Question

Differentiate the following w.r.t. x : `sin^-1((1 - x^2)/(1 + x^2))`

Answer 1

Let y = `sin^-1((1 - x^2)/(1 + x^2))`
Put x = tanθ.
Then θ = tan^–1x
∴ y = `sin^-1((1 - tan^2θ)/(1 + tan^2θ))`
= sin^–1(cos2θ)

= `sin^-1[sin(pi/2 - 2θ)]`

= `pi/(2) - 2θ`

= `pi/(2) - 2tan^-1x`
Differentiating w.r.t. x, we get
`"dy"/"dx"= "d"/"dx"(pi/2 - 2tan^-1x)`

= `"d"/"dx"(pi/2) - 2"d"/"dx"(tan^-1x)`

= `0 - 2 xx (1)/(1 + x^2)`

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