Given,

`y = tanh^-1(x)+ ln(sqrt(1-x^2))`

so we have to find the y'

so,

`y' =(tanh^-1(x)+ ln(sqrt(1-x^2)))'`

`=(tanh^-1(x))'+(ln(sqrt(1-x^2)))'`

as we know

`(tanh^-1(x))' =1/(1-x^2)`

and so we have to find out

`(ln(sqrt(1-x^2)))' `

let `u =sqrt( 1-x^2)`

it can be solved by the following,

`(df)/dx = (df)/(du) * (du)/dx`

so,

`d/dx (ln(sqrt(1-x^2)))= d/(du) ln(u) * (du)/dx`

`= 1/u * d/dx (sqrt( 1-x^2))`

`=1/sqrt( 1-x^2) * (-x/sqrt( 1-x^2))`

`= -x/( 1-x^2)`

now,as

`y'=(tanh^-1(x)+ ln(sqrt(1-x^2)))'`

`=1/( 1-x^2) +(-x/( 1-x^2))`

`= (1-x)/( 1-x^2)`

`=1/(1+x)`

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