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