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