What is the derivative of

y = (tanx) / (1+secx)

First, the the derivative of y=lnA is y' = A' / A

let y = (tanx) / (1+secx), implies ln y=ln [ (tanx) / (1+secx) ]

(lny) ' = (ln [ (tanx) / (1+secx) ]) ' so y' / y = [ (tanx) / (1+secx) ]) ' / (tanx) / (1+secx)

(tanx) / (1+secx) ' = [ (tanx) ' (1+secx) - (tanx) (1+secx) '] / (1+secx) ²

(tanx) ' = 1 + tan²x and (1+secx) '] = secx tanx (check lesson)

so

(tanx) / (1+secx) ' = (1 + tan²x) (1+secx) - (tan²xsecx) / (1+secx) ²

= (1 + secx + tan²x) / (1+secx) ²

y' / y = [ (tanx) / (1+secx) ]) ' / (tanx) / (1+secx) =

y' / y = [ (1 + secx + tan²x) / (1+secx) ² ] / (tanx) / (1+secx)

y' / y = (1 + secx + tan²x) (1+secx) / (tanx) (1+secx) ²

so y' = y. [ (1 + secx + tan²x) (1+secx) / (tanx) (1+secx) ² ]

but y = (tanx) / (1+secx),

y' = (tanx) / (1+secx) [ (1 + secx + tan²x) (1+secx) / (tanx) (1+secx) ² ]

finally the derivative of (tanx) / (1+secx) is given by

y' = (1 + secx + tan²x)