What is the derivative of "arcsec"(e^(2x))?

1 Answer
May 19, 2018

d/dx[arccos(1/e^(2x))]=-1/[sqrt(1-(1/e^(2x))^2)]*[-2/e^(2x)]

=[2/e^(2x)]/[sqrt(1-(1/e^(2x))^2)]=[2]/[e^(2x)*sqrt(1-(1/e^(2x))^2)]

Explanation:

there two different method to derive arcsec(e^(2x))

i will use one of them

arcsec(e^(2x))=arccos(1/e^(2x))

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

d/dx[arccos(1/e^(2x))]=-1/[sqrt(1-(1/e^(2x))^2)]*[-2/e^(2x)]

=[2/e^(2x)]/[sqrt(1-(1/e^(2x))^2)]=[2]/[e^(2x)*sqrt(1-(1/e^(2x))^2)]