Question

DIfferentiate `tan^-1((sqrt(1 + x^2) - 1)/x) w.r.t. tan^-1(sqrt((2xsqrt(1 - x^2))/(1 - 2x^2)))`.

Answer 1

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)`