Question

Differentiate `tan^-1 (sqrt((3 - x)/(3 + x)))` w.r.t. x.

Answer 1

Let y = `tan^-1(sqrt((3 - x)/(3 + x)))`

Put x = 3 cos 2θ  

∴ θ = `1/2 cos^-1(x/3)`

∴ y = `tan^-1[sqrt((3 - 3 cos 2θ)/(3 + 3 cos 2θ))]`

= `tan^-1[sqrt((3(1 - cos 2θ))/(3(1 + cos 2θ)))]`

= `tan^-1[sqrt((2 sin^2θ)/(2 cos^2θ))]`

y = `tan^-1(sqrt(tan^2θ)) = tan^-1(tanθ)`

y = θ = `1/2 cos^-1(x/3)`

Differentiate w.r.t. x, we get

`dy/dx = 1/2 . d/dx[cos^-1(x/3)]`

= `1/2[-1/sqrt(1 - (x/3)^2)]d/dx(x/3)`

= `-1/2 xx 1/sqrt((9 - x^2)/9) xx 1/3`

= `-1/2 xx 1/(sqrt(9 - x^2)/3) xx 1/3`

∴ `dy/dx = - 1/(2sqrt(9 - x^2))`