Question

Differentiate `sin^-1((2x)/(1 + x^2))w.r.t. cos^-1((1 - x^2)/(1 + x^2))`

Answer 1

Let u = `sin^-1((2x)/(1 + x^2))` and

v = `cos^-1((1 - x^2)/(1 + x^2))`

Then we want to find `"du"/"dv"`.
Put x = tanθ.
Then θ = tan^–1x.
u = `sin^-1((2tanθ)/(1 + tanθ))`
= sin^–1(sin2θ)
= 2θ
= 2tan^–1x
∴ `"du"/"dx" = 2"d"/"dx"(tan^-1x)`

= `2 xx (1)/(1 + x^2)`

= `(2)/(1 + x^2)`
Also, v = `cos^-1((1 - tan^2θ)/(1 + tan^2θ))`
= cos^–1(cos 2θ)
= 2θ
= 2 tan^–1x
∴ `"dv"/"dx" = 2"d"/"dx"(tan^-1x)`

= `2 xx (1)/(1 + x^2)`

= `(2)/(1 + x^2)`

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

= `(((2)/(1 + x^2)))/(((2)/(1 + x^2))`
= 1.
Alternative Method :
Let u = `sin^-1((2x)/(1 + x^2)) and v = cos^-1((1 - x^2)/(1 + x^2))`

Then we want to find `"du"/"dv"`
Put x = tanθ.
Then u = `sin^-1((2tanθ)/(1 + tanθ))`
= sin^–1 (sin2θ) 
= 2θ
and
v = `cos^-1((1 - tan^2θ)/(1 + tan^2θ))`
= cos^–1 (cos2θ) 
= 2θ
∴  u = v
Differentiating both sides w.r.t. v, we get
`"du"/"dv"` = 1.