Given

`y=sinh^-1 (tanx)`

to find

`y' = (sinh^-1 (tanx))'

let `u=tanx`

so ,

`y' = d/(du) (sinh^-1 (u)) d/dx tanx`

`=1/sqrt(u^2+1) sec^2x`

`=(sec^2 x)/sqrt(tan^2+1)`

Given

`y=sinh^-1 (tanx)`

to find

`y' = (sinh^-1 (tanx))'

let `u=tanx`

so ,

`y' = d/(du) (sinh^-1 (u)) d/dx tanx`

`=1/sqrt(u^2+1) sec^2x`

`=(sec^2 x)/sqrt(tan^2+1)`