2*sqrt(1-(1/x)^2)]`

Explanation:

show that

`y=arcsecx=1/[arccosx]=arccos(1/x)`

`d/dx[arccosu]=1/sqrt(1-u^2)*u'`

`dy/dx=-1/[sqrt(1-(1/x)^2)]*[-1/x^2]`

`dy/dx=1/[x^2*sqrt(1-(1/x)^2)]`