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Find `bb(dy/dx)` in the following:
`y = tan^(-1) ((3x -x^3)/(1 - 3x^2)), - 1/sqrt3 < x < 1/sqrt3`
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y = `tan^-1 ((3x - x^3)/(1 - 3x^2))`
Putting x = tan θ,
∴ y = `tan^-1 ((3 tan theta - tan^3 theta)/(1 - 3 tan^2 theta))`
= tan−1 (tan 3 θ)
= 3 θ
= 3 tan−1 x ...[because θ = tan−1 x]
On differentiating with respect to x,
`dy/dx = 3 d/dx tan^-1 x`
`dy/dx = 3 xx 1/(1 + x^2)`
`dy/dx = 3/(1 + x^2)`
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