Question

Differentiate `tan^-1[(sqrt(1+"x"^2)-sqrt(1-"x"^2))/(sqrt(1+"x"^2) + sqrt(1-"x"^2))]`with respect to cos⁻¹x².

Answer 1

Let u = tan⁻¹`[(sqrt(1+"x"^2) - sqrt(1-"x"^2))/(sqrt(1+"x"^2)+ sqrt(1-"x"^2))]` and v = cos-¹ x²

Putting  x²  = cos θ in u

`"u" = tan^-1 [(sqrt(1+costheta) - sqrt(1-costheta))/(sqrt(1-costheta) + sqrt(1-costheta))]`

`⇒"u" = tan^-1 [(sqrt(2+cos^2  theta/2) -sqrt(2 sin^2  theta/2))/(sqrt(2 cos^2  theta/2) + sqrt(2 sin^2  theta/2))]`

`⇒"u" = tan^-1[(cos  theta/2-sin  theta/2)/(cos  theta/2 + sin  theta/2)]`

`⇒ "u" = tan^-1[(1-tan  theta/2)/(1+ tan  theta/2)]`

`⇒ "u" = tan^-1[(1-tan  theta/2)/(1+ tan  theta/2)]`

`⇒ "u" = tan^-1[tan(pi/4 - theta/2)]`

`⇒ "u" = pi/4 - theta/2`

`⇒ "u"=pi/4 -(cos^-1"x"^2)/2`

`⇒ "du"/"dx"= 0 - 1/2 ((-1)/sqrt(1-"x"^4)(2"x"))`

`⇒ "du"/"dx" = "x"/sqrt(1-"x"^4)`

Also, v = cos^-1x²

`"du"/"dv" = -1/sqrt(1-"x"^4) (2"x") = (-2"x")/sqrt(1-"x"^4)`

`"du"/"dv" = ("du"/"dz")/("dv"/"dz")`

`= ("x"/sqrt(1-"x"^4))/((-2"x")/sqrt(1-"x"^4))`

`= (-1)/2`