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

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Answer 1 · 2016-10-25T16:21:02

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

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

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

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

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

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

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

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

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

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