Question

Differentiate `tan^-1((x)/(sqrt(1 - x^2))) w.r.t. sec^-1((1)/(2x^2 - 1))`.

Answer 1

Let u = `tan^-1((x)/(sqrt(1 - x^2)))` and

v = `sec^-1((1)/(2x^2 - 1))`.
Then we want to find `"du"/"dv"`.
Put x = cosθ.
Thenθ = cos^–1x.

∴ u = `tan^-1((cosθ)/(sqrt(1 - cos^2θ)))`

= `tan^-1((cosθ)/(sinθ))`

= `tan^-1 (cotθ)`

= `tan^-1[tan(pi/2 - θ)]`

= `pi/(2) - θ`

= `pi/(2) - cos^-1x`

∴ `"du"/"dx" = "d"/"dx"(pi/2) - "d"/"dx"(cos^-1x)`

= `0 - (-1)/(sqrt(1 - x^2)) = (1)/(sqrt( - x^2)`

v = `sec^-1((1)/(2x^2 - 1))`
= cos^–1(2x² – 1)
= cos^–1(2 cos²θ – 1)
= cos^–1 (cos2θ)
= 2θ
= 2 cos^–1x
∴ `"dv"/"dx" = 2."d"/"dx"(cos^-1x)`

= `(-2)/(sqrt(1 - x^2)`

∴ `"du"/"dv" = (("du"/"dx"))/(("dv"/"dx")`

= `(1)/sqrt(1 - x^2) xx sqrt(1 - x^2)/(-2)`

= `-(1)/(2)`.