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

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Answer 1 · 2018-05-19T22:04:19

#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)]#

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)]#