How do you differentiate #f(theta)=sectheta/(1+sectheta)#?

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Answer 1 · 2016-10-25T16:21:02

#f'(theta)=(secthetatantheta)/(1+sectheta)^2#

#for y=u/v => (dy)/(dx)=(v(du)/(dx)-u(dv)/(dx))/v^2#

#f(theta)=sectheta/(1+sectheta)#

#u=sectheta=>(du)/(d(theta))=secthetatantheta#

#v=1+sectheta=>(dv)/(d(theta))=secthetatantheta#

#f'(theta)=((1+sectheta)(secthetatantheta)-sectheta(secthetatantheta))/(1+sectheta)^2#

#f'(theta)=(secthetatantheta+sec^2thetatantheta-sec^2thetatantheta)/(1+sectheta)^2#

#f'(theta)=(secthetatantheta+cancel(sec^2thetatantheta-sec^2thetatantheta))/(1+sectheta)^2#

#f'(theta)=(secthetatantheta)/(1+sectheta)^2#