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