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Question
DIfferentiate `tan^-1((sqrt(1 + x^2) - 1)/x) w.r.t. tan^-1(sqrt((2xsqrt(1 - x^2))/(1 - 2x^2)))`.
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Solution
Let u = `tan^-1((sqrt(1 + x^2) - 1)/(x))`
and
v = `tan^-1((2xsqrt(1 - x^2))/(1 - 2x^2))`
Then we want to find `"du"/"dv"`
u = `tan^-1((sqrt(1 + x^2) - 1)/(x))`
Put x = tanθ
Then θ = tan–1 x
and
`(sqrt(1 + x^2) - 1)/(x) = (sqrt(1 + tan^2θ) - 1)/tanθ`
= `(secθ - 1)/(tanθ)`
= `((1)/(cosθ) - 1)/((sinθ/cosθ)`
= `(1 - cosθ)/(sinθ)`
= `(2sin^2(θ/2))/(2sin(θ/2)cos(θ/2))`
= `tan(θ/2)`
∴ u = `tan^-1[tan(θ/2)] = θ/(2) = (1)/(2)tan^-1x`
∴ `"du"/"dx" = (1)/(2)"d"/"dx"(tan^-1x)`
= `(1)/(2) xx (1)/(1 + x^2)`
= `(1)/(2(1 + x^2)`
v = `tan^-1((2xsqrt(1 - x^2))/(1 - 2x^2))`
Put x = sin θ.
Then θ = sin–1x
and
`(2xsqrt(1 - x^2))/(1 - 2x^2)`
= `(2sinθsqrt(1 - sin^2θ))/(1 - 2sin^2θ)`
= `(2sinθcosθ)/(1 - 2sin^2θ)`
= `(sin2θ)/(cos2θ)`
= tan 2θ
∴ v = tan−1(tan2θ)
= 2θ
= 2sin−1x
∴ `"dv"/"dx" = 2"d"/"dx"(sin^-1x)`
= `2 xx (1)/sqrt(1 - x^2) = (2)/sqrt(1 - x^2)`
∴ `"du"/"dx" = (("du"/"dx"))/(("dv"/"dx")`
= `([(1)/(2(1 + x^2))])/(((2)/sqrt(1 - x^2))`
= `(1)/(2(1 + x^2)) xx sqrt(1 - x^2)/(2)`
= `sqrt(1 - x^2)/(4(1 + x^2)`
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