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Question
If y = `tan^-1((2x)/(1 - x^2))`, x ∈ (−1, 1) then `("d"y)/("d"x)` = ______.
Options
`(-2)/(1 + x^2)`
1
`(2)/(1 + x^2)`
`1/(1 + x^2)`
Solution
If y = `tan^-1((2x)/(1 - x^2))`, x ∈ (−1, 1) then `("d"y)/("d"x)` = `bbunderline((2)/(1 + x^2))`.
Explanation:
Put x = tan θ, `(-pi)/4 < theta < pi/4`
∴ tan−1 x
`therefore y = tan^-1 ((2 tan theta)/(1 - tan^2theta))`
∴ y = tan−1 (tan 2 θ)
∴ y = 20
∴ y = 2 tan−1 x
`therefore dy/dx = 2/(1 + x^2)`
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